Add preprints files (part 2)

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 preprints/A-Counterexample-to-Kaplanskys-Direct-Finiteness-Conjecture-in-Characteristic-Two-September-23-2026/paper.pdf filter=lfs diff=lfs merge=lfs -text
 preprints/A-Counterexample-to-Kaplanskys-Direct-Finiteness-Conjecture-in-Odd-Characteristic-September-26-2026/paper.pdf filter=lfs diff=lfs merge=lfs -text
 preprints/A-counterexample-to-Kaplanskys-quasitrace-conjecture-September-23-2026/paper.pdf filter=lfs diff=lfs merge=lfs -text
+preprints/A-Counterexample-to-Kuroshs-Division-Ring-Problem-September-23-2026/paper.pdf filter=lfs diff=lfs merge=lfs -text
+preprints/A-counterexample-to-Naimarks-problem-in-ZFC-September-24-2026/naimark-counterexample-zfc.pdf filter=lfs diff=lfs merge=lfs -text
+preprints/A-Counterexample-to-Rysers-Covering-Conjecture-September-23-2026/paper.pdf filter=lfs diff=lfs merge=lfs -text
+preprints/A-counterexample-to-Sidorenkos-conjecture-September-23-2026/paper.pdf filter=lfs diff=lfs merge=lfs -text
+preprints/A-counterexample-to-Tachikawas-second-conjecture-September-23-2026/paper.pdf filter=lfs diff=lfs merge=lfs -text
+preprints/A-counterexample-to-the-coarse-Novikov-conjecture-September-23-2026/paper.pdf filter=lfs diff=lfs merge=lfs -text
+preprints/A-counterexample-to-the-Colin-de-Verdiere-chromatic-conjecture-September-23-2026/paper.pdf filter=lfs diff=lfs merge=lfs -text
+preprints/A-Counterexample-to-the-Group-Ring-Determinant-Conjecture-September-23-2026/paper.pdf filter=lfs diff=lfs merge=lfs -text
+preprints/A-Counterexample-to-the-Infinite-Matroid-Packing-Covering-Conjecture-September-24-2026/paper.pdf filter=lfs diff=lfs merge=lfs -text
+preprints/A-counterexample-to-the-Monge-ansatz-for-the-three-marginal-Coulomb-cost-September-25-2026/paper.pdf filter=lfs diff=lfs merge=lfs -text
+preprints/A-counterexample-to-the-nearby-Lagrangian-conjecture-September-23-2026/paper.pdf filter=lfs diff=lfs merge=lfs -text
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+preprints/A-cubic-lower-bound-for-border-determinantal-complexity-of-the-permanent-September-24-2026/A-cubic-lower-bound-for-border-determinantal-complexity-of-the-permanent-September-24-2026.pdf filter=lfs diff=lfs merge=lfs -text
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+preprints/A-degenerate-counterexample-to-the-critical-number-Arnold-bound-September-23-2026/paper.pdf filter=lfs diff=lfs merge=lfs -text
+preprints/A-density-uniform-condensate-bound-for-dilute-Bose-gases-September-27-2026/paper.pdf filter=lfs diff=lfs merge=lfs -text
+preprints/A-dimension-free-logarithmic-Sobolev-inequality-for-subgaussian-log-concave-measures-September-23-2026/paper.pdf filter=lfs diff=lfs merge=lfs -text
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+preprints/A-direct-proof-of-the-complete-Crouzeix-inequality-September-26-2026/paper.pdf filter=lfs diff=lfs merge=lfs -text
+preprints/A-directional-zero-one-law-for-finite-range-dependent-random-environments-October-5-2026/directional-zero-one-finite-range.pdf filter=lfs diff=lfs merge=lfs -text

preprints/A-Counterexample-to-Kuroshs-Division-Ring-Problem-September-23-2026/README.md

@@ -0,0 +1,18 @@
+# [A Counterexample to Kurosh’s Division-Ring Problem](paper.pdf)
+
+**Author:** OpenAI
+
+**Date:** September 23, 2026
+
+## Citation
+
+```bibtex
+@misc{OAI:A-Counterexample-to-Kuroshs-Division-Ring-Problem-September-23-2026,
+  author = {{OpenAI}},
+  title = {{A Counterexample to Kurosh's Division-Ring Problem}},
+  howpublished = {OpenAI Math Release preprint
+                  \href{https://github.com/openai/math/blob/main/preprints/A-Counterexample-to-Kuroshs-Division-Ring-Problem-September-23-2026/paper.pdf}{OAI:A-Counterexample-to-Kuroshs-Division-Ring-Problem-September-23-2026}},
+  year = {2026}
+}
+```
+

preprints/A-Counterexample-to-Kuroshs-Division-Ring-Problem-September-23-2026/build/sections/application.tex

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+\section{The countable construction and the actual center}
+\label{sec:countable-construction}
+
+We now apply Theorem~\ref{thm:lifting} repeatedly.  Recall its
+full conclusion: from a finite-dimensional central division algebra $A/k$
+and a tuple of free series $X$ whose constant term generates $A$ and whose
+two linear coefficients are independent modulo inner directions, it
+produces such a tuple over a countable extension $k'$ and a central
+division algebra of strictly larger degree.  It preserves every old
+polynomial identity and gives a monic algebraic relation for any chosen
+series $w$.  The countability, generation, and two-direction conclusions
+are all needed to iterate this theorem.
+
+The resulting division algebras need not embed in one another.  We use
+an ultraproduct to retain their polynomial relations, and then prove
+separately that the resulting algebra has infinite dimension over its
+actual center.
+
+\subsection{Scheduling all polynomial expressions}
+
+Start with
+\[
+ k_0=\overline{\mathbb Q},\qquad A_0=k_0,\qquad
+ X^{(0)}=(z_1,z_2).
+\]
+Its constant pair $(0,0)$ generates $k_0$ as a unital $k_0$-algebra.
+The inner-direction space is zero, and its two linear coefficients
+$(1,0)$ and $(0,1)$ are independent.  Thus it satisfies the hypotheses
+of Theorem~\ref{thm:lifting}.
+
+Inductively we obtain countable fields
+\[
+ k_0\subseteq k_1\subseteq k_2\subseteq\cdots,
+\]
+finite-dimensional central division algebras $A_i/k_i$ of strictly
+increasing degrees, and pairs
+\[
+ X^{(i)}\in A_i\langle\!\langle z_1,z_2\rangle\!\rangle^2
+\]
+satisfying the same generation and transverse-independence hypotheses.
+Write $(x_i,y_i)=X^{(i)}_0$ for the constant pair; in particular,
+\begin{equation}
+ A_i=k_i[x_i,y_i].
+ \label{eq:stage-generation}
+\end{equation}
+
+Here is a schedule that ensures algebraicity of every expression
+needed in the limit.  When $k_i$ has been constructed, enumerate the
+countable set $k_i\langle x,y\rangle$ as
+$(p_{i,j})_{j\geq0}$.  Enumerate the pairs $(i,j)\in\mathbb N^2$ by
+increasing $i+j$, in a fixed order within each diagonal.  If $(i_n,j_n)$
+is the $n$th pair, beginning at $n=0$, then $i_n\leq n$.
+At stage $n$, apply the lifting theorem to
+\[
+ w=p_{i_n,j_n}(X^{(n)}).
+\]
+This expression is defined over $k_n$, since $k_{i_n}\subseteq k_n$.
+The identity $w-p_{i_n,j_n}(X^{(n)})=0$ is preserved, so the new relation
+for the lifted $w$ gives a nonzero monic polynomial
+\[
+ q_{i_n,j_n}(t)\in k_{n+1}[t]
+\]
+such that
+\begin{equation}
+ q_{i_n,j_n}\bigl(p_{i_n,j_n}(X^{(n+1)})\bigr)=0.
+ \label{eq:retained-formal-relation}
+\end{equation}
+Every later lifting preserves this polynomial identity, now regarded
+over its coefficient field $k_{n+1}$.  Taking constant terms shows that
+the corresponding identity holds at $(x_h,y_h)$ for every $h\geq n+1$.
+Each polynomial over each $k_i$ is treated at a finite stage.
+
+\subsection{A two-generated algebraic division ring}
+
+Fix a nonprincipal ultrafilter $\mathcal U$ on $\mathbb N$ and form
+(see \cite{Keisler2010} for the general construction)
+\[
+ \mathcal D=\prod_{i\geq0} A_i\big/\mathcal U.
+\]
+Here two sequences represent the same element when their equality set
+belongs to $\mathcal U$.  This is a division ring: a nonzero class has
+nonzero coordinates on a set in $\mathcal U$, and taking their
+coordinatewise inverses on that set gives its inverse.
+
+Put $k_\infty=\bigcup_i k_i$.  An element $c\in k_j$ determines a class
+in $\mathcal D$ by taking its image in the central field $k_i\subset A_i$
+for $i\geq j$ and choosing arbitrary coordinates at the finitely many
+earlier indices.  These classes are independent of the early choices
+and of $j$.  They define a central field embedding
+\[
+ k_\infty\hookrightarrow\mathcal D:
+\]
+the field operations agree at every sufficiently large index, and a
+nonzero field element stays nonzero at all those indices.  This
+construction uses only the inclusions of coefficient fields; it does
+not require embeddings between the algebras $A_i$.
+
+Let
+\[
+ x=[(x_i)_i],\qquad y=[(y_i)_i],\qquad
+ R=k_\infty[x,y]\subseteq\mathcal D.
+\]
+Every $r\in R$ has the form $p(x,y)$ for some
+$p\in k_j\langle x,y\rangle$.  The schedule and
+\eqref{eq:retained-formal-relation} provide a nonzero monic polynomial
+$q\in k_h[t]$, for a suitable $h$, such that
+$q(p(x_i,y_i))=0$ at every sufficiently large index.  Therefore
+$q(r)=0$ in $\mathcal D$.  Every element of $R$ is algebraic over
+$k_\infty$.
+
+This already makes $R$ a division ring.  For $0\ne r\in R$, choose a
+nonzero relation of least degree
+\[
+ a_0+a_1r+\cdots+a_nr^n=0,
+ \qquad a_j\in k_\infty.
+\]
+Its constant coefficient is nonzero: otherwise multiplication by
+$r^{-1}$ in the ambient division ring $\mathcal D$ would give a shorter
+relation.  Consequently
+\[
+ r^{-1}=-a_0^{-1}(a_1+a_2r+\cdots+a_nr^{n-1})\in R.
+\]
+Since $k_\infty$ is countable and $R$ is generated by two elements as an
+algebra, $R$ is countable.  Its characteristic is zero.
+
+\subsection{Infinite dimension and passage to the center}
+
+We first prove
+\begin{equation}
+ [R:k_\infty]=\infty.
+ \label{eq:infinite-auxiliary-dimension}
+\end{equation}
+For each $i$ and $\ell\geq0$, let $V_{i,\ell}$ be the $k_i$-span of the
+words in $x_i,y_i$ of length at most $\ell$, including the empty word
+$1$.  If $V_{i,\ell}=V_{i,\ell+1}$, then $V_{i,\ell}$ is stable under
+right multiplication by $x_i$ and $y_i$ and contains every word.
+By \eqref{eq:stage-generation}, it equals $A_i$.  Until this happens,
+the dimensions increase strictly, starting from $1$.
+
+Fix $\ell$.  The dimensions $[A_i:k_i]$ tend to infinity, so
+\[
+ \dim_{k_i}V_{i,\ell}\geq\ell+1
+\]
+for every sufficiently large $i$.  There are only finitely many
+subsets of size $\ell+1$ in the finite list of words of length at most
+$\ell$.  The ultrafilter therefore selects one fixed subset
+$w_0,\ldots,w_\ell$ that is linearly independent at a set of stages
+belonging to $\mathcal U$.
+
+These same words are independent over the embedded $k_\infty$ in $R$.
+Indeed, the coefficients of a putative relation
+\[
+ \sum_{j=0}^{\ell} c_j w_j(x,y)=0
+\]
+all belong to one field $k_h$.  Equality in the ultraproduct holds on
+a set in $\mathcal U$.  Intersecting it with the independence set and
+the tail $i\geq h$ gives a nonempty set where stage independence
+forces every $c_j$ to vanish.  Thus $R$ has arbitrarily large finite
+linearly independent sets, proving
+\eqref{eq:infinite-auxiliary-dimension}.
+
+Finally let $F=Z(R)$ be the actual center.  It contains $k_\infty$ and
+is algebraic over $k_\infty$, since every element of $R$ is.  In
+particular, every element of $R$ is algebraic over $F$.  Also
+\[
+ R=k_\infty[x,y]=F[x,y]=F(x,y),
+\]
+because $F\subseteq R$ and $R$ is already a division ring.
+
+We use finite descent of the multiplication table and generator
+coordinates; compare the proof of the local-finiteness criterion in
+\cite[Section~4]{DeoBienHai2019}. Suppose, for a contradiction, that
+$[R:F]=n<\infty$. Choose an
+$F$-basis $b_1=1,b_2,\ldots,b_n$ of $R$, and write
+\[
+ b_i b_j=\sum_{h=1}^n c_{ijh}b_h,
+ \qquad
+ x=\sum_{h=1}^n u_hb_h,
+ \qquad
+ y=\sum_{h=1}^n v_hb_h,
+\]
+with all coefficients in $F$.  Let $E$ be the subfield of $F$
+generated over $k_\infty$ by these finitely many coefficients.
+Algebraicity gives $[E:k_\infty]<\infty$.  The space
+\[
+ B=\sum_{h=1}^n E b_h
+\]
+contains $1,x,y$ and is closed under multiplication by the displayed
+structure constants.  It is therefore a $k_\infty$-subalgebra of $R$
+containing $k_\infty[x,y]=R$, so $B=R$.  This implies
+\[
+ [R:k_\infty]\leq n[E:k_\infty]<\infty,
+\]
+contrary to \eqref{eq:infinite-auxiliary-dimension}.
+
+Thus the countable characteristic-zero division ring $D=R$ is
+algebraic over its actual center $F$, while
+\[
+ D=F[x,y]=F(x,y),\qquad [D:F]=\infty.
+\]
+The finite set $\{x,y\}$ witnesses the failure of local finiteness.
+
+This proves Theorem~\ref{thm:main}, subject only to Theorem~\ref{thm:lifting},
+which we now prove.

preprints/A-Counterexample-to-Kuroshs-Division-Ring-Problem-September-23-2026/build/sections/correction.tex

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+\section{Solving the free-series equations}
+\label{sec:free-correction}
+
+We now use the response and cyclic-sum results to keep the output of a
+substitution conjugate to its constant term.  The input retains two prescribed
+linear coefficients.  This is the point at which free words, rather than
+commuting parameters, enter the construction essentially.
+
+Work over a common split scalar field $E$ in which the preceding evaluations
+and spectral projections are defined.  All series substitutions below have
+small constant terms and are interpreted by the coefficientwise valuation
+convergence established earlier.  In particular, the letters of a free series
+commute with its matrix coefficients; the letters do not commute with each
+other.
+
+Put $m=11$ and index the blocks by $i\in\mathbb Z/m\mathbb Z$.
+Let $P_i$ be jointly independent small generic pairs of $b\times b$
+matrices.  Set $R=\operatorname{diag}(P_i)$, a pair of $mb\times mb$
+matrices.  The old coefficient algebra acts on a separate factor of dimension
+$d$.  Write $\lambda_{i,h}$ and $\pi_{i,h}$ for the distinct eigenvalues
+and spectral projections of $w(P_i)$.  Eigenvalues belonging to different
+blocks are distinct.  For a substitution $Z$ with constant term $R$, let
+$S_{i,h}(Z)$ denote the fixed-root Schur expression at $\lambda_{i,h}$,
+using the constant projection $\pi_{i,h}$ in the $i$-th block.
+
+We recall precisely the response information needed here.  In each block
+choose one eigenvalue label from every part of the active partition, giving
+$a$ selected labels.  The selected loop-response map is
+\[
+ D_i:M_b(E)^2\longrightarrow E^a.
+\]
+Put $L_i=\operatorname{im}D_i$ and $H_i=E^a/L_i$.
+Proposition~\ref{prop:compatible-relations} supplies a vector space
+$G=\mathcal G$ of dimension $g$ and coordinate isomorphisms
+\[
+ p_i:G\longrightarrow H_i^*.
+\]
+For a generic fixed input arrow $U$ on $i\leftarrow j$, varying its
+opposite arrow $V$ gives two selected responses
+\[
+ m_i(U,V)\in E^a,
+ \qquad m_j(V,U)\in E^a.
+\]
+The annihilator of these paired responses in $H_i^*\oplus H_j^*$ is
+exactly
+\begin{equation}
+ \{(p_i(c),p_j(c)):c\in G\}.
+ \label{eq:correction-paired-annihilator}
+\end{equation}
+The identifications $p_i$ are compatible for the whole collection of blocks
+and for all the generic arrows used below.
+
+The transfer result supplies the following additional facts.  If all Schur
+expressions vanish in degrees less than $\ell$, their degree-$\ell$
+coefficients are scalar on the projected eigenspaces.  The resulting tuples
+belong to their active spaces and are determined by their selected entries.
+Denote the selected error for a word $\omega$ of length $\ell$ in block
+$i$ by $e_{i,\omega}\in E^a$.  The cyclic-sum result says that, for every
+orbit $\mathcal O$ of words under cyclic rotation,
+\begin{equation}
+ \sum_{\omega\in\mathcal O}\sum_i
+ p_i(c)\bigl(e_{i,\omega}+L_i\bigr)=0
+ \qquad(c\in G).
+ \label{eq:correction-cyclic-input}
+\end{equation}
+These statements hold for every partial substitution under consideration,
+including after an extension of its scalar field.  The preceding results
+therefore supply all the hypotheses used in the next construction.
+
+\begin{proposition}
+\label{prop:eleven-block-correction}
+Prescribe generic input-arrow pairs on the edges
+\[
+ \begin{array}{c|c}
+ \text{letter}&\text{offsets }i-j\text{ of edges }i\leftarrow j\\ \hline
+ s_1&1,2\\
+ s_2&3,4.
+ \end{array}
+\]
+Under the response and cyclic-sum hypotheses above, there is a pair
+$Z(s_1,s_2)\in M_{mb}(E)\langle\!\langle s_1,s_2\rangle\!\rangle^2$
+with constant term $R$ and exactly these prescribed linear coefficients,
+for which
+\[
+ S_{i,h}(Z)=0\qquad\text{for every }i,h.
+\]
+It can be constructed by finite linear systems.  At step $\ell\ge3$ the
+new unknowns are coefficients of degree $\ell-1$ on the reverse edges
+and coefficients of degree $\ell$ on the diagonal blocks.
+\end{proposition}
+
+\begin{proof}
+None of the four forward offsets is zero.  The sum of two forward offsets
+lies between $2$ and $8$, and is therefore nonzero modulo $11$.
+Thus a path of one or two forward steps cannot return to its initial block.
+Constant input steps preserve each block.  The path expansions for the
+output and for its Schur expressions consequently show that the prescribed
+constant and linear terms, with zero higher coefficients, solve all the
+Schur equations through degree two.
+
+Suppose a partial substitution solves them in degrees less than
+$\ell\ge3$.  Add a correction $V B$ on a reverse edge $j\leftarrow i$,
+where $B$ is a word of length $\ell-1$, and add arbitrary degree-$\ell$
+loop corrections.  At the block-diagonal constant term, an off-diagonal
+input variation gives an output variation in the same off-diagonal block.
+Its compression on any fixed root block is zero.  Thus the reverse
+correction changes no Schur equation through degree $\ell-1$.
+
+Its contribution in degree $\ell$ must pair it with a single prescribed
+linear arrow.  There is a unique forward edge that returns along this
+reverse edge: the reverse offsets $10,9,8,7$ correspond respectively to
+$1,2,3,4$.  If that edge has coefficient $U$ and color $s_c$, the
+contributions are
+\begin{equation}
+ \begin{array}{c|c|c}
+ \text{block}&\text{word}&\text{selected response}\\ \hline
+ i&s_cB&m_i(U,V)\\
+ j&Bs_c&m_j(V,U).
+ \end{array}
+ \label{eq:correction-leading-pair}
+\end{equation}
+This is exactly the two-arrow Schur calculation defining the paired
+responses.  A previously present coefficient of degree at least two
+cannot interact with the new degree-$\ell-1$ coefficient in degree
+$\ell$.  Two new corrections have total degree at least
+$2(\ell-1)>\ell$.  The degree-$\ell$ variation is therefore linear.
+Loop corrections give $D_i$ independently at every vertex and every
+word of length $\ell$.
+
+We compute the annihilator of this finite correction map.  Let
+$f_{i,\omega}\in(E^a)^*$ be a collection of functionals annihilating its
+image.  The loop columns first imply that $f_{i,\omega}$ annihilates
+$L_i$.  Write its induced functional on $H_i$ uniquely as
+\[
+ f_{i,\omega}=p_i(c_{i,\omega}),\qquad c_{i,\omega}\in G.
+\]
+Here and below we use the same notation for the induced functional and
+its pullback to $E^a$.  By
+\eqref{eq:correction-paired-annihilator} and
+\eqref{eq:correction-leading-pair}, annihilation of the reverse columns
+is equivalent to
+\begin{equation}
+ c_{i,s_cB}=c_{j,Bs_c}
+ \quad\text{on every edge }i\leftarrow j\text{ of color }c.
+ \label{eq:correction-edge-equality}
+\end{equation}
+The two offsets of either color are consecutive integers, say $a,a+1$.
+Applying \eqref{eq:correction-edge-equality} to both edges out of $j$
+gives
+\[
+ c_{j+a,s_cB}=c_{j+a+1,s_cB}.
+\]
+As $j$ ranges over $\mathbb Z/11\mathbb Z$, this makes the value
+independent of the vertex.  Equation~\eqref{eq:correction-edge-equality}
+then identifies its value with that on the rotated word $Bs_c$.
+Iterating rotations shows that the annihilator consists exactly of one
+element of $G$ per cyclic orbit of words.  The converse follows from
+the same paired-annihilator identity.
+
+Equation~\eqref{eq:correction-cyclic-input} says that every such
+annihilator kills the degree-$\ell$ error.  Finite-dimensional linear
+duality therefore puts the negative error in the image of the correction
+map.  Choose a correction solving all selected equations.  Lower degrees
+remain solved, so transfer and the active partition imply that every
+nonselected equation vanishes in degree $\ell$ as well.
+
+Continue inductively.  A coefficient of degree $n$ can change only at
+steps $n$ and $n+1$, so the resulting free series is well-defined.
+Every fixed Schur coefficient is eventually zero and remains zero.
+The construction also covers $g=0$, when the annihilator is zero.
+For a periodic word the orbit consists of its distinct rotations; no
+multiplicity is inserted into the annihilator calculation or into
+\eqref{eq:correction-cyclic-input}.
+\end{proof}
+
+\begin{figure}[tbp]
+\centering
+\input{figures/eleven-block-offsets}
+\caption{The eleven-block input pattern. Indices lie in $\mathbb Z/11\mathbb Z$.
+The left panel shows the offset-one cycle; the right panel specifies all
+four forward offsets, translated at every vertex. Each forward arrow carries a
+generic pair of input matrices. At order $\ell\ge3$, a reverse correction
+has degree $\ell-1$; degree-$\ell$ loop corrections are not drawn.}
+\label{fig:eleven-blocks}
+\end{figure}
+
+Figure~\ref{fig:eleven-blocks} separates the generating cycle from the
+full local arrow pattern. The role of the four offsets is now explicit.  Avoiding returns of length
+one or two starts the induction without changing the linear data.
+Consecutive offsets within each color make its only obstructions
+independent of the vertex.  The remaining obstructions are exactly the
+cyclic sums already proved to vanish.
+
+\subsection{Fixed roots and Schur complements}
+
+The following elementary fact identifies the spectral equations with a
+single polynomial identity. Its coefficient ring need not be commutative.
+Let $E$ be a field, let $\mathfrak a=(s_1,s_2)$ be the augmentation ideal
+of the free algebra $E\langle s_1,s_2\rangle$, and put
+\[
+  T_N=E\langle s_1,s_2\rangle/\mathfrak a^{N+1}
+  \qquad (N\geq 0).
+\]
+Elements of $E$ are central in $T_N$. Reduction modulo $\mathfrak a$
+is called the constant term.
+
+\begin{lemma}\label{lem:schur-polynomial}
+Let $W\in M_d(T_N)$ have constant term
+\[
+  W_0=\sum_{h=1}^{e}\lambda_h\pi_h,
+  \qquad
+  \pi_h\pi_j=\delta_{hj}\pi_h,
+  \qquad
+  \sum_{h=1}^{e}\pi_h=I_d,
+\]
+where the $\lambda_h\in E$ are distinct and the $\pi_h\in M_d(E)$
+are nonzero projections. Set
+\[
+  q(t)=\prod_{h=1}^{e}(t-\lambda_h),
+  \qquad \bar\pi_h=I_d-\pi_h.
+\]
+On the complementary summand $\bar\pi_h T_N^d$, the operator
+\[
+  C_h=\bar\pi_h(W-\lambda_h I_d)\bar\pi_h
+\]
+is invertible. Define the Schur expression on $\pi_h T_N^d$ by
+\[
+  S_h=\pi_h(W-\lambda_h I_d)\pi_h
+      -\pi_hW\bar\pi_h C_h^{-1}\bar\pi_hW\pi_h.
+\]
+Then $q(W)=0$ if and only if $S_h=0$ for every $h$.
+\end{lemma}
+
+\begin{proof}
+The constant term of $C_h$ is
+$\sum_{j\ne h}(\lambda_j-\lambda_h)\pi_j$, which is invertible on
+$\bar\pi_h E^d$. A matrix over $T_N$ with invertible constant term is
+invertible: after factoring out its constant term, the inverse is a
+finite geometric series in a matrix with entries in $\mathfrak a$.
+This proves the assertion about $C_h$. All modules below are right
+$T_N$-modules, with matrices acting on column vectors from the left.
+
+Suppose first that every $S_h$ vanishes. Define
+$K_h:\pi_hT_N^d\longrightarrow T_N^d$ by
+\[
+  K_hu=u-C_h^{-1}\bar\pi_hW\pi_hu,
+\]
+where the second term lies in the complementary summand. Then
+$\pi_hK_hu=u$, and direct block multiplication gives
+\[
+  (W-\lambda_h I_d)K_hu=S_hu=0.
+\]
+The map $K=\sum_hK_h\pi_h$ has constant term $I_d$, so it is
+invertible. The displayed kernel equations imply
+$WK=KW_0$. Consequently
+\[
+  q(W)=Kq(W_0)K^{-1}=0.
+\]
+
+Conversely, suppose $q(W)=0$. Define
+\[
+  P_h=\prod_{j\ne h}
+       \frac{W-\lambda_j I_d}{\lambda_h-\lambda_j}.
+\]
+These products are unambiguous because their factors are polynomials
+in the same matrix with central scalar coefficients. Polynomial
+interpolation modulo $q$ shows that the $P_h$ are spectral
+idempotents; in particular,
+\[
+  (W-\lambda_h I_d)P_h=0,
+  \qquad P_h\equiv\pi_h\pmod{\mathfrak a}.
+\]
+Let $\iota_h:\pi_hT_N^d\longrightarrow T_N^d$ be inclusion. The
+endomorphism $A_h=\pi_hP_h\iota_h$ of $\pi_hT_N^d$ has constant
+term the identity, and is therefore invertible. Thus
+\[
+  \widetilde K_h=P_h\iota_hA_h^{-1}
+\]
+satisfies
+$\pi_h\widetilde K_h=\operatorname{id}_{\pi_hT_N^d}$ and
+$(W-\lambda_h I_d)\widetilde K_h=0$. The complementary component
+of the latter equation determines its remaining component uniquely:
+\[
+  \bar\pi_h\widetilde K_h
+    =-C_h^{-1}\bar\pi_hW\pi_h.
+\]
+Its $\pi_h$ component now gives $S_h=0$.
+\end{proof}
+
+For a matrix over $E\langle\!\langle s_1,s_2\rangle\!\rangle$,
+apply Lemma~\ref{lem:schur-polynomial} after reduction to $T_N$.
+It follows that the fixed-root Schur equations hold through degree
+$N$ exactly when the identity $q(W)=0$ holds through degree $N$.
+In particular, this equivalence applies at every finite stage of a
+formal correction, with the same polynomial $q$ throughout.

preprints/A-Counterexample-to-Kuroshs-Division-Ring-Problem-September-23-2026/build/sections/cyclic-obstruction.tex

@@ -0,0 +1,406 @@
+\section{Cyclic relations on the leading free-word errors}
+\label{sec:cyclic-obstructions}
+
+The space \(\mathcal G\) controls the linear corrections available
+at several blocks. We must show that the errors to be corrected
+satisfy its relations for individual cyclic orbits of free words.
+Commuting scalar substitutions would combine too many words. We
+instead substitute matrices, repeat each block, and attach hubs
+whose derivative columns identify the possible left nullspace.
+
+Traces of words and their cyclic symmetry are classical in matrix
+invariant theory; see \cite[\S4]{Procesi1976}. Here the response
+relations must first be proved to annihilate every matrix trace test.
+An explicit directed-cycle calculation then isolates each cyclic orbit.
+
+Fix \(m\geq1\) independent generic size-\(b\) points \(P_i\).
+Write \(I_i\) for their selected root labels, so \(|I_i|=a\), and
+let \(\mathcal G\subseteq\bigoplus_i H_{P_i}^*\) be the space
+of Proposition~\ref{prop:compatible-relations}. As before, all
+scalars may be extended to a common field containing the required
+splitting data.
+
+\begin{theorem}[Cyclic compatibility of the error]
+\label{thm:cyclic-compatibility}
+Let \(Z(s_1,s_2)\) be a pair of formal series in two free letters,
+with constant term \(\operatorname{diag}(P_1,\ldots,P_m)\).
+Suppose that the fixed-root Schur expressions \(S_{i,h}\), for
+every block \(i\) and every root label \(h\), vanish in every
+degree below \(\ell\), where \(\ell\geq1\).
+By transfer, their degree-\(\ell\) coefficients are scalar on
+their root spaces; write
+\[
+ [s^\omega]S_{i,h}=e_{i,h,\omega}\pi_{i,h},
+ \qquad |\omega|=\ell.
+\]
+For each word the scalar tuple at block \(i\) belongs to its
+active space. For every cyclic orbit \(\mathcal O\) of words of
+length \(\ell\) and every \(c\in\mathcal G\),
+\begin{equation}
+ \sum_{\omega\in\mathcal O}
+ \sum_{i=1}^m\sum_{h\in I_i}
+ c_{i,h}e_{i,h,\omega}=0.
+ \label{eq:cyclic-error-relation}
+\end{equation}
+The orbit sum runs over distinct words, including when a word has
+a nontrivial rotational stabilizer.
+\end{theorem}
+
+We prove the theorem in three steps. First, arbitrary matrix
+substitutions turn the free-word errors into matrices on a repeat
+factor. Second, a derivative test shows that every relation in
+\(\mathcal G\) annihilates their traces. Finally, directed-cycle
+matrices isolate each cyclic orbit.
+
+\begin{lemma}[Splitting repeated blocks]
+\label{lem:repeated-splitting}
+Fix \(L\geq1\) and arbitrary scalar matrices
+\(S_1,S_2\in M_L(E)\). Substitute \(s_c=\delta S_c\) into
+\(Z\), with a commuting Taylor parameter \(\delta\), and add
+an order-\(\delta^\ell\) input loop \(Y_i\) in each old block
+\(i\). Here \(Y_i\) is an arbitrary pair in
+\(M_b(E)\otimes M_L(E)\). Put
+\[
+ B_{i,h}=\sum_{|\omega|=\ell}
+ e_{i,h,\omega}S^\omega+D_{i,h}(Y_i),
+\]
+where \(D_{i,h}\) is applied entrywise to the repeat-factor
+coefficients. One can choose the \(Y_i\) so that every \(B_{i,h}\),
+for all root labels \(h\), has simple spectrum. For such a choice,
+after a finite splitting extension the leaf output has branches
+\[
+ \lambda_{i,h,j}(\delta)
+ =\lambda_{i,h}+\delta^\ell\theta_{i,h,j}
+    +O(\delta^{\ell+1}),\qquad 1\leq j\leq L,
+\]
+each scalar of multiplicity \(t\), where \(\theta_{i,h,j}\)
+are the eigenvalues of \(B_{i,h}\). Their spectral projections
+are integral in \(\delta\) and reduce to
+\(\pi_{i,h}\otimes\Pi_{i,h,j}\), where
+\(\Pi_{i,h,j}\) are the rank-one spectral projections of
+\(B_{i,h}\).
+\end{lemma}
+
+\begin{proof}
+Matrix substitution is legitimate coefficientwise in \(\delta\).
+The leading Schur expression on the repeated \((i,h)\)-space is
+\(I_t\otimes B_{i,h}\): the original errors give the first sum,
+and an order-\(\delta^\ell\) input loop contributes its first
+response. All lower orders are zero.
+
+For each label \(h\), \(D_{i,h}\) is a nonzero linear functional
+on the input directions. Its matrix-valued extension is therefore
+onto \(M_L(E)\). The discriminant of \(B_{i,h}\), as a polynomial
+in the loop coordinates, is consequently not identically zero.
+There are finitely many labels, so the product of these discriminants
+is nonzero. Over the infinite field in use, or a scalar extension,
+choose loops at which their product is nonzero. This choice is made
+for all labels, not merely the selected ones. For \(L=1\) the
+simple-spectrum condition is automatic.
+
+Separate the old distinct roots by a formal conjugation equal to
+the identity at \(\delta=0\). The separated \((i,h)\)-block is
+\begin{equation}
+ \lambda_{i,h}I+\delta^\ell(I_t\otimes B_{i,h})
+ +O(\delta^{\ell+1}).
+ \label{eq:old-cluster-expansion}
+\end{equation}
+Here is an exact comparison with the fixed-root Schur expression.
+Write \(M=W-\lambda_{i,h}I\) in the fixed root/complement
+decomposition, with blocks \(M_{uv}\). Its complementary block
+\(M_{22}\) is invertible. If \(F\) is the separated root block
+minus \(\lambda_{i,h}I\), its invariant columns can be written
+\(\binom QR\), with \(Q=I+O(\delta)\), \(R=O(\delta)\), and
+\(M\binom QR=\binom QR F\). Eliminating the lower row gives
+\[
+ \bigl(M_{11}-M_{12}M_{22}^{-1}M_{21}\bigr)Q
+   =\bigl(Q-M_{12}M_{22}^{-1}R\bigr)F.
+\]
+The left parenthesis is the fixed-root Schur expression. Both
+outer factors relating it to \(F\) reduce to identity at zero.
+Consequently it and \(F\) have the same first nonzero coefficient.
+Its vanishing below degree \(\ell\) forces the same vanishing
+for \(F\), and its degree-\(\ell\) coefficient is
+\(I_t\otimes B_{i,h}\). This proves
+\eqref{eq:old-cluster-expansion}.
+
+Within this block divide the change from \(\lambda_{i,h}I\) by
+\(\delta^\ell\). Its constant term is \(I_t\otimes B_{i,h}\),
+with \(L\) distinct eigenvalues. A second formal conjugation,
+integral in \(\delta\), separates these slopes into blocks of
+size \(t\). Across all old labels there are exactly
+\[
+ mLe_b=e_{bmL}
+\]
+such blocks. Their characteristic polynomials are pairwise coprime
+over the Laurent fraction field: different old roots are already
+distinct at order zero, and distinct slopes separate the blocks
+inside one old root. The minimal-degree bound at size \(bmL\)
+is \(e_{bmL}\). Each separated block contributes at least one
+to that degree, so each contributes exactly one and is scalar.
+The two integral conjugations also give integral spectral
+projections with the stated limits.
+\end{proof}
+
+Keep a choice as in Lemma~\ref{lem:repeated-splitting}. Adjoin
+\(L+1\) independent generic size-\(b\) hub blocks, indexed by
+\(p=1,\ldots,L\) and by \(*\), with spectra disjoint from the
+old blocks. The individual hub \(p\) sends an input arrow to
+repeat position \(p\) of every old block \(i\), with fixed generic
+value \(U_{i,p}\). The common hub \(*\) sends the same generic
+value \(U_{i,*}\) to every repeat position in old block \(i\).
+There are no reverse arrows at this test input. Arrow values for
+different pairs of old blocks and hubs can be chosen independently
+and to satisfy the surjectivity conditions of
+Lemma~\ref{lem:pair-quotient-surjectivity}.
+
+Figure~\ref{fig:hub-test} shows the distinction between the individual
+hubs and the common hub.
+
+\begin{figure}[tbp]
+\centering
+\input{figures/repeated-leaves-hubs}
+\caption{The hub test at $\delta=0$, for one old block $i$ and $L=3$.
+Hub $p$ sends a generic matrix pair only to position $p$; the common hub
+sends the same pair to all positions. Every old block uses these same
+$L+1$ hubs. At $\delta\ne0$ the leaf input may couple repeat positions.
+The individual hubs eliminate off-diagonal dual entries, and the common
+hub identifies the diagonal functionals.}
+\label{fig:hub-test}
+\end{figure}
+
+Treat the entire deformed leaf array as one block and order it
+before the hubs. Over the \(\delta\)-Laurent field this is a star
+with maximal root count. Put
+\[
+ N=mL+L+1.
+\]
+Select the \(L\) branches over each old selected label and the
+selected labels of all hubs, for a total of \(Na\) output rows.
+The rank of their derivative matrix is at most
+\(\rho_{bN}\leq Na-g\). This bound follows from the original
+trace-minor identities, evaluated coefficientwise at the present
+Taylor input; it does not require the specialized input to be
+generic.
+
+\newpage
+\begin{lemma}[The limiting derivative columns]
+\label{lem:limiting-columns}
+The following derivative columns at the preceding star have finite
+limits at \(\delta=0\).
+\begin{enumerate}
+\item A loop \(Y\otimes T\) inside old block \(i\) has limit
+\[
+ D_{i,h}(Y)\operatorname{tr}(\Pi_{i,h,j}T)
+\]
+on row \((i,h,j)\), and zero on all other rows. Hub loops give
+their own loop responses.
+\item For individual hub \(p\), a reverse arrow of value \(V\)
+from repeat position \(r\) in old block \(i\) has limit
+\begin{align*}
+ &m_{i,h}(P_i,Q_p;U_{i,p},V)
+      \operatorname{tr}(\Pi_{i,h,j}E_{pr})
+       &&\text{on row }(i,h,j),\\
+ &\mathbf 1_{p=r}\,
+   m_{p,k}(Q_p,P_i;V,U_{i,p})
+       &&\text{on hub row }(p,k),
+\end{align*}
+and zero elsewhere. Here \(E_{pr}\) is the repeat-factor matrix
+unit with row \(p\) and column \(r\).
+\item For the common hub, use arrow \(U_{i,*}\) and replace the
+repeat-factor matrix by
+\[
+ \sum_{p=1}^L E_{pr}.
+\]
+The hub contribution is the single response
+\(m_{*,k}(Q_*,P_i;V,U_{i,*})\).
+\item The derivative along the whole input \(\delta\)-curve,
+multiplied by \(\delta^{1-\ell}\), has limit
+\(\ell\theta_{i,h,j}\) on leaf rows and zero on hub rows.
+\end{enumerate}
+The matrix consisting of a finite spanning collection of these
+limiting columns has rank at most \(Na-g\).
+\end{lemma}
+
+\begin{proof}
+Over the Laurent field, apply Lemma~\ref{lem:star-derivatives} with
+the whole leaf array as one block. For loop derivatives, the
+bounded leaf projection tends to
+\(\pi_{i,h}\otimes\Pi_{i,h,j}\). Compression of the ordinary
+first response gives \(D_{i,h}(Y)\) on the multiplicity space,
+and taking trace divided by \(t\) gives the first formula.
+
+For paired derivatives on a leaf, the Schur inverse is on the hub
+at a root \(\lambda_{i,h,j}(\delta)\); that root stays away from
+the hub spectrum at \(\delta=0\). For paired derivatives on a
+hub, the inverse is on the leaf array at a hub root, which stays
+away from every old leaf root. These inverses are regular at
+\(\delta=0\). No inverse at the difference of two approaching
+leaf roots occurs in this calculation.
+
+At the reduction, the leaf input is diagonal between old blocks
+and repeated inside each old block. A forward arrow at position
+\(p\) followed by a reverse arrow from position \(r\) inserts
+\(E_{pr}\) on the leaf repeat factor. At the hub end, the repeat
+indices must match, giving \(\mathbf 1_{p=r}\). The ordinary
+paired-response formula and trace divided by \(t\) now give the
+second formula. Summing forward positions gives the common-hub
+formula. The bounded projections and the distant-root inverses
+justify all these limits coefficientwise.
+
+The curve column lies in the span of input derivative columns over
+the Laurent field by the chain rule. The expansion in
+Lemma~\ref{lem:repeated-splitting} gives
+\[
+ \delta^{1-\ell}\frac{d}{d\delta}
+       \lambda_{i,h,j}(\delta)
+ =\ell\theta_{i,h,j}+O(\delta).
+\]
+Hub roots are constant on the curve. Thus this column also has a
+finite limit. Choose bases for the finitely many loop and reverse
+direction spaces and include the curve column. Every
+\((Na-g+1)\)-minor of the resulting matrix vanishes over the
+Laurent field; since its entries have finite limits, these minors
+remain zero after specialization. This proves the last assertion.
+\end{proof}
+
+We now use the individual hubs to remove off-diagonal repeat-factor
+weights and the common hub to make all diagonal weights agree.
+The rank bound will then show that no element of \(\mathcal G\)
+has been lost.
+
+\begin{lemma}[The left nullspace]
+\label{lem:hub-nullspace}
+The left nullspace of the limiting matrix in
+Lemma~\ref{lem:limiting-columns}, including its curve column,
+has dimension \(g\). Its projection to the old blocks identifies
+it with \(\mathcal G\): for each \(c\in\mathcal G\), its leaf
+weights are \(c_{i,h}\) on every row \((i,h,j)\), independently
+of \(j\), and its hub weights are the uniquely determined
+compatible functionals. Consequently
+\begin{equation}
+ \sum_i\sum_{h\in I_i}c_{i,h}\operatorname{tr}(B_{i,h})=0
+ \qquad(c\in\mathcal G).
+ \label{eq:error-matrix-trace}
+\end{equation}
+\end{lemma}
+
+\begin{proof}
+Let \(\mu_{i,h,j}\) be leaf row weights in the left nullspace
+and let \(\nu_p,\nu_*\) be the hub weights. Define
+\[
+ T_{i,h}=\sum_{j=1}^L\mu_{i,h,j}\Pi_{i,h,j}\in M_L(E).
+\]
+The loop columns imply
+\[
+ \sum_{h\in I_i}D_{i,h}(Y)
+          \operatorname{tr}(T_{i,h}T)=0
+ \qquad\text{for every }Y,T.
+\]
+Thus, for every matrix entry \((r,p)\), the coordinate vector
+\( ((T_{i,h})_{rp})_{h\in I_i}\) defines a functional in
+\(H_{P_i}^*\). The hub weights similarly belong to their own
+loop annihilators.
+
+Fix an individual hub \(p\) and a reverse position \(r\ne p\).
+There is no hub response. Since
+\[
+ \operatorname{tr}(T_{i,h}E_{pr})=(T_{i,h})_{rp},
+\]
+the reverse columns say that the functional with these
+\((r,p)\)-entries kills the leaf endpoint projection of the paired
+map. That projection is onto \(H_{P_i}\), so all these entries
+vanish. As \(p,r\) vary, every \(T_{i,h}\) is diagonal. When
+\(r=p\), the corresponding diagonal-entry functional and the
+hub functional \(\nu_p\) are exactly a pair in the relation of
+Proposition~\ref{prop:compatible-relations}.
+
+For the common hub, the leaf expression is
+\[
+ \operatorname{tr}\left(T_{i,h}\sum_pE_{pr}\right)
+   =\sum_p(T_{i,h})_{rp}=(T_{i,h})_{rr}.
+\]
+For every \(r\), this diagonal-entry functional is therefore
+paired with the same \(\nu_*\). Projection of the pair relation
+onto the hub dual space is an isomorphism, so the diagonal-entry
+functionals are independent of \(r\). Compatibility for the
+collection of old blocks and hubs identifies them across all old
+blocks. Hence
+\[
+ T_{i,h}=c_{i,h}I_L,
+ \qquad c\in\mathcal G,
+\]
+and all hub weights are uniquely determined by \(c\). Since the
+rank-one projections \(\Pi_{i,h,j}\) are mutually orthogonal
+and sum to identity, this also gives
+\(\mu_{i,h,j}=c_{i,h}\). The map from the left nullspace to
+\(\mathcal G\) is consequently injective.
+
+There are \(Na\) rows, and the limiting rank is at most \(Na-g\)
+even after the curve column is included. The nullspace therefore
+has dimension at least \(g\). Its injection into the
+\(g\)-dimensional space \(\mathcal G\) is an isomorphism.
+In particular every \(c\in\mathcal G\), not just a possible
+subspace of relations, occurs. Pairing its null vector with the
+curve column and cancelling the nonzero integer \(\ell\) yields
+\[
+ 0=\sum_i\sum_{h\in I_i}\sum_{j=1}^L
+       c_{i,h}\theta_{i,h,j}
+  =\sum_i\sum_{h\in I_i}c_{i,h}\operatorname{tr}(B_{i,h}).
+\]
+If \(g=0\), the same proof gives zero nullspace and the assertion
+about \(c\) is vacuous. If \(L=1\), the off-diagonal step has
+no entries to remove; the common-hub step still gives precisely
+the stated conclusion.
+\end{proof}
+
+\begin{proof}[Proof of Theorem~\ref{thm:cyclic-compatibility}]
+Apply Lemma~\ref{lem:hub-nullspace} to arbitrary \(S_1,S_2\)
+and to the auxiliary loops chosen in
+Lemma~\ref{lem:repeated-splitting}. The added-loop contribution
+to \eqref{eq:error-matrix-trace} is zero. Indeed, write
+\(Y_i=\sum_\alpha Y_{i,\alpha}\otimes T_{i,\alpha}\). Then
+\[
+ \sum_{h\in I_i}c_{i,h}\operatorname{tr}(D_{i,h}(Y_i))
+ =\sum_\alpha\operatorname{tr}(T_{i,\alpha})
+      \sum_{h\in I_i}c_{i,h}D_{i,h}(Y_{i,\alpha})=0,
+\]
+because \(c_i\) annihilates the loop image. We conclude that,
+for arbitrary matrix substitutions,
+\begin{equation}
+ \sum_{|\omega|=\ell}
+ \left(\sum_i\sum_{h\in I_i}c_{i,h}e_{i,h,\omega}\right)
+ \operatorname{tr}(S^\omega)=0.
+ \label{eq:all-matrix-trace-tests}
+\end{equation}
+Auxiliary field extensions impose no restriction: this equality
+has its coefficients in the original field, and extension is
+injective.
+
+We use a matrix-cycle test of the kind appearing in the constructive
+trace argument of \cite[Proposition~4.1]{KlepSpenko2014}, recording the
+rotational multiplicity explicitly. Fix a word
+\(\omega_0=c_0c_1\cdots c_{\ell-1}\), with letters
+\(c_j\in\{1,2\}\), and take \(L=\ell\). On vertices indexed
+modulo \(\ell\), set
+\[
+ S_c=\sum_{j:\,c_j=c}E_{j,j+1}\qquad(c=1,2).
+\]
+For a word \(\omega\) of length \(\ell\), its matrix trace
+counts the starting vertices at which one reads \(\omega\)
+around this directed cycle. It is zero unless \(\omega\) lies
+in the cyclic orbit \(\mathcal O\) of \(\omega_0\); for every
+distinct word in that orbit its value is the same positive integer
+\[
+ \frac{\ell}{|\mathcal O|}.
+\]
+This is the rotational stabilizer size and includes periodic and
+constant-letter words. Substitution into
+\eqref{eq:all-matrix-trace-tests}, followed by cancellation of that
+integer in characteristic zero, gives
+\eqref{eq:cyclic-error-relation}.
+\end{proof}
+
+The leading errors now satisfy every cyclic relation imposed by
+\(\mathcal G\). The colored correction graph will make these
+exactly the annihilators of its order-by-order correction map.

preprints/A-Counterexample-to-Kuroshs-Division-Ring-Problem-September-23-2026/build/sections/division-realization.tex

@@ -0,0 +1,240 @@
+\section{Realization in a central division algebra}
+\label{sec:division-realization}
+
+The preceding construction solved equations in split matrices.  We now
+place its constant blocks and prescribed arrows in one central division
+algebra.  Each subsequent correction is the solution of a finite linear
+system over its center.  This permits descent without any assertion that
+solutions of nonlinear equations descend.
+
+Throughout this section $k$ has characteristic zero and contains all roots
+of unity, and $A$ is a central division algebra of degree $d$ over $k$.
+The integer $b$ is the block size chosen earlier in the spectral case.
+In the case where $q_0(w)=0$ already holds as a free-series identity, any
+positive integer $b$ may be used.  Put $m=11$.
+
+\subsection{The two division-algebra factors}
+
+Choose independent central indeterminates $\alpha,\beta$ and put
+$k_0=k(\alpha,\beta)$.  Let $B_b$ be the symbol algebra over $k_0$
+generated by $u,v$ with
+\[
+ u^b=\alpha,\qquad v^b=\beta,\qquad vu=\zeta_buv,
+\]
+where $\zeta_b$ is a primitive $b$-th root of unity.  For $b=1$ this
+means $B_1=k_0$.  By the construction in Section~\ref{subsec:generic-division}, $B_b$
+is central simple of degree $b$.  Lemma~\ref{lem:symbol-division}
+shows that $A\otimes_k B_b$ is division and remains division under
+pure rational and central Laurent scalar extensions.  We will use
+these facts twice in adjoining the cyclic block factor.
+
+Fix a $k_0$-basis $(e_a)_{a=1}^{b^2}$ of $B_b$.  Over $k_0$ introduce
+the following jointly independent variables:
+\[
+ x_{c,a,j},\qquad y_{r,c,a,j},
+ \qquad
+ \begin{matrix}
+ c\in\{1,2\},\quad 1\le a\le b^2,\\
+ j\in\mathbb Z/m\mathbb Z,\quad r\in\{1,2,3,4\}.
+ \end{matrix}
+\]
+Let $L_0$ be their rational function field and let $\sigma$ increase
+the index $j$ by one, fixing $k_0$.  This automorphism has order exactly
+$m$.  Put $H_0=L_0^{\langle\sigma\rangle}$.  Artin's fixed-field theorem
+for a finite group of automorphisms gives a cyclic Galois extension
+$L_0/H_0$ of degree $m$; see \cite[Corollary~3.11, p.~38]{MilneFT2022}.
+
+Choose a further independent central indeterminate $a_*$ and form the
+cyclic algebra $C$ over $F_0=H_0(a_*)$ with coefficient field
+$L_0(a_*)$ and generator $\tau$, subject to
+\begin{equation}
+ \tau^m=a_*,\qquad \tau c\tau^{-1}=\sigma(c)
+ \quad(c\in L_0).
+ \label{eq:cyclic-realization-relations}
+\end{equation}
+It has vector-space decomposition
+$C=\bigoplus_{r=0}^{m-1}L_0(a_*)\tau^r$.
+After extending scalars to a field containing $L_0(a_*)$, represent
+$c\in L_0$ by the diagonal with entries $\sigma^{-i}(c)$ and
+$\tau$ by the cyclic forward shift, whose wrap entry is $a_*$.
+These matrices satisfy \eqref{eq:cyclic-realization-relations} and
+span all matrix units.  This gives the usual splitting of $C$ and
+shows that it is central simple of degree $m$; see
+\cite[Construction~2.5.1 and Proposition~2.5.2]{GilleSzamuely2006}.
+Our conjugation convention uses the inverse of their Galois generator.
+
+Define
+\[
+ \widetilde B_0=(B_b\otimes_{k_0}F_0)\otimes_{F_0}C,
+ \qquad D_0=A\otimes_k\widetilde B_0.
+\]
+Then $D_0$ is central simple of degree $dbm$ over $F_0$ by the tensor-product and scalar-extension properties of
+central simple algebras
+\cite[Chapter~IV, Corollary~2.8 and Proposition~2.15]{MilneCFT}.
+It is also division. Indeed
+\[
+ D_L=(A\otimes_k B_b)\otimes_{k_0}L_0
+\]
+is division by the rational-extension argument.  Extend $\sigma$ to
+$D_L$ by fixing the old algebra factors.  The skew Laurent division
+ring $D_L((t;\sigma))$ receives a unital homomorphism from $D_0$
+by $\tau\mapsto t$ and $a_*\mapsto t^m$.  The central element
+$t^m$ is transcendental over $H_0$.  Simplicity again makes this map
+injective, and finite dimension makes $D_0$ division.
+
+Finally put
+\[
+ F=F_0((\varepsilon)),\qquad
+ \widetilde B=\widetilde B_0\otimes_{F_0}F,\qquad
+ D=D_0\otimes_{F_0}F.
+\]
+The central Laurent argument shows that $D$ is a central division
+algebra over $F$ of degree $dbm$.  Input coefficients will lie in
+$\widetilde B$, so that they commute with the old factor $A$.
+
+\subsection{One realization of all generic data}
+
+For $r\in\mathbb Z/m\mathbb Z$, let
+\[
+ \mathcal V_r=
+ \bigl(B_b\otimes_{k_0}L_0\bigr)\tau^r
+ \otimes_{H_0}F
+ \ \subseteq\ \widetilde B.
+\]
+This notation uses the evident coefficient embedding and scalar
+extension; equivalently it is the $F$-span of that graded summand in
+$\widetilde B_0$.  Upon splitting, $\mathcal V_r$ consists exactly
+of all $b\times b$ entries at offset $r$ in the $m$-block matrix.
+In fact $L_0\otimes_{H_0}E\simeq E^m$ over a splitting field $E$,
+and $B_b$ becomes $M_b(E)$.  Multiplication by $\tau^r$ moves these
+independent diagonal entries to offset $r$, with nonzero wrap factors.
+In particular $\mathcal V_0$ gives all diagonal blocks, and
+$\mathcal V_{-r}$ gives all reverse entries for offset $r$.
+
+Define the two constant inputs and four arrow pairs by
+\[
+ P_c^0=\varepsilon\sum_a x_{c,a,0}e_a,
+ \qquad
+ U_{r,c}=\left(\sum_a y_{r,c,a,0}e_a\right)\tau^r
+ \quad(c=1,2).
+\]
+The prescribed initial substitution is the pair
+\begin{equation}
+ Z_c^{\mathrm{init}}(s_1,s_2)
+ =P_c^0+(U_{1,c}+U_{2,c})s_1
+             +(U_{3,c}+U_{4,c})s_2.
+ \label{eq:division-initial-input}
+\end{equation}
+After splitting, the diagonal blocks of $(P_1^0,P_2^0)$ are exactly
+the required independent small generic pairs.  The two input-component
+entries at every forward edge are independent generic entries, jointly
+with all the diagonal data.
+
+Here independence is over the old split coefficient field, which is the
+field used in the generic rank tests.  To check it, first split $A$ and
+$B_b$ using constants independent of the variables $x,y$.  The $m$
+conjugates of each coefficient use the disjoint sets of variables with
+different index $j$.  The split symbol basis is a matrix basis.
+Permuting those variables, changing that basis, and multiplying entries
+by powers of $a_*$ preserve their algebraic independence.  This verifies
+all generic determinant conditions simultaneously.  No independence
+over the larger invariant field $H_0$ is asserted or needed.
+
+We may perform the spectral calculations after extending $F$ to a
+split Laurent field and then to the finite extensions needed for
+eigenvalues.  Its constant coefficient field contains the old algebraic
+closure and the rational variables just chosen.  The matrix constant
+terms remain $\varepsilon$-small.  Arrow coefficients need not be
+small, since they occur only with positive free-word degree.
+
+\subsection{Linear descent of the correction}
+
+Assume first that $q_0(w)\ne0$.  Let $W_0=w(P_1^0,P_2^0)\in D$
+and choose its monic minimal polynomial
+\begin{equation}
+ q(t)\in F[t].
+ \label{eq:fixed-central-polynomial}
+\end{equation}
+We keep this polynomial fixed throughout the correction.  It is
+irreducible, because $F[W_0]$ is a finite-dimensional commutative
+domain, and separable because the characteristic is zero.  In a
+splitting field its roots are exactly the distinct eigenvalues of the
+constant output.  For completeness, left multiplication by $W_0$ on
+$D$ has characteristic polynomial a power of $q$, since $D$ is a
+vector space over $F[W_0]$.  After splitting, that left multiplication
+is a direct sum of copies of the action of the split matrix $W_0$.
+Every root of $q$ therefore occurs in that matrix, and no other
+eigenvalue occurs.  Separability makes its action semisimple.
+
+We solve $q(w(Z))=0$ in free series over $D$.  Initially take
+\eqref{eq:division-initial-input}.  Its split Schur equations vanish
+through degree two, so Lemma~\ref{lem:schur-polynomial} gives the same
+assertion for the polynomial relation.  Suppose inductively that a
+partial substitution solves this relation through degree $\ell-1$,
+where $\ell\ge3$.  Allow a new coefficient in
+$\mathcal V_{-r}^2$ at every word of length $\ell-1$, for
+$r=1,2,3,4$, and a new coefficient in $\mathcal V_0^2$ at every word
+of length $\ell$.  These are finite-dimensional $F$-spaces.
+
+Impose every coefficient equation of
+\begin{equation}
+ q(w(Z))=0\pmod{(s_1,s_2)^{\ell+1}}.
+ \label{eq:finite-central-correction}
+\end{equation}
+This is a finite affine linear system over $F$ in the allowed new
+coordinates.  Indeed two new insertions have degree at least
+$2(\ell-1)>\ell$.  At any fixed free-word degree, only boundedly
+many positive-degree insertions occur in the evaluation of the old
+series; the remaining insertions are from the $\varepsilon$-small
+constant input.  The finitely many previously fixed coefficients have
+a finite lower valuation bound.  Thus the sums defining each
+coefficient of \eqref{eq:finite-central-correction}, and its linear
+part in the new coordinates, converge in $D$ and are defined over $F$.
+Choosing an $F$-basis of $D$ turns them into finitely many ordinary
+scalar linear equations.
+
+After scalar extension, each permitted input space becomes exactly
+the collection of reverse edges or loops used in
+Proposition~\ref{prop:eleven-block-correction}.  The current partial
+substitution is a valid partial split solution by
+Lemma~\ref{lem:schur-polynomial}.  The proposition supplies a split
+correction, and that lemma converts its Schur vanishings back to all
+the equations in \eqref{eq:finite-central-correction}.  In particular
+the new degree-$\ell-1$ reverse coefficients preserve the already
+solved equations: their diagonal Schur variation at that degree is
+zero.  We have therefore solved the whole finite linear system after
+a field extension.  A matrix and its augmented matrix have the same
+ranks before and after extension, so the system already has a solution
+over $F$.
+
+Choose such a solution at every step.  The coefficient stabilization
+in Proposition~\ref{prop:eleven-block-correction} produces
+\[
+ Z\in\widetilde B\langle\!\langle s_1,s_2\rangle\!\rangle^2,
+ \qquad q(w(Z))=0,
+\]
+with the prescribed constant and linear terms.  Only finite linear
+systems have been descended.  Neither the roots of $q$ nor the split
+correction coefficients need belong to $F$.
+
+Since the input coefficients commute with $A$, evaluation at $Z$
+defines a substitution homomorphism
+\[
+ A\langle\!\langle z_1,z_2\rangle\!\rangle
+ \longrightarrow D\langle\!\langle s_1,s_2\rangle\!\rangle.
+\]
+It is well-defined by the same coefficientwise valuation bound: at
+fixed output degree, only boundedly many nonconstant input factors
+occur.  The bound also justifies multiplication and regrouping of the
+convergent coefficient sums.  Hence every old polynomial identity on
+the tuple and the chosen series is preserved by this substitution.
+
+If $q_0(w)=0$ already holds, use the very same division algebra and
+the initial substitution \eqref{eq:division-initial-input}, with any
+$b\ge1$.  No spectral correction is required; the old relation
+$q_0(w)$ remains zero under the substitution homomorphism.  Both cases
+produce a central division algebra of degree $11bd>d$, with identical
+generic forms for the constant and the two linear input coefficients.
+The next step restores generation and two transverse directions using
+these coefficients.  That step is needed in the unconstrained case
+as well as in the spectral case.

preprints/A-Counterexample-to-Kuroshs-Division-Ring-Problem-September-23-2026/build/sections/formal-generic.tex

@@ -0,0 +1,406 @@
+\section{Formal evaluations and generic spectra}\label{sec:formal-generic}
+
+We begin the proof of Theorem~\ref{thm:lifting} by constructing a
+substitution $Z=(Z_1,Z_2)$ for the two free letters. Its coefficients will commute
+with the old algebra $A$, so applying it to $(X,w)$ preserves their
+polynomial identities. The matrix calculation focuses on $w$;
+Sections~\ref{sec:division-realization} and~\ref{sec:generation}
+realize the substitution in a division algebra of larger degree and
+restore generation and both transverse directions for $X(Z)$.
+
+Throughout the lifting argument, $k$ is a field of characteristic zero
+containing all roots of unity, $A$ is a central division algebra of degree
+$d$ over $k$, and
+\[
+ w\in A\langle\!\langle z_1,z_2\rangle\!\rangle.
+\]
+The letters commute with $A$, but not with each other.  Thus an element
+is a family $(a_\omega)_\omega$ indexed by finite words, written
+$\sum_\omega a_\omega z^\omega$, and multiplication uses concatenation
+of words.  Fix an algebraic closure $K$ of $k$ and a splitting
+$A\otimes_k K\simeq M_d(K)$; this uses the splitting and
+scalar-extension properties of central simple algebras
+\cite[Chapter~IV, Example~2.14(a) and Proposition~2.15]{MilneCFT}.
+
+If $P=(P_1,P_2)$ is a pair of $n\times n$ indeterminate matrices, put
+\[
+ W_n(P)=\sum_\omega a_\omega\otimes P^\omega
+       \in M_{dn}(K[[P]]).
+\]
+Here $K[[P]]$ is the ordinary commutative power series ring in the
+$2n^2$ entries.  We omit the subscript $n$ when the size is clear.
+The coefficient matrices act on the first tensor factor.  In
+particular, every input arrow between blocks is a pair of scalar
+matrices acting on the second factor.  It is not an arbitrary pair of
+matrices on both factors.
+
+In the spectral case we seek a substitution $Z=(Z_1,Z_2)$, with
+constant term $R$, for which $W(Z)$ is conjugate to a split semisimple
+constant output $W(R)$. If $\Lambda$ is the set of distinct eigenvalues
+of $W(R)$, this gives
+\[
+ q(W(Z))=0,\qquad q(t)=\prod_{\lambda\in\Lambda}(t-\lambda).
+\]
+The division realization will identify $q$ with the minimal polynomial
+over its central coefficient field. If $w$ already satisfies the
+minimal polynomial of $w(0)$, substitution retains that relation and
+the spectral correction is unnecessary.
+
+For the spectral construction, we will find matrix sizes with a common
+eigenvalue multiplicity, disjoint spectra at independent inputs, and
+scalar spectral blocks under commuting Taylor perturbations. These
+properties will make the eigenvalue responses of
+Section~\ref{sec:responses} well-defined. We first specify the
+evaluations and specializations used to establish them and to evaluate
+triangular auxiliary inputs later.
+
+\subsection{Convergence and specialization}\label{subsec:formal-conventions}
+
+Let $E$ be a complete discretely valued field containing $K$, with
+valuation $v$, such that $v(c)=0$ for $c\in K^\times$.  We use
+$E=C((\varepsilon))$, or a finite extension of such a field, and
+normalize $v(\varepsilon)=1$.  Matrix valuations mean the minimum of
+the entry valuations.  A matrix is \emph{small} if this minimum is
+positive.  A small generic point is
+$P=\varepsilon P^\circ$, where all entries of $P^\circ$ are
+algebraically independent over the previously specified constant
+field.  Several independent points can use the same $\varepsilon$.
+
+The finite extensions used to split spectra are separable in
+characteristic zero. The valuation extends to each such extension,
+which remains complete and discretely valued
+\cite[Theorem~7.38 and Remark~7.43]{MilneANT}; with the normalization
+$v(\varepsilon)=1$, its value group may be $\frac1e\mathbb Z$ for a
+positive integer $e$.
+
+\begin{lemma}[Admissible evaluations]\label{lem:admissible-evaluation}
+Suppose a constant input pair $R$ is block upper triangular with $b$
+blocks and small diagonal blocks.  Its strictly upper blocks may have
+arbitrary entries in $E$.  Let $H$ be a matrix pair of formal series
+of positive order in finitely many commuting or free variables.
+Then $W(R+H)$ is defined coefficientwise by convergent sums in $E$.
+The same assertion holds for products, finite determinants, and
+derivatives of fixed order of finitely many such entry series.
+
+These evaluations preserve formal entry identities.  They also
+remain valid coefficientwise in an additional parameter $\eta$ if
+the diagonal blocks have entries in $E[[\eta]]$ with small constant
+terms, while the strictly upper blocks and the coefficients of $H$
+have entries in $E((\eta))$.
+\end{lemma}
+
+\begin{proof}
+Fix a coefficient of degree $r$ in the variables of $H$.  An input
+path contributing to this coefficient uses at most $r$ positive-order
+terms of $H$.  Between these insertions it follows the constant upper
+triangular pair.  In each of the at most $r+1$ resulting intervals,
+there are at most $b-1$ strictly upper steps.  Thus at most
+$(r+1)(b-1)$ constant factors are strictly upper.  All remaining
+constant factors come from small diagonal blocks.
+
+Only finitely many coefficients of $H$ can occur at the chosen degree.
+Their valuations, and those of the finitely many strictly upper
+blocks, have a common lower bound.  If the original word has length
+$L$, its contribution consequently has valuation at least
+$cL-C_r$, where $c>0$ and $C_r$ is independent of the word.  There
+are finitely many words of each length, so the sum converges.  The
+same estimate applies after inserting the finitely many extra factors
+needed for a product or a derivative of fixed order.  Completeness
+allows regrouping of these sums, proving compatibility with
+multiplication and with identities.
+
+For the last assertion, first fix both the degree $r$ in $H$ and an
+$\eta$-coefficient.  The bounded number of strictly upper steps and
+of $H$-insertions bounds their total negative $\eta$-order.  Hence
+only a bounded number of positive-order diagonal $\eta$-terms can
+contribute, and only finitely many of their coefficients are used.
+Every other factor is a small constant diagonal term.  The preceding
+valuation estimate therefore applies again.  The resulting assertion
+is coefficientwise; no interchange of the two Laurent topologies is
+being asserted.
+\end{proof}
+
+We will use inverses of finite matrices as rational operations after
+evaluation.  The following elementary distinction is important.
+Although $(\varepsilon-Q)^{-1}$ has a formal expansion at $Q=0$ over
+$E$, that expansion need not converge after $Q=\varepsilon Q^\circ$.
+Accordingly, when transferring an identity involving an inverse, we
+first multiply by its determinant.  We evaluate the resulting
+numerator by Lemma~\ref{lem:admissible-evaluation}, and divide only
+when the evaluated determinant is nonzero.  Formal inverse expansions
+are used to define coefficients; they are not presumed to converge at
+every subsequent specialization.
+
+\begin{lemma}[Detection by independent residues]\label{lem:generic-detection}
+Let
+\[
+ F(Q)=\sum_\alpha c_\alpha Q^\alpha\in E[[Q_1,\ldots,Q_N]],
+ \qquad \inf_\alpha v(c_\alpha)>-\infty.
+\]
+Adjoin elements $Q_1^\circ,\ldots,Q_N^\circ$ whose residues are
+algebraically independent over the residue field of $E$, and complete
+with respect to the extended valuation.  Then
+$F(\varepsilon Q^\circ)$ is defined, and is nonzero whenever $F$ is
+nonzero.
+\end{lemma}
+
+\begin{proof}
+The valuations $v(c_\alpha)+|\alpha|$ tend uniformly to infinity
+with $|\alpha|$.  Their least value is therefore attained by a
+nonempty finite set of monomials.  After scaling to this valuation,
+the residue of their sum is a nonzero polynomial in the independent
+residues of the $Q_j^\circ$.  This proves both assertions.
+\end{proof}
+
+In applications, the coefficients of a cleared numerator are finite
+sums of products of base-point scalars and matrix-entry series.
+The path estimate gives a uniform valuation bound in the auxiliary
+coordinates.  Lemma~\ref{lem:generic-detection} therefore identifies
+formal identities in these coordinates with identities at independent
+small generic points.  The base-point valued field is constructed
+first; the auxiliary residues are adjoined afterward.  This order
+keeps their independence valid even after the base spectrum has been
+split.  The argument can be applied separately at each Taylor degree.
+
+We record two further field conventions.  Evaluation
+\[
+ K[[P]]\longrightarrow C(P^\circ)((\varepsilon)),
+ \qquad P\longmapsto\varepsilon P^\circ,
+\]
+is injective, by homogeneous degree or by
+Lemma~\ref{lem:generic-detection}.  It extends to the fraction field,
+and a finite splitting field embeds into a finite algebraic extension
+of the target.  Also, if $R$ is a small constant pair, then
+\begin{equation}
+K[[P]]\longrightarrow E[[\Delta]],\qquad P\longmapsto R+\Delta
+ \label{eq:generic-taylor-embedding}
+\end{equation}
+is injective: reduction modulo the valuation ideal sends the image of
+$f$ to $f(\Delta)$.  We use this observation for small block diagonal
+$R$.  Denominators in the fraction field remain nonzero, so the map
+extends to fraction fields.
+
+Finally, the Taylor substitution $P\mapsto P+\eta v$ maps
+$\operatorname{Frac}K[[P]]$ into
+$\operatorname{Frac}K[[P]][[\eta]]$, with reduction the identity.
+It extends to a finite separable extension $\Omega$ with values in
+$\Omega[[\eta]]$: choose a primitive element and lift its defining
+equation using its nonzero derivative.  The lift reducing to the
+chosen primitive element is unique, by coefficient recursion.  This
+convention lets us move a base point while retaining its labeled
+eigenvalue branches.
+
+\subsection{Generic division evaluations}\label{subsec:generic-division}
+
+Generic matrix evaluations have a useful divisibility property for
+their eigenvalue multiplicities.  We obtain it by first evaluating
+inside a division algebra.
+
+For $n\ge1$, choose independent central variables $\alpha,\beta$
+and a primitive $n$th root of unity $\zeta_n\in k$.  Over
+$k_0=k(\alpha,\beta)$ let $B_n$ be generated by $x,y$ with
+\[
+ x^n=\alpha,\qquad y^n=\beta,\qquad yx=\zeta_nxy.
+\]
+For $n=1$ this means $B_1=k_0$.  The reduced monomials
+$x^iy^j$, $0\le i,j<n$, form a basis.  Indeed they span by the
+relations, and after adjoining $n$th roots of $\alpha$ and $\beta$
+their clock-and-shift matrices are linearly independent.  These
+matrices span $M_n$, so $B_n$ is central simple of degree $n$.
+This is the standard symbol construction
+\cite[Corollary~2.5.5(1)]{GilleSzamuely2006}, with the inverse
+root of unity for our convention.
+
+\begin{lemma}\label{lem:symbol-division}
+The central simple algebra $A\otimes_k B_n$ is division.  Its pure
+rational scalar extensions and its central Laurent series extensions
+are division as well.
+\end{lemma}
+
+\begin{proof}
+Start with $A((x))$, where $x$ commutes with $A$, and let $\sigma$
+fix $A$ and send $x$ to $\zeta_nx$.  The skew Laurent series ring
+$A((x))((y;\sigma))$, with $ya=\sigma(a)y$, is a division ring by
+the leading-coefficient construction of inverses.  The elements
+$x^n,y^n$ are central and algebraically independent.  The defining
+relations give a unital map of $A\otimes_k B_n$ into this division
+ring.  Central simplicity makes the map injective.  A finite
+dimensional domain over a field is division, since multiplication by
+a nonzero element is an injective and hence surjective linear map.
+
+For a central division algebra $D/F$, the rational extension $D(t)$
+embeds in the division ring $D((t))$.  Iterating proves the rational
+assertion.  Because $D$ has a finite $F$-basis,
+$D\otimes_F F((\varepsilon))=D((\varepsilon))$, which is again
+division by leading coefficients.
+\end{proof}
+
+Choose a basis $b_1,\ldots,b_{n^2}$ of $B_n$, and let
+$c_{ij}$, $i=1,2$, $j=1,\ldots,n^2$, be independent variables.
+Evaluate $w$ at
+\[
+ P_i=\varepsilon\sum_j c_{ij}b_j.
+\]
+This gives an element of a central division algebra of degree $dn$
+over
+\[
+ F_n=k(\alpha,\beta,(c_{ij}))((\varepsilon)).
+\]
+After splitting $A$ and $B_n$ by constants independent of the $c_{ij}$,
+the $b_j$ become a basis of $M_n$.  The change from the $c_{ij}$ to
+the entries of $P_i/\varepsilon$ is therefore an invertible linear
+change of independent coordinates.  The resulting evaluation is a
+small generic scalar matrix evaluation in the sense fixed above.
+
+Let $e_n$ denote the minimal-polynomial degree of this element.  Its
+minimal polynomial is irreducible and separable over $F_n$.
+Consequently $W(P)$ is semisimple over a splitting field, and every
+distinct root has the same multiplicity
+\begin{equation}
+t_n=\frac{dn}{e_n}.
+ \label{eq:generic-multiplicity}
+\end{equation}
+For completeness, if $D/F$ has degree $N$ and $a\in D$ has minimal
+polynomial $q$ of degree $e$, then left multiplication by $a$ has
+characteristic polynomial $q^{N^2/e}$: view $D$ as a vector space
+over $F[a]$.  After splitting $D=M_N$, left multiplication acts by
+the matrix $a$ on each of its $N$ columns.  Its characteristic
+polynomial is therefore the $N$th power of the matrix characteristic
+polynomial.  Comparing root multiplicities gives $N/e$ in the
+matrix representation.  This also proves that $N/e$ is an integer.
+
+Linear independence of $1,a,\ldots,a^{e-1}$ is preserved by field
+extension.  Thus the same $e_n,t_n$ describe $W(P)$ over a splitting
+field of $\operatorname{Frac}K[[P]]$.
+
+\subsection{Nonconstant roots and stable multiplicity}
+\label{subsec:stable-multiplicity}
+
+Let $q_0$ be the minimal polynomial of $w(0)$ over $k$.  If
+$q_0(w)=0$ as a free series, the subsequent spectral construction is
+unnecessary; the unconstrained case of the lifting argument will use
+this existing relation.  In the rest of this section assume
+$q_0(w)\ne0$.
+
+A nonzero homogeneous part of $q_0(w)$ is detected by scalar matrices
+of every sufficiently large size.  To see this, choose a word of
+length $L$ with nonzero coefficient and use a path of $L+1$ vertices,
+placing its successive letters on the consecutive matrix-unit
+arrows.  The $(1,L+1)$ entry detects precisely that coefficient.
+Padding by zero blocks works for all larger sizes.  It follows that
+$q_0(W(P))\ne0$ at every generic size $n\ge n_0$, for some $n_0$.
+
+\begin{lemma}\label{lem:nonconstant-roots}
+For $n\ge n_0$, every distinct generic eigenvalue of $W(P)$ has a
+nonzero differential in the input entries.  Independent small generic
+points of sizes at least $n_0$ have disjoint spectra.  Their spectra
+also avoid the constant spectrum of $w(0)$.
+\end{lemma}
+
+\begin{proof}
+Use the division evaluation over $F_n$.  Each derivation
+$\partial/\partial c_{ij}$ acts coefficientwise on Laurent series
+and extends uniquely to algebraic extensions in characteristic zero.
+Uniqueness implies that it commutes with automorphisms of a splitting
+field over $F_n$.  If all these derivations killed one root, they
+would kill all its Galois conjugates.  Irreducibility makes these all
+the distinct roots.  Thus every coefficient of the reduced
+characteristic polynomial would have all its input derivatives zero.
+
+The coefficient of $\varepsilon^r$ in such a characteristic
+coefficient is a homogeneous polynomial of degree $r$ in the
+$c_{ij}$.  In characteristic zero a polynomial whose partial
+derivatives all vanish is constant.  All terms of positive
+$\varepsilon$-degree therefore vanish.  The characteristic polynomial
+is the characteristic polynomial of $w(0)$.  Semisimplicity then
+implies $q_0(W(P))=0$, a contradiction.
+
+Splitting $A,B_n$ uses constants independent of the inputs.  The
+invertible basis change described above, together with the nonzero
+factor $\varepsilon$, transports the differential assertion to
+ordinary matrix-entry coordinates.  For independent points $P,Q$,
+derivations in the $P$-coordinates kill the roots belonging to $Q$.
+They cannot kill a root belonging to $P$.  No root can therefore
+belong to both spectra.  The same argument excludes a constant root.
+\end{proof}
+
+Choose $p\ge n_0$ for which the positive integer $t_p$ is minimal
+among all $t_n$, $n\ge n_0$, and put
+\[
+ t=t_p,\qquad \mathcal S=\{pN:N\ge1\}.
+\]
+
+\begin{lemma}\label{lem:stable-multiplicity}
+For every $N\ge1$,
+\[
+ t_{pN}=t,\qquad e_{pN}=Ne_p.
+\]
+\end{lemma}
+
+\begin{proof}
+Specialize a size-$pN$ point to a diagonal sum of $N$ independent
+size-$p$ points.  Lemma~\ref{lem:nonconstant-roots} gives $Ne_p$
+distinct eigenvalues there.  The generic bound on minimal-polynomial
+degree survives this specialization: it is expressed by the vanishing
+of all $(e_{pN}+1)$-minors of the matrix whose columns are the
+vectorizations of $1,W,\ldots,W^{e_{pN}}$.  These are formal entry
+identities covered by Lemma~\ref{lem:admissible-evaluation}.
+Consequently $e_{pN}\ge Ne_p$.  Minimality of $t$ gives the opposite
+inequality,
+$e_{pN}=dpN/t_{pN}\le dpN/t=Ne_p$.
+\end{proof}
+
+We henceforth use sizes in $\mathcal S$.  At size $n$, the minimal
+polynomial always has degree at most $e_n=dn/t$ at every admissible
+specialization.  The following consequence controls all commuting
+Taylor deformations used below.
+
+\begin{lemma}[Scalar branches at maximal root count]
+\label{lem:scalar-branches}
+Let $V(T)$ be a matrix over a commuting formal Taylor ring $E[[T]]$.
+Suppose $V(0)$ is split semisimple with $e$ distinct eigenvalues and
+the minimal-polynomial degree of $V(T)$ over
+$\operatorname{Frac}E[[T]]$ is at most $e$.  Then a formal change of
+basis reducing to the identity separates $V(T)$ into $e$ scalar
+blocks.  The sizes of these blocks are their multiplicities at
+$T=0$.
+\end{lemma}
+
+\begin{proof}
+Use the eigenspace decomposition at zero.  Off-diagonal blocks can be
+removed recursively in total Taylor degree: the equation for the
+block between eigenvalues $\lambda_i,\lambda_j$ is multiplication
+by $\lambda_i-\lambda_j$, a nonzero scalar.  We obtain blocks
+$V_i(T)$ reducing to $\lambda_i I$.  Their characteristic polynomials
+have pairwise unit resultants, because their reductions have different
+roots.  Their minimal polynomials over the Taylor fraction field are
+therefore pairwise coprime.  The minimal-polynomial degree of their
+direct sum is the sum of their degrees.  There are $e$ nonempty
+blocks and the total degree is at most $e$, so every block has
+minimal-polynomial degree one.  Each $V_i(T)$ is scalar.  Its scalar
+entry belongs to $E[[T]]$, since the block entries do.
+\end{proof}
+
+In particular, the roots at a small semisimple point with $e_n$
+distinct roots have labeled scalar Taylor branches.  The splitting
+field of the generic point embeds in their Taylor fraction field by
+\eqref{eq:generic-taylor-embedding}; a labeling is chosen by reduction
+to the roots at the base.
+
+Let $\rho_n$ be the generic rank of the input Jacobian of
+\[
+ \operatorname{tr}W,\operatorname{tr}W^2,\ldots,
+ \operatorname{tr}W^{dn}.
+\]
+Its minors are formal entry identities, so $\rho_n$ bounds the
+corresponding specialized ranks.  At a maximal-root-count point the
+scalar branches satisfy
+\[
+ d\operatorname{tr}W^j
+   =jt\sum_{h=1}^{e_n}\lambda_h^{j-1}\,d\lambda_h.
+\]
+The first $e_n$ rows give an invertible Vandermonde matrix.  Hence
+the trace Jacobian and the eigenvalue Jacobian have the same rank
+there, including over a commuting Taylor fraction field.  We have
+therefore obtained both ingredients needed for the response
+calculations: scalar branches and specialization-stable rank bounds.

preprints/A-Counterexample-to-Kuroshs-Division-Ring-Problem-September-23-2026/build/sections/generation.tex

@@ -0,0 +1,310 @@
+\section{Restoring generation and two transverse directions}
+\label{sec:generation}
+
+The preceding construction gives a substitution $Z(s_1,s_2)$ inside
+the centralizer of $A$ in a larger central division algebra.  It
+preserves the old polynomial identities and algebraizes the chosen
+series.  To complete Theorem~\ref{thm:lifting}, we must
+make the new constant tuple generate the larger algebra and retain
+two independent directions modulo inner derivations. We first use the
+original transverse directions to prove generation of the generic
+diagonal product and to control the arrows between its blocks.
+Translating the first free letter by a new central parameter then joins
+the blocks into a full matrix algebra. A single rank calculation shows
+that both directions remain independent modulo inner derivations.
+
+Throughout this section, brackets with a tuple are taken componentwise.
+Thus, for $C=(C_1,\ldots,C_u)$ in an algebra $B$, write
+\[
+ J(C)=\{([T,C_1],\ldots,[T,C_u]):T\in B\}.
+\]
+Polynomial relations on $X_0$ detect the two input directions to first
+order.
+
+\begin{lemma}[Derivatives of relations]
+\label{lem:derivatives-of-relations}
+Let $A$ be a finite-dimensional central simple algebra over a field
+$k$ of characteristic zero.  Let
+$X\in A\langle\!\langle z_1,z_2\rangle\!\rangle^u$ have constant
+tuple $X_0$ generating $A$, and suppose its two linear coefficients
+$X_1,X_2$ are independent modulo $J(X_0)$.  Define
+\[
+ \theta:k\langle x_1,\ldots,x_u\rangle\longrightarrow A,
+ \qquad \theta(f)=f(X_0),\qquad I=\ker\theta.
+\]
+Then the map taking $f\in I$ to the coefficients of $z_1,z_2$ in
+$f(X)$ is onto $A^2$.
+\end{lemma}
+
+\begin{proof}
+Denote these two coefficient maps by $\delta_1,\delta_2$.  Each
+satisfies
+\[
+ \delta_j(fg)=\theta(f)\delta_j(g)+\delta_j(f)\theta(g).
+\]
+Since $\theta$ is onto and $I$ is a two-sided ideal, the image
+$M=(\delta_1,\delta_2)(I)$ is an $A$-bimodule in $A^2$.
+
+Left and right multiplication generate $\operatorname{End}_k(A)$;
+see \cite[Chapter~IV, Corollary~2.9]{MilneCFT}.
+Indeed, after splitting $A$, multiplication by matrix units on the
+left and right gives every matrix unit on the vector space $A$;
+the resulting isomorphism
+$A\otimes_k A^{\mathrm{op}}\simeq\operatorname{End}_k(A)$ descends
+by finite-dimensional linear algebra.  Applying these endomorphism
+matrix units to $M\subseteq A\otimes_k k^2$ shows that
+$M=A\otimes_k V$ for a subspace $V\subseteq k^2$.
+If $V\ne k^2$, some nonzero $(a,b)\in k^2$ annihilates $V$.
+The map $a\delta_1+b\delta_2$ then vanishes on $I$ and descends to
+a derivation of $A$.
+
+Every $k$-linear derivation of $A$ is inner
+\cite[Theorem~8]{Jacobson1937Derivations}. For completeness, after
+splitting $A=M_d(E)$, a derivation $\partial$ is implemented by
+\[
+ T=\sum_i\partial(E_{i1})E_{1i}.
+\]
+Differentiating $E_{kl}E_{i1}=\delta_{li}E_{k1}$, multiplying by
+$E_{1i}$, and summing gives $\partial(E_{kl})=[T,E_{kl}]$.
+The equations for an implementing element on a fixed $k$-basis of
+$A$ form a linear system over $k$.  Solvability after scalar
+extension implies solvability over $k$.
+It follows that $aX_1+bX_2\in J(X_0)$, a contradiction.
+\end{proof}
+
+In particular, there are $f_1,f_2\in I$ whose derivative pairs are
+$(1,0)$ and $(0,1)$.  These relations recover the two generic input
+matrices from small evaluations of the tuple.
+
+\begin{lemma}[Generic blocks]
+\label{lem:generic-generating-blocks}
+Under the hypotheses of Lemma~\ref{lem:derivatives-of-relations},
+write $d=\deg A$ and split $A$ into $d\times d$ matrices.  Fix a
+positive integer $b$ and finitely many independent small generic input pairs
+\[
+ P_i=\varepsilon P_i^{\circ},\qquad 1\le i\le r,
+\]
+of $b\times b$ matrices, with jointly independent entries in the
+$P_i^{\circ}$.  Work over a common splitting Laurent field.
+Then each tuple $C_i=X(P_i)$ generates $M_{db}$, and the diagonal
+tuple $C=\operatorname{diag}(C_i)$ generates the full product of
+these matrix algebras.
+
+For distinct $i,j$, the output derivative of a generic input arrow
+$U=(U_1,U_2)$ from block $j$ to block $i$ is nonzero modulo
+\[
+ \{C_iT-TC_j:T\in M_{db}\}.
+\]
+More precisely, the image of the arrow derivatives in this quotient
+has dimension at least $b^2$.
+\end{lemma}
+
+\begin{proof}
+Let $E$ contain the splitting data and independent input coordinates.
+Consider integral elements of the $E((\varepsilon))$-algebra
+generated by $X(P_i)$.  Their reductions modulo $\varepsilon$
+contain $M_d(E)\otimes1$, because $X_0$ generates the old algebra.
+Lemma~\ref{lem:derivatives-of-relations} also gives
+\[
+ \varepsilon^{-1}f_j(X(P_i))
+ \equiv I_d\otimes P_{i,j}^{\circ}\pmod\varepsilon.
+\]
+A generic pair generates $M_b(E)$.  To see the relevant determinant
+is nonzero, specialize one matrix to a diagonal matrix with distinct
+entries and the other to a matrix with every entry nonzero.
+Interpolation and sandwiching give all matrix units.  Thus the
+reductions generate $M_d(E)\otimes M_b(E)=M_{db}(E)$.
+Choose finitely many integral algebra elements whose reductions
+form a matrix basis.  Their determinant is a unit, so they remain
+independent over $E((\varepsilon))$, proving full generation.
+
+The representations defined by different $C_i$ have distinct kernels.
+Indeed,
+\[
+ \operatorname{tr}f_1(C_i)
+  =\varepsilon d\operatorname{tr}P_{i,1}^{\circ}
+    +O(\varepsilon^2),
+\]
+and the leading coefficients differ.  Equal kernels would induce an
+isomorphism of the full matrix images carrying every polynomial
+evaluation in one to the other.  Such an isomorphism preserves trace:
+the trace of left multiplication is matrix size times matrix trace.
+This contradicts the display.
+The kernels are therefore distinct maximal two-sided ideals.
+Evaluation is onto their finite product.  Explicitly, for each pair
+$i\ne j$ choose a polynomial vanishing in block $j$ and equal to
+the identity in block $i$.  The product over $j\ne i$ vanishes in
+all other blocks and is the identity in block $i$; multiplying by
+a lift of any matrix gives that product coordinate.
+
+For the last assertion, compare the inner columns and arrow columns
+at $P_i=P_j=0$.  The inner image is
+$J(X_0)\otimes M_b$, of dimension $(d^2-1)b^2$.  The arrow derivative
+is
+\[
+ X_1\otimes U_1+X_2\otimes U_2.
+\]
+The original transverse independence implies that the combined map
+has rank $(d^2+1)b^2$ at this reduction.  Its generic rank is at
+least this number.  At a generic pair, the inner map has rank at
+most $(db)^2=d^2b^2$; in fact it is injective, since a nonzero
+intertwiner between the two simple representations would identify
+their kernels.  Subtracting this generic inner rank leaves at least
+$b^2$ independent arrow directions in the quotient.  A generic
+fixed arrow consequently has nonzero quotient class.
+\end{proof}
+
+All determinant conditions in Lemma~\ref{lem:generic-generating-blocks}
+hold on the same independent block and arrow coordinates used in
+the division realization.  Only finitely many blocks and arrows are
+involved, so their nonzero minors can be imposed simultaneously.
+In particular, the argument does not require replacing that realization
+by a different choice of data.
+
+We next isolate the translation argument.  Its hypotheses concern
+only the constant and linear coefficients; higher-order corrections
+may be arbitrary.
+
+\begin{lemma}[A translation preserving two directions]
+\label{lem:translation-two-directions}
+Let $E$ be a field of characteristic zero, let $n\ge1$ be an integer,
+let $m=11$, and index
+the blocks of $M_{mn}(E)$ by $\mathbb Z/m\mathbb Z$.
+Suppose
+\[
+ Y(s_1,s_2)\in M_{mn}(E)
+       \langle\!\langle s_1,s_2\rangle\!\rangle^u
+\]
+has constant tuple $C=\operatorname{diag}(C_i)$ generating
+$\prod_iM_n(E)$.  Suppose its two linear coefficients have the
+following supports:
+\[
+ (Y_1)_{ij}=0\ \text{unless }i-j\in\{1,2\},
+ \qquad
+ (Y_2)_{ij}=0\ \text{unless }i-j\in\{3,4\}.
+\]
+Assume every displayed edge coefficient is nonzero modulo the
+cross-inner image $\{C_iT-TC_j:T\in M_n(E)\}$.
+For a new central Laurent parameter $\gamma$, define
+\[
+ Y'(s'_1,s'_2)=Y(\gamma+s'_1,s'_2).
+\]
+Then $Y'_0=Y(\gamma,0)$ generates $M_{mn}(E((\gamma)))$, and
+$Y'_1,Y'_2$ are independent modulo $J(Y'_0)$.
+\end{lemma}
+
+\begin{proof}
+The substitution is well-defined.  In the coefficient of a fixed
+new word, a contribution from an old word of length $r$ contains
+$r$ minus the new word length factors of $\gamma$.  Thus only
+finitely many terms contribute at each $\gamma$ degree.  Coefficients
+converge in $E[[\gamma]]$, and the substitution respects products.
+
+Let $B=\prod_iM_n(E)$ and let $I_C$ be the kernel of evaluation of
+the free algebra at $C$.  First-letter differentiation on $I_C$
+has image a $B$-bimodule in $M_{mn}(E)$.  Its projection onto every
+first-color edge is nonzero.  Otherwise the derivative into that
+cross bimodule would descend to a derivation $\delta:B\to M_n(E)$,
+where the left and right actions are through the distinct endpoint
+factors $i,j$.  Such a derivation is inner.  In fact, for the central
+idempotent $e_i$ of factor $i$, differentiating $e_i a=a e_i$ gives
+\[
+ \delta(a)=a_iT-Ta_j,\qquad T=\delta(e_i).
+\]
+Applied to the generators, this contradicts the transverse edge
+hypothesis.  Since each full rectangular block is a simple bimodule
+for its two matrix factors, the derivative image contains the full
+block on every first-color edge.
+
+Divide the corresponding evaluations of elements of $I_C$ at
+$Y(\gamma,0)$ by $\gamma$.  Their reductions supply all these edge
+blocks.  Ordinary evaluations already supply the full diagonal
+product.  The offset-one edges form a directed cycle through all
+vertices, so products supply every matrix block.  A finite set of
+such elements with independent reductions therefore proves that
+$Y(\gamma,0)$ generates the full matrix algebra.
+
+It remains to check the two directions.  Put $N=mn$ and
+$J_0=J(C)$.  Since $C$ generates $B$, its commutant consists of the
+$m$ block scalars.  Hence $\dim J_0=N^2-m$.  Choose $N^2-m$ inner
+columns with independent reductions spanning $J_0$.  For a block
+scalar $c=\operatorname{diag}(c_i I_n)$, the additional column
+\[
+ \gamma^{-1}[c,Y(\gamma,0)]
+\]
+is integral and reduces to $[c,Y_1]$.  These columns have rank
+$m-1$ modulo $J_0$.  Indeed, $[c,Y_1]\in J_0$ forces $c_i=c_j$
+on each offset-one edge, by its transverse hypothesis, and the
+cycle forces all $c_i$ to agree.
+
+The two shifted linear coefficients reduce to $Y_1,Y_2$.  Suppose
+\[
+ aY_1+bY_2\in J_0+\{[c,Y_1]:c\text{ is block scalar}\}.
+\]
+At a second-color edge, only $bY_2$ contributes outside $J_0$,
+so $b=0$.  On the directed offset-one cycle, the relation then
+forces $a=c_i-c_j$ on each edge.  Summing gives $ma=0$, whence
+$a=0$ in characteristic zero.  Thus these two reductions are
+independent modulo the enlarged inner-column space.
+
+We have found $N^2-m+(m-1)+2=N^2+1$ independent reductions among
+inner columns and the two tangents.  They lift to independent
+columns over $E((\gamma))$.  Full generation of $Y'_0$ implies
+$\dim J(Y'_0)=N^2-1$, so the two tangent classes are independent.
+\end{proof}
+
+Apply the lemma to $Y=X(Z)$.  Lemma~\ref{lem:generic-generating-blocks}
+supplies its diagonal-product and edge hypotheses, while the
+eleven-block construction supplies precisely the two color supports.
+Consequently the translated tuple generates after splitting and has
+two independent tangent classes.  These are finite rank conditions,
+so they also hold over the division algebra's own center.
+The central Laurent extension by $\gamma$ preserves division.
+Both the old polynomial identities and the new algebraic relation
+are preserved by the convergent substitution.
+
+\subsection{Returning to a countable center}
+
+The complete Laurent fields used so far need not be countable.
+We finish by retaining only the scalar data required by the result.
+Use the notation of Section~\ref{sec:division-realization}:
+$F_0=H_0(a_*)$ is the construction field, $D_0$ is the central
+division algebra over it, and
+\[
+ L=F_0((\varepsilon))((\gamma))
+\]
+is the full shifted Laurent center. The field $F_0$ is countable:
+$L_0$ is generated by finitely many indeterminates over the countable
+field $k(\alpha,\beta)$, and $H_0$ is a subfield of $L_0$.
+The algebra $D_0$ has degree $dbm$, and $D_0\otimes_{F_0}L$
+is the central division algebra just constructed.
+
+Choose an $F_0$-basis of $D_0$. Let $k'\subseteq L$ be the field
+generated by $F_0$, the coefficients of the monic polynomial $q$
+(taken to be $q_0$ when $q_0(w)=0$),
+and every scalar coordinate in this basis of every coefficient
+of the translated series $X'$ and $w'$. There are countably many
+free words and finitely many coordinates per coefficient, so $k'$
+is countable.  Put
+\[
+ A'=D_0\otimes_{F_0} k'.
+\]
+This is central simple of degree $dbm$.  Its scalar extension to
+$L$ is division, so $A'$ embeds in a division ring and has no zero
+divisors.  A finite-dimensional algebra without zero divisors is
+a division algebra.  Thus $A'/k'$ has the required central division
+property and strictly larger degree.
+
+All coefficients of $X',w'$ and $q$ lie in this smaller model.
+The identities descend by its injective scalar extension to $L$.
+Generation and transverse independence descend as well, since
+the ranks of the finite coordinate matrices defining them do not
+change under field extension.  This proves every conclusion of
+Theorem~\ref{thm:lifting}.
+
+In the unconstrained case $q_0(w)=0$, the same argument applies
+without the spectral corrections: use the same eleven-block
+generic constant and colored linear terms, retain the relation
+$q_0$, and carry out the translation and countable restriction
+above.  Thus that case also increases degree while preserving
+generation and both directions.

preprints/A-Counterexample-to-Kuroshs-Division-Ring-Problem-September-23-2026/build/sections/global-defect.tex

@@ -0,0 +1,298 @@
+\section{The common space of spectral relations}
+\label{sec:global-defect}
+
+The active space describes the possible responses at one block. We now
+identify the relations that remain when many independent blocks are used
+together. The essential point is that the derivatives at different
+triangular inputs belong to one common space. A bound on their ranks
+separately would not give the relations needed for the free-series
+construction.
+
+We retain the preceding notation. The allowed sizes are
+\(\mathcal S=\{pN:N\geq1\}\); at size \(n\in\mathcal S\), there are
+\(e_n=dn/t\) distinct generic roots, each of multiplicity \(t\).
+The active space has dimension \(a_n\), and \(\rho_n\) is the generic
+rank of the eigenvalue differential, equivalently of the trace
+differential. In particular \(\rho_n\leq a_n\).
+We use the replication identity \(a_{pN}=Na_p\) and the following
+consequence of the preceding attainability argument. There are a size
+\(r\in\mathcal S\) and an integer \(K_0\geq1\) such that the loop
+responses at a generic size-\(r\) point, together with paired responses
+through \(K_0\) independent generic size-\(r\) partners, span its
+active space. In each pairing, either arrow may be fixed generically
+and the opposite arrow varied. All spaces below are extended to a
+common splitting field when several independent points are involved.
+
+\begin{lemma}[Derivatives at a star]
+\label{lem:star-derivatives}
+Let allowed-size blocks, evaluated by the preceding formal
+specialization conventions, have split semisimple outputs with
+disjoint spectra and the maximal total number of roots. Add input
+arrows only from designated
+hub blocks to designated leaf blocks, with no arrows between hubs or
+between leaves. The resulting input has the same roots and
+multiplicities as its diagonal part. A loop derivative at a block is
+its loop response \(D\), with zero entries at all other blocks. If
+\(U\) is a forward arrow between a leaf \(P\) and a hub \(Q\), a
+derivative in its reverse arrow \(V\) has entries
+\[
+ \bigl(m_h(P,Q;U,V)\bigr)_h,
+ \qquad
+ \bigl(m_k(Q,P;V,U)\bigr)_k
+\]
+at these two blocks and zero entries at the others.
+\end{lemma}
+
+\begin{proof}
+Order the leaves before the hubs. The input and its evaluated output
+are block upper triangular. The diagonal output blocks are the
+original ones; since their spectra are disjoint, the output is similar
+to their direct sum. In particular it is semisimple with the stated
+roots and multiplicities.
+
+Differentiate \(\operatorname{tr}(W^j)\). A closed block path
+contributing to a reverse-arrow derivative must use that reverse
+arrow once and its matching forward arrow once. There is no other
+way to return to the starting block in this star. Its contribution
+is therefore exactly the mixed-arrow contribution at the diagonal
+input. Loop derivatives use no off-diagonal return. The previously
+proved mixed-response formula gives the displayed pairs after
+converting trace derivatives to root derivatives. This conversion is
+invertible on all distinct roots: its entries are
+\(jt\lambda_h^{j-1}\), and the corresponding Vandermonde determinant
+is nonzero. The same reasoning applies to specialized blocks whenever
+the stated maximal-count and disjointness hypotheses hold.
+\end{proof}
+
+\begin{lemma}[A stable upper bound for the differential rank]
+\label{lem:tail-defect}
+One can choose an arbitrarily large allowed size \(b\) and integers
+\(a=a_b\) and \(g\geq0\), such that
+\[
+ \rho_b=a-g,
+ \qquad
+ \rho_{bM}\leq Ma-g \quad(M\geq1).
+\]
+At a generic size-\(b\) point choose one root from each part of the
+active partition. Projection onto these \(a\) selected roots
+identifies the active space with the full coordinate space. The
+quotient of that coordinate space by the loop-response image has
+dimension \(g\).
+\end{lemma}
+
+\begin{proof}
+For \(N\geq1\), put
+\[
+ g_N=Na_r-\rho_{rN}.
+\]
+Replication and \(\rho_{rN}\leq a_{rN}\) show that \(g_N\geq0\).
+For \(N\geq K_0\), take \(K_0\) of \(N\) independent size-\(r\)
+blocks as hubs and the remainder as leaves. By
+Lemma~\ref{lem:star-derivatives} and the attainability property, the
+projection of the derivative image onto the leaf active spaces is
+surjective. Reverse-arrow variables belonging to different leaves
+are independent. Thus
+\[
+ \rho_{rN}\geq(N-K_0)a_r,
+ \qquad g_N\leq K_0a_r.
+\]
+Here the specialized rank is bounded by the generic rank by the
+formal minor identities. The finitely many smaller values of \(N\)
+do not affect boundedness.
+
+A bounded sequence of nonnegative integers has an eventual minimum
+that is attained arbitrarily late. Choose \(N_0\geq K_0\) at such
+an occurrence, and write
+\[
+ g=g_{N_0},\qquad
+ g_N\geq g\ (N\geq N_0),\qquad
+ b=rN_0,\qquad a=N_0a_r.
+\]
+Then \(a=a_b\), \(\rho_b=a-g\), and
+\[
+ \rho_{bM}=Ma-g_{N_0M}\leq Ma-g.
+\]
+No eventual constancy of \(g_N\) is required. Finally, the active
+partition says that projection to one representative of each part
+is an isomorphism on the active space. The loop image has rank
+\(\rho_b\), so its quotient has dimension \(g\).
+\end{proof}
+
+Fix \(b,a,g\) as in Lemma~\ref{lem:tail-defect}. For each generic
+size-\(b\) point \(P\), let \(E_P\) be the \(a\)-dimensional
+space of selected-root coordinates, and let
+\[
+ L_P=\operatorname{im}D_P\subseteq E_P,
+ \qquad H_P=E_P/L_P.
+\]
+Thus \(\dim H_P=g\). We regard \(H_P^*\) as the space of
+covectors on \(E_P\) annihilating all loop responses.
+
+\begin{lemma}[Surjectivity at both ends of a pair]
+\label{lem:pair-quotient-surjectivity}
+Let \(P,Q\) be independent generic size-\(b\) points. For a fixed
+generic arrow \(U\) between them, varying the reverse arrow \(V\)
+gives a joint paired-response map into \(H_P\oplus H_Q\) whose
+projection to either summand is surjective. This holds whichever
+orientation is fixed.
+\end{lemma}
+
+\begin{proof}
+Specialize \(P\) and \(Q\) to diagonal sums of \(N_0\)
+independent size-\(r\) points. For each little block in \(P\), the
+\(N_0\geq K_0\) blocks in \(Q\), together with their components
+of the fixed generic arrow, provide enough partners to span its
+active space after loop columns are included. The reverse-arrow
+entries touching different little blocks of \(P\) vary independently.
+Compression at a root of one little block sees only the reverse
+entries at that block. Consequently loops and paired columns span
+all \(N_0a_r=a\) active coordinates at the \(P\)-end.
+
+The resulting nonzero rank minor persists at generic \(P,Q,U\).
+The labeled root data and the pair inverses are regular at the
+disjoint diagonal specialization, so this is precisely the
+determinant specialization justified earlier. Passing to the loop
+quotient gives surjectivity onto \(H_P\). The same proof works at
+the \(Q\)-end and with the other orientation fixed. The finitely
+many nonzero minor conditions hold simultaneously on a nonempty
+generic set.
+\end{proof}
+
+The next lemma supplies the common subspace promised at the start
+of this section. Its formal-coordinate argument is the reason we
+obtain compatible relations between pairs of blocks.
+
+\begin{lemma}[A common tangent space]
+\label{lem:common-tangent}
+Fix \(M\geq1\) independent generic size-\(b\) points and their
+selected roots. At inputs with this diagonal and arbitrary strictly
+upper block arrows, the collective span of all selected-root
+derivative columns has dimension at most \(Ma-g\). The span of
+all loop columns and all two-block paired columns has dimension
+exactly \(Ma-g\).
+\end{lemma}
+
+\begin{proof}
+Put \(R=Ma-g\). For \(M=1\) the assertion is the definition of
+the loop rank, so suppose \(M\geq2\). Use one block as a hub,
+ordered last. The loops span \(\bigoplus_i L_{P_i}\), of dimension
+\(M(a-g)\). Modulo that space, reverse derivatives project
+surjectively onto each of the \(M-1\) leaf quotients by
+Lemma~\ref{lem:pair-quotient-surjectivity}. These projections vary
+independently. The derivative rank at this star is therefore at
+least
+\[
+ M(a-g)+(M-1)g=R.
+\]
+The generic upper bound in Lemma~\ref{lem:tail-defect} is \(R\).
+Thus at a generic strictly upper-arrow array \(\mathcal U\) the
+selected-root derivative rank equals \(R\).
+
+Work over the field containing \(\mathcal U\) and the diagonal
+splitting data. Let \(F(H)\) be the formal Taylor map of the
+selected root shifts under unrestricted input-entry perturbations
+\(H\). The base has the maximal root count, so the scalar branch
+construction applies. Every \((R+1)\)-minor of \(dF\) is zero,
+and some \(R\)-minor is a unit at \(H=0\). Choose the corresponding
+\(R\) output coordinates and use them, together with the remaining
+input coordinates, as new formal source coordinates \((y,z)\).
+The formal inverse theorem gives
+\[
+ F(y,z)=(y,G(y,z)).
+\]
+The vanishing \((R+1)\)-minors now imply
+\(\partial G/\partial z=0\). In characteristic zero, coefficientwise
+differentiation gives \(G(y,z)=G(y)\). At any formal source point
+where \(F=0\), all derivative columns therefore belong to the fixed
+graph
+\[
+ \{(v,dG(0)v):v\in E^R\}.
+\]
+
+Shifting only the strictly upper arrows leaves every root unchanged.
+Apply the preceding conclusion to these shifts: their Jacobian
+columns all belong to this one graph, not merely to spaces of the
+same dimension.
+
+It remains to pass from formal shifts at \(\mathcal U\) to all
+upper-arrow arrays. Let \(J(U)\) be the selected Jacobian at such
+an array. It is polynomial in \(U\). Indeed a trace derivative
+has just one differentiated input step, and every other
+off-diagonal step is strictly upper; a contributing block path
+therefore uses a bounded number of upper arrows. The remaining
+diagonal sums converge by smallness. The conversion from trace
+derivatives to root derivatives has coefficients depending only
+on the fixed diagonal roots, so it preserves polynomial dependence.
+
+Every covector annihilating the preceding graph annihilates the
+polynomial matrix \(J(\mathcal U+S)\) as a formal identity in
+the upper-arrow shift \(S\). Substitution \(S=V-\mathcal U\)
+is valid in this polynomial identity. Hence it annihilates every
+coefficient of \(J(V)\). The span of those coefficients, formed
+over the original diagonal field, has dimension at most \(R\),
+since scalar extension preserves rank. It contains every specialized
+column and proves the collective-span assertion.
+
+Loop columns occur at the diagonal input. A paired column supported
+on two blocks occurs at the base with just the corresponding single
+upper arrow, by Lemma~\ref{lem:star-derivatives}. Allowing both
+arrows to vary makes the collection independent of which orientation
+is called forward. All these columns therefore lie in the common
+space. The one-hub star already gives \(R\) independent columns
+among them, proving equality.
+\end{proof}
+
+\begin{proposition}[Compatible dual relations]
+\label{prop:compatible-relations}
+For any finite nonempty collection of independent generic size-\(b\)
+points \(P_i\), there is a \(g\)-dimensional space
+\[
+ \mathcal G_{\{P_i\}}\subseteq\bigoplus_i H_{P_i}^*
+\]
+with the following properties.
+\begin{enumerate}
+\item Projection to each \(H_{P_i}^*\) is an isomorphism.
+\item It annihilates all joint paired responses, with both arrows
+arbitrary.
+\item For a fixed generic arrow between \(P_i,P_j\), the
+annihilator in \(H_{P_i}^*\oplus H_{P_j}^*\) of the variable
+reverse-arrow image is exactly the restriction of
+\(\mathcal G_{\{P_i\}}\).
+\item On adjoining independent blocks, restriction of the larger
+space is the original space after scalar extension.
+\end{enumerate}
+For one block this space is \(H_P^*\). These statements include
+\(g=0\).
+\end{proposition}
+
+\begin{proof}
+Define \(\mathcal G\) as the annihilator of all loop and joint
+paired columns in the selected coordinate space. By
+Lemma~\ref{lem:common-tangent}, it has dimension \(g\); annihilation
+of loops identifies it with a subspace of \(\bigoplus_i H_{P_i}^*\).
+If an element has zero component at one vertex \(i\), its pairing
+with columns on \((i,j)\) and the surjectivity onto \(H_{P_j}\)
+force its component at every other vertex \(j\) to vanish.
+Projection onto any one component is therefore injective, and hence
+an isomorphism by dimension.
+
+For a fixed generic arrow on \((i,j)\), the reverse-image in
+\(H_{P_i}\oplus H_{P_j}\) has dimension at least \(g\), since
+its projection onto either summand is surjective. It is killed by
+the \(g\)-dimensional restriction of \(\mathcal G\). The latter's
+annihilator has dimension \(2g-g=g\), so these two spaces coincide.
+This proves the exact pairwise assertion.
+
+Finally, restriction from a larger collection annihilates every old
+loop and pair column. It is injective, because projection to any
+retained vertex is injective, and both spaces have dimension \(g\).
+It is therefore onto. For a single block the same statement follows
+from its quotient dimension. All dimension arguments remain valid
+when \(g=0\).
+\end{proof}
+
+We henceforth write \(\mathcal G\) for these compatible spaces,
+specifying the blocks when needed. We have identified all linear
+relations shared by the loop and paired corrections. The next
+section proves that the leading free-word errors satisfy the
+corresponding cyclic relations.

preprints/A-Counterexample-to-Kuroshs-Division-Ring-Problem-September-23-2026/build/sections/introduction.tex

@@ -0,0 +1,139 @@
+\section{Introduction}\label{sec:introduction}
+
+A division ring $D$ is \emph{algebraic over its center} $F=Z(D)$ if every
+$a\in D$ satisfies a nonzero polynomial in $F[t]$. For a subset $S\subseteq D$,
+write $F(S)$ for the division subring generated by $F\cup S$.
+The ring is \emph{locally finite} if
+\[
+ [F(S):F]<\infty\qquad\text{for every finite subset }S\subseteq D.
+\]
+The Kurosh problem for division rings asks whether algebraicity over the
+center implies local finiteness. We give a negative answer.
+
+All rings and subrings in this paper have the same identity. If $k$ is a
+central subfield and $x,y$ belong to a ring, $k[x,y]$ denotes the unital
+$k$-algebra generated by $x,y$; the generators need not commute.
+The free polynomial algebra on two formal generators is denoted
+$k\langle x,y\rangle$.
+
+\begin{theorem}\label{thm:main}
+There is a countable division ring $D$ of characteristic zero, with
+center $F=Z(D)$, and elements $x,y\in D$ such that
+\[
+ D=F[x,y]=F(x,y),\qquad [D:F]=\infty,
+\]
+while every element of $D$ is algebraic over $F$.
+\end{theorem}
+
+The same pair $\{x,y\}$ therefore witnesses the failure of local finiteness.
+The construction does not prescribe the center. It first produces an
+algebraic division ring over a countable central subfield, and a separate
+finite-descent argument proves infinite dimension over the actual center.
+The annihilating polynomial is allowed to depend on the element.
+
+\subsection{Historical context}
+
+Kurosh formulated his question in 1941 in connection with Burnside's
+finiteness problem for groups. He asked whether a division ring that is
+algebraic and finitely generated as an algebra over a central subfield
+must be finite-dimensional over that subfield
+\cite[Problem~(K), p.~240]{Kurosh1941}. The subfield in this formulation
+need not be the full center, and finite generation means algebra
+generation, not merely division generation.
+
+The broader problem for algebraic algebras has a different history.
+Golod's construction, based on the Golod--Shafarevich estimates, gives
+finitely generated infinite-dimensional nil algebras
+\cite[Example~1]{Golod1964}. Their unitizations are algebraic algebras
+with nonzero nilpotents, so they do not answer the division-ring question.
+
+There are positive results under additional hypotheses. Jacobson's
+bounded-degree theorem gives a polynomial identity, and Kaplansky's
+theorem makes algebraic algebras satisfying a polynomial identity
+locally finite; see \cite[Theorems~3 and~5]{Jacobson1952}.
+Bell, Drensky, and Sharifi later proved that a division ring whose
+elements satisfy nonzero polynomial relations of uniformly bounded degree $d$,
+with coefficients on the left in a possibly noncentral subfield, has
+dimension at most $d^2$ over its center
+\cite[Theorem~1.3]{BellDrenskySharifi2013}. Our construction imposes
+element-dependent relations without a uniform degree bound.
+
+Amitsur proved that a division algebra finitely generated as an algebra
+over an uncountable central field is automatically algebraic
+\cite[Corollary~$2'$]{Amitsur1956}. This conclusion is algebraicity,
+not local finiteness. Countability plays a different role here: it
+allows us to schedule all polynomial expressions in the construction.
+Goodearl and Letzter's recent work \cite{GoodearlLetzter2026} treats
+nonfinite generation as an algebra for classes of transcendental skew
+fields and explicitly distinguishes that question from the algebraic
+Kurosh problem. Theorem~\ref{thm:main} concerns algebraic division rings
+and their actual centers.
+
+The proof combines standard division-algebra constructions with matrix
+and formal-series calculations. Symbol and cyclic algebras provide
+division-ring models for the generic block data
+\cite[Section~2.5]{GilleSzamuely2006}. Traces organize cyclic word
+relations \cite[Section~4]{Procesi1976}, while inner derivations describe
+infinitesimal conjugation \cite[Theorem~8]{Jacobson1937Derivations}.
+The lifting argument supplies the additional response, compatibility,
+and generation statements needed to preserve two independent deformation
+directions while imposing successive algebraic relations.
+
+\subsection{The construction}
+
+The main ingredient is Theorem~\ref{thm:lifting}. Its input is a tuple of
+formal series in two noncommuting variables over a finite-dimensional
+central division algebra. The constant tuple generates the coefficient
+algebra, and the two linear coefficients are independent modulo changes
+coming from conjugation. The theorem makes any prescribed series
+algebraic over a larger countable central field. At the same time it
+increases the degree of the division algebra, preserves all previous
+polynomial identities, and restores both generation and the two independent
+directions. These last requirements allow the construction to continue.
+
+For the prescribed series $w$, the immediate goal is to find a formal
+substitution $Z=(Z_1,Z_2)$ for which $w(Z)$ satisfies the minimal
+polynomial of the chosen constant output $w(Z(0))$. The input $Z(0)$
+is a nonzero generic input, small with respect to an auxiliary valuation,
+not the original zero input.
+After splitting
+the coefficient algebra into matrices, we take the constant input to be
+a diagonal sum of independent generic matrix pairs. Varying a diagonal
+block gives a first eigenvalue response; adding an arrow to another block
+and an arrow back gives a paired response. Sections~\ref{sec:formal-generic}
+and~\ref{sec:responses} establish the formal evaluation rules and show that
+these responses partition the eigenvalue labels into linked groups.
+
+A transfer theorem puts each first unsolved noncommutative word coefficient
+in the same response space. It also turns proportional responses into
+functional relations between eigenvalue branches
+(Sections~\ref{sec:word-transfer} and~\ref{sec:linkage-replication}).
+The relations within each block then separate from a fixed-dimensional
+space of relations among blocks. To test the latter, we repeat each block
+and attach auxiliary blocks, called hubs. Individual hubs distinguish
+repeat positions; a common hub compares them. This matrix test proves
+that every leading error satisfies the required cyclic sums of word
+coefficients (Sections~\ref{sec:global-defect} and~\ref{sec:cyclic-obstructions}).
+
+An eleven-block arrangement of two colors supplies corrections whose
+only obstructions are those cyclic sums. We solve the equations one word
+length at a time in Section~\ref{sec:free-correction}. A cyclic division
+algebra realizes these matrix calculations over a division ring, and a
+Taylor translation restores generation and the two independent directions
+(Sections~\ref{sec:division-realization} and~\ref{sec:generation}).
+The corrections are solutions of finite affine linear systems; this is
+what permits their descent from a splitting field to the division
+algebra's center. If $w$ already satisfies its constant term's minimal
+polynomial, the spectral correction is unnecessary, but realization
+and translation are still needed to increase degree and restore the
+iteration hypotheses.
+
+Section~\ref{sec:countable-construction} first explains how the lifting
+theorem yields Theorem~\ref{thm:main}. The fields introduced during the
+iteration are countable, so every polynomial expression can be scheduled.
+A division-ring ultraproduct retains the resulting relations. Its
+two-generated subalgebra is already a division ring, and finite word
+sets prove that its dimension over the central coefficient field is
+infinite. The actual-center argument concludes this deduction. The
+remaining sections prove the lifting theorem and all its required
+compatibilities.

preprints/A-Counterexample-to-Kuroshs-Division-Ring-Problem-September-23-2026/build/sections/lifting.tex

@@ -0,0 +1,81 @@
+\section{The lifting theorem}\label{sec:lifting}
+
+The construction proceeds through finite-dimensional central division
+algebras. At each step we impose an algebraic relation on one prescribed
+element while preserving the freedom to enlarge the algebra at the next step.
+Formal power series in two noncommuting variables record that freedom.
+
+For a $k$-algebra $A$, write
+\[
+ A\langle\!\langle z_1,z_2\rangle\!\rangle
+\]
+for the ring of formal series in words in $z_1,z_2$, with coefficients
+in $A$ commuting with the variables. Multiplication concatenates words
+and multiplies their coefficients in their original order. The ring is
+complete for the filtration by word length. For a tuple
+$X=(X^{(1)},\ldots,X^{(u)})$ of such series, let $X_0\in A^u$ be its
+constant tuple and let $X_1,X_2\in A^u$ be its coefficients of the two
+individual letters. Thus the subscripts $1,2$ denote letters, not degrees.
+Set
+\begin{equation}\label{eq:inner-tangent}
+ J_A(X_0)=\{([T,X_0^{(1)}],\ldots,[T,X_0^{(u)}]):T\in A\},
+ \qquad [T,a]=Ta-aT.
+\end{equation}
+The classes of $X_1,X_2$ modulo this subspace describe changes of the tuple
+that cannot be obtained by infinitesimal conjugation.
+
+\begin{theorem}[Algebraic lifting]\label{thm:lifting}
+Let $k$ be a countable field of characteristic zero containing every root
+of unity, and let $A$ be a finite-dimensional central division algebra
+over $k$. Suppose
+$X\in A\langle\!\langle z_1,z_2\rangle\!\rangle^u$ satisfies:
+\begin{enumerate}[label=\textup{(\roman*)}]
+\item $X_0$ generates $A$ as a $k$-algebra;
+\item $X_1,X_2$ have linearly independent classes in $A^u/J_A(X_0)$.
+\end{enumerate}
+For every $w\in A\langle\!\langle z_1,z_2\rangle\!\rangle$, there exist
+a countable field extension $k'/k$, a central division algebra $A'/k'$ of
+strictly larger degree, and series
+\[
+ X'\in A'\langle\!\langle s_1,s_2\rangle\!\rangle^u,
+ \qquad w'\in A'\langle\!\langle s_1,s_2\rangle\!\rangle
+\]
+with the following properties:
+\begin{enumerate}[label=\textup{(\alph*)}]
+\item Every noncommutative polynomial identity over $k$ satisfied by the
+tuple $(X,w)$ is satisfied by $(X',w')$ as an identity of formal series.
+\item The constant tuple $X'_0$ generates $A'$ over $k'$, and the classes
+of $X'_1,X'_2$ in $(A')^u/J_{A'}(X'_0)$ are linearly independent.
+\item There is a monic nonzero polynomial $q\in k'[t]$ such that $q(w')=0$.
+\end{enumerate}
+Here the degree of a central division algebra $B/F$ is $\sqrt{[B:F]}$.
+\end{theorem}
+
+Theorem~\ref{thm:lifting} is the central construction. Its application to
+the Kurosh problem uses $u=2$ and addresses one polynomial in the two
+generators at a time. The fields are allowed to grow; they need not be
+algebraic extensions of their predecessors. The final step will therefore
+identify the actual center separately and prove infinite dimension over it.
+
+Figure~\ref{fig:response-interfaces} separates the two tasks within a
+lifting step. The response calculation makes the prescribed series
+algebraic; the final translation restores the hypotheses needed to repeat
+the step. The two transverse directions must survive together, not merely
+one at a time.
+
+\begin{figure}[tbp]
+\centering
+\input{figures/response-interfaces}
+\caption{The spectral case of Theorem~\ref{thm:lifting}.
+The two displayed classes at each end are linearly independent in the
+indicated quotient. The middle boxes describe the matrix calculation:
+allowable eigenvalue changes control the first unsolved word errors,
+and the cyclic relations are precisely the obstructions to correction.
+Realization and translation return a generating tuple over a division
+algebra of strictly larger degree. If $w$ already satisfies the minimal
+polynomial of $w(0)$, the middle spectral correction is omitted, but
+realization and translation remain necessary. No embedding tower of
+the successive division algebras is asserted.}
+\label{fig:response-interfaces}
+\end{figure}
+\clearpage

preprints/A-Counterexample-to-Kuroshs-Division-Ring-Problem-September-23-2026/build/sections/linkage.tex

@@ -0,0 +1,317 @@
+\section{Functional linkage and replication}
+\label{sec:linkage-replication}
+
+We now show that the dimension of the active response space is additive
+under generic direct sums.  The main point is to turn the linear relations
+between linked responses into identities between eigenvalue branches.
+These identities will continue to control responses through partners of
+every allowed size.
+
+Retain the progression $\mathcal S=\{pN:N\geq 1\}$ and the common
+eigenvalue multiplicity $t$.  Thus, at size $n\in\mathcal S$, the number
+of distinct generic eigenvalues is $e_n=dn/t$.  We use the following
+previously established facts.  At a direct sum of independent small
+generic inputs of sizes in $\mathcal S$, the spectra are disjoint and
+have the maximal number of distinct eigenvalues.  Their commuting
+Taylor deformations therefore have scalar spectral blocks, by
+Lemma~\ref{lem:scalar-branches}.  The response identities admit the
+specializations of Lemma~\ref{lem:active-specialization}, and their
+leading terms satisfy Proposition~\ref{prop:word-transfer}.
+
+For a generic input $P$, let $\Omega_P$ be its spectral splitting
+field, and choose a constant input direction $v$ such that
+$D_h(P;v)\neq 0$ for every eigenvalue label $h$.  Recall the active-space
+partition $\mathcal P_P$, characterized by
+\begin{equation}
+ \mathcal T_P
+ =\left\{c\in\Omega_P^{e_n}:
+       \frac{c_h}{D_h(P;v)}
+       \text{ is constant on each member of }\mathcal P_P\right\}.
+ \label{eq:linkage-partition}
+\end{equation}
+Labels in one member of this partition are called linked.  In particular,
+$a_n=\dim_{\Omega_P}\mathcal T_P$ is the number of its members.
+The argument below uses the full coefficientwise definition of
+$\mathcal T_P$: it contains the loop responses $D(P;Y)$ and the
+coefficient vectors of $m(P,Q;U,V)$ for every partner size in
+$\mathcal S$.
+
+\subsection{A formal identity for linked branches}
+
+\begin{proposition}[Functional linkage]
+\label{prop:functional-linkage}
+Let $P$ be a small generic input of size $p$, and let $h,h'$ be linked
+labels at $P$.  Let $C_1,\ldots,C_s$ be small generic inputs of sizes in
+$\mathcal S$, jointly independent with $P$, where $s=0$ is allowed.
+Put $R=\operatorname{diag}(P,C_1,\ldots,C_s)$.
+Over a common spectral splitting field $E$, denote by
+$\Lambda_j^R(\Delta)$ the scalar eigenvalue branch of $W(R+\Delta)$
+whose value at zero is $\lambda_j(P)$, for $j=h,h'$.
+
+Define
+\[
+ g_j(\eta)=\lambda_j(P+\eta v)-\lambda_j(P)
+ \in\Omega_P[[\eta]],\qquad j=h,h'.
+\]
+Then $g_{h'}'(0)\neq 0$, and
+\begin{equation}
+ \Lambda_h^R(\Delta)-\lambda_h(P)
+ =\bigl(g_h\circ g_{h'}^{-1}\bigr)
+       \bigl(\Lambda_{h'}^R(\Delta)-\lambda_{h'}(P)\bigr).
+ \label{eq:functional-linkage}
+\end{equation}
+Here $g_{h'}^{-1}$ denotes the inverse for composition.  In particular,
+the univariate series in this identity depends only on $P,h,h'$ and $v$,
+and is unchanged when auxiliary blocks are adjoined.
+\end{proposition}
+
+\begin{proof}
+We first explain why the response relations persist while $P$ moves.
+By the Taylor convention in Section~\ref{subsec:formal-conventions},
+substitution $P\mapsto P+\eta v$ gives homomorphisms
+\[
+ \operatorname{Frac}K[[P]]
+ \longrightarrow \bigl(\operatorname{Frac}K[[P]]\bigr)[[\eta]],
+ \qquad
+ \Omega_P\longrightarrow\Omega_P[[\eta]],
+\]
+both reducing to the identity at $\eta=0$. The second uniquely
+extends the first through the finite separable extension
+$\Omega_P/\operatorname{Frac}K[[P]]$. Thus every nonzero element
+has image with invertible constant term, and the eigenvalues and
+projections map to their uniquely labeled Taylor deformations.
+
+For linked labels, Equation~\eqref{eq:linkage-partition} gives
+\begin{equation}
+ D_{h'}(P;v)c_h-D_h(P;v)c_{h'}=0
+ \label{eq:linked-response-relation}
+\end{equation}
+for every loop response and every coefficient vector of every paired
+response.  Apply the preceding homomorphism coefficientwise in the
+partner coordinates.  The resulting identities are exactly
+Equation~\eqref{eq:linked-response-relation} at the moved input
+$P+\eta v$, with the moved eigenvalues, projections, and $D$-factors.
+They remain valid in the triangular specializations used in
+Proposition~\ref{prop:word-transfer}.  More explicitly, first clear the
+finitely many auxiliary resolvent determinants.  At each Taylor degree,
+the number of strictly upper steps is bounded.  This permits their
+entries to have Laurent coefficients in $\eta$, while the diagonal
+entries are expanded at the original small inputs.  Coefficientwise
+convergence and specialization are then supplied by
+Lemma~\ref{lem:active-specialization}.  Thus the transfer argument
+applies over $E((\eta))$ with the transported response relation.
+
+Write $v_R$ for $v$ supported in the $P$-block and put
+$R(\eta)=R+\eta v_R$.  At $R(\eta)+H$, impose the equation
+\begin{equation}
+ \Lambda_{h'}^{R(\eta)}(H)=\lambda_{h'}(P+\eta v).
+ \label{eq:fixed-linked-branch}
+\end{equation}
+Choose linear coordinates $H=u v_R+H_{\mathrm{rest}}$, with
+$H_{\mathrm{rest}}$ in a fixed complementary space of input directions.
+The derivative of the left side in $u$ at $H=0$ is
+$D_{h'}(P+\eta v;v)$, a unit in $E[[\eta]]$.  Formal implicit
+substitution therefore eliminates $u$.  In particular, the quotient
+of $E[[\eta,H]]$ by Equation~\eqref{eq:fixed-linked-branch} is
+\begin{equation}
+ E[[\eta,H_{\mathrm{rest}}]],
+ \qquad
+ E[[\eta,H_{\mathrm{rest}}]]
+ \lhook\joinrel\longrightarrow
+ E((\eta))[[H_{\mathrm{rest}}]].
+ \label{eq:linkage-quotient-injection}
+\end{equation}
+The eliminated series has zero constant term in $H_{\mathrm{rest}}$.
+
+On this quotient, scalar-block separation implies that the fixed-root
+Schur expression $S_{h'}$ at $R(\eta)$ vanishes.  We claim that $S_h$
+also vanishes over the right-hand ring in
+Equation~\eqref{eq:linkage-quotient-injection}.  Induct on total degree
+in $H_{\mathrm{rest}}$.  Both Schur expressions vanish in degree zero.
+If they vanish below degree $l$, the commuting-parameter conclusion of
+Proposition~\ref{prop:word-transfer}, together with the transported
+relation~\eqref{eq:linked-response-relation}, gives scalar leading
+coefficients $[S_j]_l=c_{j,l}\pi_j$ for $j=h,h'$, with
+\[
+ D_{h'}(P+\eta v;v)c_{h,l}
+   =D_h(P+\eta v;v)c_{h',l}=0.
+\]
+Here the $c_{j,l}$ are scalar homogeneous polynomials in the remaining
+parameters over $E((\eta))$; the projections are the moved target
+projections. The first factor is nonzero, so $c_{h,l}=0$ and
+$[S_h]_l=0$.
+
+It follows that the $h$-branch is fixed as well:
+\[
+ \Lambda_h^{R(\eta)}(H)=\lambda_h(P+\eta v)
+\]
+on the quotient.  For clarity, the implication from $S_h=0$ uses the
+scalar blocks of Lemma~\ref{lem:scalar-branches}: the complementary
+block at the fixed root is invertible, and the determinant of the full
+matrix at that root is a unit times the $t$-th power of the shift of
+the $h$-branch.  The quotient is a domain, so the shift is zero.
+Both branch series and the eliminated coordinate are regular in
+$\eta$.  The injection in
+Equation~\eqref{eq:linkage-quotient-injection} therefore descends this
+equality to $E[[\eta,H_{\mathrm{rest}}]]$.
+
+Finally make the invertible linear coordinate change
+$\Delta=H+\eta v_R$.  Equation~\eqref{eq:fixed-linked-branch} becomes
+\[
+ \Lambda_{h'}^R(\Delta)-\lambda_{h'}(P)=g_{h'}(\eta).
+\]
+Since $g_{h'}'(0)=D_{h'}(P;v)\neq0$, it eliminates $\eta$ as
+$g_{h'}^{-1}(\Lambda_{h'}^R(\Delta)-\lambda_{h'}(P))$.
+Substitution in the equality for the $h$-branch gives
+Equation~\eqref{eq:functional-linkage}.
+\end{proof}
+
+\subsection{All partner sizes and the dimension formula}
+
+We next pass from an identity near a direct sum to an identity for
+every coefficient defining the generic active space.  This step is
+necessary because a bound proved using a fixed collection of partners
+would not bound $a_n$.
+
+\begin{proposition}[Replication of active dimensions]
+\label{prop:active-replication}
+For every positive integer $N$,
+\[
+ a_{pN}=N a_p.
+\]
+\end{proposition}
+
+\begin{proof}
+Put $n=pN$ and choose independent small generic inputs
+$P_1,\ldots,P_N$ of size $p$.  Write
+$R_1=\operatorname{diag}(P_1,\ldots,P_N)$ and use
+$x=R_1+\Delta_1$ to label the generic size-$n$ spectrum by its branches
+at $R_1$.
+
+This substitution embeds $K[[x]]$ in $E[[\Delta_1]]$.  Indeed, for
+an image of a series over $K$, the Taylor coefficients are uniformly
+integral in the small parameter $\varepsilon$.  Substituting
+$\Delta_1=\varepsilon X-R_1$, with independent coordinates $X$,
+is therefore convergent and recovers the original series at
+$\varepsilon X$.  Homogeneous expansion at that point detects every
+nonzero series.  The embedding extends to the fraction field.  All
+generic spectral roots are supplied by the scalar branches at $R_1$,
+so it also embeds a spectral splitting field $\Omega_x$ into
+$\operatorname{Frac}E[[\Delta_1]]$, with these labels.  The branches
+and spectral projections themselves are regular at $\Delta_1=0$.
+
+Fix a block $i$ and labels $h,h'$ in one member of
+$\mathcal P_{P_i}$.  Choose a constant direction $v_i$ supported in
+that block, nonzero under every $D_h(P_i;\cdot)$.  Apply
+Proposition~\ref{prop:functional-linkage} first at $R_1$.
+Differentiating its identity gives proportional loop responses at
+$x$.  Now adjoin an independent generic partner $Q$ of any fixed
+size in $\mathcal S$, and apply the same proposition to
+$\operatorname{diag}(R_1,Q)$.  Its univariate series is the same
+one.  At $\operatorname{diag}(x,Q)$, perturb the opposite input
+blocks by $uU$ and $zV$.  Either arrow alone preserves the spectrum,
+so the first derivatives in $u$ and $z$ vanish.  The coefficient of
+$uz$ in the branch is the paired response $m_h(x,Q;U,V)$.
+The mixed chain rule and the separate first derivative in $v_i$
+give
+\begin{equation}
+ D_{h'}(x;v_i)m_h(x,Q;U,V)
+   -D_h(x;v_i)m_{h'}(x,Q;U,V)=0.
+ \label{eq:replicated-pair-relation}
+\end{equation}
+The two $D$-factors have nonzero reductions at $\Delta_1=0$.
+
+We must verify this relation coefficientwise in $Q$, rather than only
+at its generic value.  Cross multiplication has already removed the
+response ratio.  Multiply further by the finitely many determinants
+of $W(Q)-\lambda_h(x)$ and $W(Q)-\lambda_{h'}(x)$ needed to clear
+the auxiliary inverses.  At each fixed $\Delta_1$-degree, the resulting
+series in the independent coordinates of $Q$ has coefficients
+uniformly bounded below in $\varepsilon$-valuation.  The branch and
+projection coefficients do not involve $Q$; the other factors are
+ordinary matrix series with two prescribed arrow steps and a fixed
+number of Taylor insertions.  Their path expansions give precisely
+this bound, as in Lemma~\ref{lem:admissible-evaluation}.
+Evaluation at small generic $Q$ therefore detects every coefficient.
+Equation~\eqref{eq:replicated-pair-relation} is consequently a
+formal identity in those coordinates.  Injectivity of the embedding
+of $\Omega_x$ descends it to the generic size-$n$ field.  Finally,
+the cleared determinants have nonzero constant terms in $Q$, so
+they can be canceled in $\Omega_x[[Q]]$.
+
+This argument applies separately to every partner size with the same
+labeling of $x$ and the same $D$-factors.  It therefore gives a
+relation on every coefficient vector used to define $\mathcal T_x$.
+Together with the corresponding loop relations, it says that
+$c_h/D_h(x;v_i)$ is constant on each old partition member within
+block $i$, for every $c\in\mathcal T_x$.  These partitions have
+$Na_p$ members in total, and all their normalizing factors are
+nonzero.  Thus $a_n\leq Na_p$.
+
+For the opposite inequality, use
+Lemma~\ref{lem:active-specialization} for each $P_i$, choosing all
+partners jointly independent.  At $R_1$, support each loop direction,
+and both arrows of each paired response, in its chosen block $i$.
+The responses then coincide with the original responses on that
+block and vanish on every other block.  They have rank $Na_p$.
+Evaluate the same finite collection first at $R_1+\Delta_1$.
+All branch data and partner resolvents are regular at
+$\Delta_1=0$, since the partner spectra are disjoint from that of
+$R_1$.  A generic rank bound smaller than $Na_p$ would make every
+minor of that order vanish over the embedded generic field, and hence
+also at the reduction.  The displayed specialization contradicts this.
+Therefore $a_n\geq Na_p$.
+\end{proof}
+
+\subsection{Partners of one common size}
+
+The dimension formula permits us to use a fixed matrix size on both
+ends of every paired response.  We will also need to fix one arrow
+and vary only its reverse.
+
+\begin{corollary}[Equal-size partners]
+\label{cor:equal-size-partners}
+There are $r\in\mathcal S$ and a positive integer $K_0$ with the
+following property.  Let $P,Q_1,\ldots,Q_{K_0}$ be independent small
+generic inputs of size $r$.  For generic fixed input arrows
+$U_j:P\leftarrow Q_j$, the loop columns and the images of
+\[
+ V\longmapsto m(P,Q_j;U_j,V),
+ \qquad V:Q_j\leftarrow P,\quad 1\leq j\leq K_0,
+\]
+span $\mathcal T_P$ after extension to the joint field of the data.
+The analogous assertion holds with the other arrow fixed.  The same
+$r$ and $K_0$ may be used for both assertions.
+\end{corollary}
+
+\begin{proof}
+Choose a finite witness list at size $p$ from
+Lemma~\ref{lem:active-specialization}, and let its partner sizes be
+$q_1,\ldots,q_M\in\mathcal S$.  Choose $r=pN$ with
+$r\geq q_j$ for every $j$.  Specialize the size-$r$ base to a direct
+sum of $N$ independent size-$p$ inputs.  For each block, use a fresh
+copy of the witness list.  Pad a size-$q_j$ partner to size $r$ by
+adjoining independent generic diagonal data of size $r-q_j$, omitting
+the padding if this difference is zero.  Give both arrows zero entries
+on the padding.  The response remains the original size-$p$ witness
+on its chosen base block and is zero on the other base blocks.
+
+The resulting columns have rank $Na_p=a_r$ by
+Proposition~\ref{prop:active-replication}.  All inputs at this
+specialization have maximal spectral count, and the partner spectra
+are disjoint from the base spectrum.  The regular-minor argument in
+the preceding proof therefore transfers this rank to generic
+size-$r$ base and partner inputs.
+
+Assign a separate partner to every paired column in this finite
+list.  At the witness specialization, choose its forward arrow to
+be the specified $U_j$ and test the variable reverse arrow on the
+specified $V_j$.  A minor of order $a_r$ is nonzero there.  Since
+paired responses are bilinear in their arrows, this minor remains
+nonzero for generic fixed forward arrows and suitable reverse-arrow
+columns.  This proves the first assertion with finitely many partners.
+The identical argument fixes the reverse arrows instead.  Increasing
+the number of independent partners to the larger of the two finite
+numbers proves both assertions with one $K_0$.  If no paired column
+is needed, one unused generic partner may be added so that $K_0>0$.
+\end{proof}

preprints/A-Counterexample-to-Kuroshs-Division-Ring-Problem-September-23-2026/build/sections/responses.tex

@@ -0,0 +1,312 @@
+\section{The space of eigenvalue responses}\label{sec:responses}
+
+Fix a size $n\in\mathcal S$, a generic input pair $P$, and a splitting
+field $\Omega_P$ of $W(P)$ over $\operatorname{Frac}K[[P]]$.
+Write its distinct roots as $\lambda_h$, with spectral projections
+$\pi_h$, $1\le h\le e_n$.  Each projection has rank $t$.
+We also use these data after embedding into the valued fields of
+Section~\ref{sec:formal-generic}.
+
+We seek the possible leading changes of the eigenvalues at $P$.
+Besides varying $P$ itself, we allow an excursion through another
+input block and a return to $P$.  Their linear span will determine
+which eigenvalues can be varied independently.
+
+\subsection{Loop and paired responses}\label{subsec:response-formulas}
+
+Upper triangular block evaluations are also the basic mechanism of
+noncommutative difference-differential calculus; see
+\cite[Proposition~2.2 and Theorem~3.11]{KaliuzhnyiVerbovetskyiVinnikov2014}.
+Here the admissible evaluations of Section~\ref{sec:formal-generic}
+let us extract scalar eigenvalue responses from those block calculations.
+
+For a perturbation $\mathcal W$ of $W(P)$, put $\pi=\pi_h$,
+$\bar\pi=1-\pi$, and
+\[
+ C_h(\mathcal W)=\bar\pi(\mathcal W-\lambda_h)\bar\pi.
+\]
+This operator acts on the complementary space. Its constant term is
+invertible, so its inverse has a unique expansion in either commuting
+or free formal parameters. Define the fixed-root Schur expression by
+\begin{equation}
+S_h(\mathcal W)=
+ \pi(\mathcal W-\lambda_h)\pi
+ -\pi\mathcal W\bar\pi
+  C_h(\mathcal W)^{-1}
+  \bar\pi\mathcal W\pi.
+ \label{eq:fixed-schur}
+\end{equation}
+This expression records the obstruction to keeping the root $\lambda_h$
+fixed. For $u$ in the image of $\pi$, the vector
+\[
+ u-C_h(\mathcal W)^{-1}\bar\pi\mathcal W\pi u
+\]
+is the unique vector projecting to $u$ whose complementary component
+under $\mathcal W-\lambda_h$ vanishes. Its remaining component is
+$S_h(\mathcal W)u$. Thus $S_h(\mathcal W)=0$ means that the fixed-root
+kernel is a graph over the original eigenspace.
+
+When the constant output is split semisimple and these equations are
+imposed for every one of its distinct roots, the kernel graphs combine
+into a formal change of basis. The output remains conjugate to its
+constant term and satisfies that constant term's minimal polynomial.
+Lemma~\ref{lem:schur-polynomial} proves the equivalence with the
+polynomial identity at every finite word order.
+
+The same Schur expression is defined near $\operatorname{diag}(P,C_0)$
+if $\lambda_h$ is absent from $W(C_0)$, embedding $\pi_h$ at the $P$
+block. In this use, equations indexed by labels at $P$ record those
+fixed roots only.
+
+For a loop direction $Y=(Y_1,Y_2)$ at $P$, define the scalar
+$D_h(P;Y)$ by
+\begin{equation}
+\pi_h W'_P(Y)\pi_h=D_h(P;Y)\pi_h.
+ \label{eq:loop-response}
+\end{equation}
+This compression is scalar: apply Lemma~\ref{lem:scalar-branches}
+to the commuting deformation $P+xY$ and take its first coefficient.
+Equivalently, $D_h(P;Y)$ is the differential of the branch
+$\lambda_h$ in direction $Y$.  By
+Lemma~\ref{lem:nonconstant-roots}, each of these linear functionals is
+nonzero.  Because $K$ is infinite, we can choose a constant direction
+$v$ over $K$ with
+\begin{equation}
+D_h(P;v)\ne0\qquad(1\le h\le e_n).
+ \label{eq:nonzero-direction}
+\end{equation}
+Indeed the forbidden directions form a finite union of proper
+$K$-linear subspaces, even though their defining coefficients belong
+to $\Omega_P$.
+
+Let $Q$ be an independent generic point of size $n'\in\mathcal S$.
+Let $U$ be an input arrow $P\leftarrow Q$ and $V$ an input arrow
+$Q\leftarrow P$; each is a pair of rectangular scalar matrices.
+Form the triangular input pair
+\[
+ T=\begin{pmatrix}P&U&0\\0&Q&V\\0&0&P\end{pmatrix}.
+\]
+Set
+\begin{equation}
+M_h(P,Q;U,V)=\pi_h\left(
+ W(T)_{13}
+ -W(T)_{12}(W(Q)-\lambda_h)^{-1}W(T)_{23}
+ \right)\pi_h.
+ \label{eq:paired-response}
+\end{equation}
+Block paths show that this is bilinear in $U,V$.  The inverse exists
+because the spectra at independent points are disjoint.
+
+\begin{lemma}\label{lem:paired-scalar}
+There is a scalar $m_h(P,Q;U,V)$ such that
+\[
+ M_h(P,Q;U,V)=m_h(P,Q;U,V)\pi_h.
+\]
+It is the coefficient of $xy$ in the branch reducing to $\lambda_h$
+for the commuting input deformation
+\[
+ \begin{pmatrix}P&xU\\yV&Q\end{pmatrix}.
+\]
+The same assertion holds at an admissible specialized pair whose
+spectra are disjoint and whose combined output is split semisimple
+with the maximal number $e_{n+n'}$ of distinct roots.
+\end{lemma}
+
+\begin{proof}
+At $x=y=0$, disjointness and Lemma~\ref{lem:stable-multiplicity}
+give $e_n+e_{n'}=e_{n+n'}$ roots.  Thus
+Lemma~\ref{lem:scalar-branches} applies.  Either arrow alone leaves
+the spectrum unchanged, so the chosen branch has no terms depending
+on $x$ alone or on $y$ alone.
+
+Compute the $xy$ coefficient of its fixed-root Schur expression.
+The direct return from $P$ through $Q$ gives $W(T)_{13}$.  Eliminating
+the $Q$ block gives the second term in
+\eqref{eq:paired-response}.  The other eigenspaces in the $P$ block
+give no further term of this order: their couplings to $\pi_h$ have
+zero constant term and no first-order term under these off-diagonal
+input perturbations.  Replacing the fixed root by the scalar branch
+subtracts its $xy$ coefficient times $\pi_h$.  The resulting Schur
+expression vanishes, because the branch eigenspace projects
+isomorphically to the base eigenspace.  This proves the formula.
+The specialized assertion uses exactly the same argument and the
+specialization-stable degree bound.
+\end{proof}
+
+\subsection{Coefficient spans and their specializations}
+\label{subsec:active-space}
+
+At $Q=0$ the matrices $W(Q)-\lambda_h$ are invertible, since no
+$\lambda_h$ belongs to the spectrum of $w(0)$.  Hence
+\eqref{eq:paired-response} has a formal expansion in the entries of
+$Q$ over $\Omega_P$.  Define
+\[
+ m_h=\frac1t\operatorname{tr}M_h,
+ \qquad m=(m_h)_{h=1}^{e_n},\qquad D=(D_h)_{h=1}^{e_n}.
+\]
+Here $D=D(P;\cdot):M_n(\Omega_P)^2\to\Omega_P^{e_n}$ is the linear
+map of loop responses. Its columns are its values on the standard
+coordinate directions in the two input matrices.
+The scalar identity of Lemma~\ref{lem:paired-scalar} holds also for
+these formal expansions, by clearing their determinants and applying
+Lemma~\ref{lem:generic-detection}.
+
+Define $\mathcal T_P\subseteq\Omega_P^{e_n}$ to be the linear span
+of the columns of $D$ and of every coefficient vector of $m$, for
+every allowed partner size $n'\in\mathcal S$ and every pair of
+coordinate arrows $U,V$.  Write
+\[
+ a_n=\dim_{\Omega_P}\mathcal T_P.
+\]
+Although the definition permits infinitely many partner sizes and
+coefficients, the ambient space is finite dimensional.
+
+\begin{lemma}[Specialization and finite witnesses]
+\label{lem:active-specialization}
+The scalar identities for $M_h$ and every linear identity expressing
+$m\in\mathcal T_P$ hold after any admissible specialization of the
+partner input for which the required inverses exist.  For identities
+on a subset of labels, only the inverses for those labels are needed.
+
+After a field extension, the columns of $D$ and finitely many paired
+responses span $\mathcal T_P$.  These responses can be taken at
+independent small generic partner points, whose sizes need not be
+equal.
+\end{lemma}
+
+\begin{proof}
+For the first assertion, apply any $\Omega_P$-linear functional
+annihilating $\mathcal T_P$ to the formal tuple $m$.  The result is
+zero.  Multiply by the finitely many determinants of
+$W(Q)-\lambda_h$ that occur.  The cleared expression is a finite
+sum of products of base-point scalars and matrix-entry series; its
+auxiliary coefficients have a uniform valuation bound.  Its formal
+identity is therefore equivalent to its identity at a small generic
+partner, by Lemma~\ref{lem:generic-detection}.  It persists under the
+admissible specializations of Lemma~\ref{lem:admissible-evaluation}.
+Divide by the specialized determinants when they are nonzero.  The
+same argument, entry by entry, proves the scalar identities for
+$M_h$.  It also proves the projected assertion, without introducing
+any spectral labels for the partner input.
+
+Choose finitely many coefficient vectors, together with columns of
+$D$, that span $\mathcal T_P$, and select a nonzero minor witnessing
+their rank.  For each column requiring a partner, use a separate set
+of partner variables.  By multilinearity, the corresponding
+determinant of the series columns has that nonzero minor as the
+coefficient of a specified product of monomials.  It is therefore a
+nonzero formal series.  Clear the column denominators.  The numerator
+is still nonzero, and its coefficients are uniformly bounded below
+in valuation.  Independent small generic evaluations keep this
+numerator and its denominators nonzero, by
+Lemma~\ref{lem:generic-detection}.  The resulting values attain the
+full dimension.  The first assertion places all of them in the
+extended active space.
+
+This argument evaluates cleared numerators, not the possibly
+divergent geometric expansions of their inverses.  The distinction
+is precisely the one made in Section~\ref{subsec:formal-conventions}.
+\end{proof}
+
+The lemma also applies after adjoining Taylor or Laurent parameters.
+One first clears determinants and then works coefficientwise as in
+Lemma~\ref{lem:admissible-evaluation}.  In particular, it permits
+triangular partners with repeated base blocks, provided the inverses
+required for the selected target roots exist in the chosen Laurent
+field.
+
+\subsection{A partition of the eigenvalue labels}
+\label{subsec:active-partition}
+
+The active space has a stronger structure than a general linear
+subspace.  Its coordinates form groups whose responses are always in
+a fixed ratio.
+
+\begin{proposition}\label{prop:active-partition}
+For the direction $v$ chosen in \eqref{eq:nonzero-direction}, there
+is a partition of $\{1,\ldots,e_n\}$ such that
+\[
+ \mathcal T_P=
+ \left\{c\in\Omega_P^{e_n}:
+ \frac{c_h}{D_h(P;v)}\text{ is constant on each part}\right\}.
+\]
+\end{proposition}
+
+\begin{proof}
+We first prove that $\mathcal T_P$ is closed under the operation
+\begin{equation}
+(a,b)\longmapsto
+ \left(\frac{a_hb_h}{D_h(P;v)}\right)_h.
+ \label{eq:weighted-product}
+\end{equation}
+Choose two finite spanning sets supplied by
+Lemma~\ref{lem:active-specialization}, using independent partner
+points for the two sets.  It suffices to prove the assertion for a
+column $a$ in the first set and a column $b$ in the second, over their
+joint coefficient field $E$.
+
+Put $P'=P+\xi v$.  Use a triangular chain with endpoints $P,P$,
+middle point $P'$, an optional point $Q_1$ between the first endpoint
+and $P'$, and an optional point $Q_2$ between $P'$ and the last
+endpoint.  A direct step represents a loop column; an excursion
+through $Q_i$ represents a paired column.  Put the arrows defining
+$a$ and $b$ on the successive steps, with no other input arrows.
+Group the internal positions as a single partner $C(\xi)$.  Its
+size belongs to $\mathcal S$.
+
+For every target label, $W(C(\xi))-\lambda_h$ is invertible over
+$E((\xi))$.  The optional blocks have spectra disjoint from $P$.
+At the middle block, the branch reducing to $\lambda_h$ moves with
+nonzero derivative $D_h(P;v)$; all other branches are already away
+from $\lambda_h$.  Thus
+\begin{equation}
+\left.\xi\bigl(W(P')-\lambda_h\bigr)^{-1}\right|_{\xi=0}
+   =\frac{\pi_h}{D_h(P;v)}.
+ \label{eq:middle-residue}
+\end{equation}
+This follows directly by scalar branch separation.
+
+By Lemma~\ref{lem:active-specialization}, the paired response through
+$C(\xi)$ belongs to $\mathcal T_P\otimes E((\xi))$.  Eliminate
+the optional blocks first in its Schur formula.  Triangularity means
+that their elimination does not change the middle diagonal block.
+All terms are regular at $\xi=0$ except the middle inverse in
+\eqref{eq:middle-residue}.  The effective first-to-middle coupling,
+compressed by $\pi_h$ at both ends and evaluated at zero, is
+$a_h\pi_h$.  For a direct step this is
+\eqref{eq:loop-response}; for an excursion it is
+\eqref{eq:paired-response}.  The corresponding middle-to-last
+coupling is $b_h\pi_h$.
+
+The singular term occurs with the minus sign in the Schur
+complement.  Multiplying by $\xi$ and reducing at zero therefore
+gives
+\[
+ \left(-\frac{a_hb_h}{D_h(P;v)}\right)_h
+       \in\mathcal T_P\otimes E.
+\]
+Taking a Laurent coefficient preserves membership in this fixed
+finite dimensional space: apply its annihilating linear functionals.
+Bilinearity and the two spanning sets prove closure under
+\eqref{eq:weighted-product} after extension to $E$, and faithful
+scalar extension proves it over $\Omega_P$.
+
+Divide coordinates by the nowhere-zero vector $D(P;v)$.  The resulting
+space $\mathcal U\subseteq\Omega_P^{e_n}$ contains $1$, since
+$D(P;v)\in\mathcal T_P$, and is closed under coordinatewise
+multiplication.  Declare two coordinates equivalent if every element
+of $\mathcal U$ has the same value on them.  By avoiding finitely
+many proper linear subspaces, choose an element of $\mathcal U$
+whose values distinguish all the equivalence classes.  Lagrange
+interpolation in this element gives the indicator function of each
+class.  These indicators span all class-constant functions.  Thus
+$\mathcal U$ is exactly that algebra of functions, proving the
+proposition.
+\end{proof}
+
+We call two labels in the same part \emph{linked}.  The number of
+parts is $a_n$.  Selecting one representative from each part
+identifies $\mathcal T_P$ with the full coordinate space on those
+representatives.  The next stage must show that these response
+relations control higher-order free-letter errors, rather than only
+the two elementary perturbations used to define them.

preprints/A-Counterexample-to-Kuroshs-Division-Ring-Problem-September-23-2026/build/sections/transfer.tex

@@ -0,0 +1,250 @@
+\section{Transfer of the first unsolved word coefficient}
+\label{sec:word-transfer}
+
+Our next objective is to control equations in free letters.  The first and
+paired responses describe matrix perturbations with one loop or two opposite
+arrows.  We show that they also contain the first unsolved coefficient of an
+arbitrary free-series perturbation.  The argument unfolds a word along a
+triangular chain and then replaces its internal states by a single auxiliary
+partner.  The only difficulty is that this partner initially has the target
+eigenvalue.  The following linear-algebra observation explains how to remove
+that singularity without changing the coefficient being computed.
+
+\begin{lemma}[An inverse between two vanishing couplings]
+\label{lem:killed-pole}
+Let $E$ be a field and let $M(\xi)\in\operatorname{End}_E(V)[[\xi]]$, where
+$V$ is finite dimensional.  Suppose that zero is a semisimple eigenvalue of
+$M_0=M(0)$, and let $\Pi$ be its spectral projection.  Suppose that
+\[
+ G=\Pi M'(0)\Pi\big|_{\ker M_0}
+\]
+is invertible.  Then $M(\xi)^{-1}$ exists over $E((\xi))$ and has a pole of
+order at most one.  Let $J$ be zero on $\ker M_0$ and equal to $M_0^{-1}$ on
+$\operatorname{im}M_0$.  If $L(\xi)$ and $B(\xi)$ are compatible matrix
+series satisfying
+\[
+ L(0)\Pi=0,\qquad \Pi B(0)=0,
+\]
+then
+\begin{equation}
+\label{eq:killed-pole-limit}
+ \lim_{\xi\to0}L(\xi)M(\xi)^{-1}B(\xi)=L(0)JB(0).
+\end{equation}
+
+Suppose additionally that $e$ is an idempotent, $f=1-e$,
+$fM_0f$ is invertible on $fV$, and $e:\ker M_0\to eV$ is an isomorphism.
+Under the same two coupling conditions,
+\begin{equation}
+\label{eq:fixed-complement-inverse}
+ L(0)JB(0)=L(0)f(fM_0f)^{-1}fB(0).
+\end{equation}
+\end{lemma}
+
+\begin{proof}
+Use $V=\ker M_0\oplus\operatorname{im}M_0$.  In this decomposition
+\[
+ M(\xi)=
+ \begin{pmatrix}
+  \xi G+O(\xi^2)&O(\xi)\\
+  O(\xi)&M_0|_{\operatorname{im}M_0}+O(\xi)
+ \end{pmatrix}.
+\]
+The lower-right block is invertible and the Schur complement of that block
+is $\xi G+O(\xi^2)$.  The upper-left block of $M(\xi)^{-1}$ is consequently
+$\xi^{-1}G^{-1}+O(1)$; its off-diagonal blocks are bounded; and its
+lower-right block tends to $M_0^{-1}|_{\operatorname{im}M_0}$.
+The kernel components of both couplings are $O(\xi)$.  Multiplying these
+four blocks proves \eqref{eq:killed-pole-limit}.
+
+For the second assertion, write $b=B(0)u$.  The condition $\Pi b=0$ places
+$b$ in $\operatorname{im}M_0$.  Solutions of $M_0x=b$ differ by
+$\ker M_0$.  Since projection to $eV$ is an isomorphism on that kernel,
+there is a unique solution with $ex=0$.  Its $f$ coordinate is
+$(fM_0f)^{-1}fb$.  All solutions have the same image under $L(0)$ because
+$L(0)$ kills $\ker M_0$.  Comparing this solution with $Jb$ proves
+\eqref{eq:fixed-complement-inverse}.
+\end{proof}
+
+We now recall exactly which response identities enter the transfer
+argument.  At a small generic point $P$ of size $n\in\mathcal S$, the
+eigenvalues $\lambda_h$ have projections $\pi_h$ of rank $t$.  The first
+response is $D_h(P;Y)\pi_h$.  A pair of arrows through an independent
+partner $Q$ gives $m_h(P,Q;U,V)\pi_h$.  The columns of the first responses
+and the coefficient vectors of all these paired responses span
+$\mathcal T_P$.  By Lemma~\ref{lem:active-specialization}, their scalarity
+and all linear containment identities persist for a block-triangular
+partner after its finitely many resolvent determinants are cleared.  For a
+set of target labels $I$, only the determinants belonging to labels in $I$
+are needed.  We use the notation
+\[
+ \mathcal T_P|_I=\{(c_h)_{h\in I}:(c_h)_h\in\mathcal T_P\}.
+\]
+These spaces and identities are extended to the coefficient field in use.
+We also fix a direction $v$ with $D_h(P;v)\ne0$ for every target label.
+
+\begin{proposition}[Transfer of a leading coefficient]
+\label{prop:word-transfer}
+Let $R=\operatorname{diag}(P,C_0)$ be a small matrix input of total size $N\in
+\mathcal S$;
+the block $C_0$ may be absent.  Let $I$ be a set of labels at
+$P$ such that $W(C_0)-\lambda_h$ is invertible for every $h\in I$.
+Work over a field $E$ containing the labeled spectral data at $P$ and the
+entries of $C_0$, with the formal evaluations allowed by
+Lemma~\ref{lem:admissible-evaluation}.  Let
+\[
+ Z(s_1,s_2)\in M_N(E)\langle\!\langle s_1,s_2\rangle\!\rangle^2
+\]
+have constant term $R$.  These are input matrices: their scalar entries
+act on the input factor and commute with the old coefficient algebra.
+
+For $h\in I$, embed $\pi_h$ at the $P$ block, put $\bar\pi_h=1-\pi_h$,
+and form
+\[
+ S_h(Z)=\pi_h(W(Z)-\lambda_h)\pi_h
+ -\pi_hW(Z)\bar\pi_h
+ \bigl(\bar\pi_h(W(Z)-\lambda_h)\bar\pi_h\bigr)^{-1}
+ \bar\pi_hW(Z)\pi_h.
+\]
+Suppose that every $S_h(Z)$ vanishes in degrees less than $\ell$, where
+$\ell\ge1$.  For each word $\omega$ of length $\ell$, there are scalars
+$c_{h,\omega}\in E$ such that
+\[
+ [S_h(Z)]_\omega=c_{h,\omega}\pi_h,
+ \qquad (c_{h,\omega})_{h\in I}\in
+        (\mathcal T_P|_I)\otimes E.
+\]
+The same conclusion holds for homogeneous degree-$\ell$ coefficients of
+perturbations in finitely many ordinary commuting parameters.
+\end{proposition}
+
+\begin{proof}
+For $\ell=1$, only the $P$-to-$P$ input coefficient contributes to the
+compressed derivative, so the assertion is the first-response identity.
+Assume $\ell\ge2$ and fix a word $\omega$ of length $\ell$.
+
+\paragraph{Unfolding the word.}
+Use $\ell+1$ ordered states.  On consecutive steps put the letters of
+$\omega$, and evaluate a word by following that path.  Thus a coefficient
+of $Z$ for a substring becomes the corresponding strictly upper block of
+the input, while every diagonal block is $R$.  The $(1,\ell+1)$ block
+extracts precisely the coefficient of $\omega$.  This substitution is a
+homomorphism on the free-series ring modulo degree $\ell+1$, and it
+therefore respects every inverse with invertible constant term.
+
+Keep only $P$ at the two endpoint states, leaving all $R$ blocks at the
+$\ell-1$ internal states.  This operation does not change the required
+compressed endpoint coefficient.  At an endpoint, constant input steps
+preserve the blocks $P,C_0$, and the constant output commutes with
+$\pi_h$.  A contributing Schur path can therefore neither enter the
+first endpoint complement from its target projection nor leave the last
+endpoint complement for its target projection.  All complement
+eliminations relevant to the coefficient occur at internal states.
+
+Denote the internal input by $Q_0$.  Its size is $(\ell-1)N$ and hence
+belongs to $\mathcal S$.  Grouping the internal
+states, write the cropped input as
+\[
+ T_0=\begin{pmatrix}
+ P&G_1&Y\\0&Q_0&G_2\\0&0&P
+ \end{pmatrix}.
+\]
+Here each entry denotes a pair of input matrices.  In the output write
+$E_1=W(T_0)_{12}$, $E_2=W(T_0)_{23}$, and $F=W(T_0)_{13}$.
+
+\paragraph{The internal target eigenspace.}
+Fix $h\in I$, set $M_0=W(Q_0)-\lambda_h$, and let $e$ be the sum of the
+fixed copies of $\pi_h$ on its internal states.  Put $f=1-e$.
+The matrix $fM_0f$ is upper triangular with invertible diagonal blocks.
+Moreover its Schur complement on $eV$ is zero: every strictly upper
+coefficient is a coefficient for a proper substring of $\omega$, and
+these vanish by hypothesis.  Its kernel is consequently the graph
+\[
+ u\longmapsto u-f(fM_0f)^{-1}fM_0e u,
+ \qquad u\in eV.
+\]
+Projection onto $eV$ is an isomorphism on this kernel, whose dimension is
+$(\ell-1)t$.  Triangularity shows that this is also the full algebraic
+multiplicity of zero.  Thus zero is semisimple.  Let $\Pi$ be its
+spectral projection.
+
+The same argument applies to the chain consisting of the first endpoint
+and the internal states, and to the chain consisting of the internal
+states and the last endpoint.  They involve only proper substrings.
+Solving their triangular kernel equations shows that the couplings
+between their zero eigenspaces vanish:
+\begin{equation}
+\label{eq:transfer-coupling-annihilation}
+ \pi_hE_1\Pi=0,\qquad \Pi E_2\pi_h=0.
+\end{equation}
+Indeed any nonzero compressed coupling would reduce the kernel dimension
+below the sum of the diagonal zero multiplicities.  This argument uses
+semisimplicity only at the target eigenvalue.
+
+\paragraph{A regularized auxiliary partner.}
+In every internal copy of $R$, replace its $P$ block by $P+\xi v$, keeping
+all arrows fixed; call the resulting input $Q(\xi)$.  Put
+$M(\xi)=W(Q(\xi))-\lambda_h$.
+The spectral projection $\Pi$ is upper triangular in the state order,
+with diagonal equal to the fixed target projections.  The graph
+identification of its range with $eV$ is upper triangular with identity
+diagonal.  It follows that
+\[
+ \Pi M'(0)\Pi\big|_{\ker M_0}
+\]
+is upper triangular in this identification, with every diagonal block
+equal to $D_h(P;v)I_t$.  It is therefore invertible.  This verifies the
+simple-pole hypothesis of Lemma~\ref{lem:killed-pole}, including the
+effect of all internal triangular arrows.
+
+Let $T(\xi)$ be the corresponding full cropped input and use $E_1(\xi)$,
+$E_2(\xi)$, $F(\xi)$ for its output blocks.  For nonzero formal $\xi$, the
+compressed endpoint expression is
+\begin{align}
+ \mathcal E_h(\xi)
+ &=\pi_h\bigl(F(\xi)-E_1(\xi)M(\xi)^{-1}E_2(\xi)\bigr)\pi_h
+ \label{eq:transfer-regularized-expression}\\
+ &=D_h(P;Y)\pi_h+M_h(P,Q(\xi);G_1,G_2).
+ \nonumber
+\end{align}
+Its tuple is scalar and belongs to the projected active space over
+$E((\xi))$.  This is exactly the specialized paired-response identity:
+the auxiliary block is upper triangular, its diagonal entries are small
+blocks with a Taylor shift, and its strictly upper entries are fixed
+finite matrices.  Clear the target resolvent determinants first and
+apply Lemma~\ref{lem:active-specialization}; restore the inverses over
+$E((\xi))$, where the preceding argument proves they exist.  No
+specialization of an undefined inverse is being used.
+
+\paragraph{Identifying the limit.}
+By \eqref{eq:transfer-coupling-annihilation} and
+Lemma~\ref{lem:killed-pole}, expression
+\eqref{eq:transfer-regularized-expression} has the finite limit
+\[
+ \pi_h\bigl(F-E_1JE_2\bigr)\pi_h
+ =\pi_h\bigl(F-E_1f(fM_0f)^{-1}fE_2\bigr)\pi_h.
+\]
+The last expression eliminates precisely the fixed complementary
+spaces in the internal states.  By the unfolding and endpoint-cropping
+argument it is $[S_h(Z)]_\omega$.  In particular the limiting inverse has
+not changed the coefficient to one computed with a different choice of
+complement.  Taking the limit in the scalar-response identities and in
+the finitely many linear equations defining $\mathcal T_P|_I$ proves
+the assertion simultaneously for all target labels.
+
+Finally consider a perturbation in commuting variables $h_1,\ldots,h_s$.
+Substitute $h_j=a_jz$ for arbitrary scalar directions $a_j$ and one free
+letter $z$.  The lower homogeneous terms still vanish, so the result
+just proved applies to the degree-$\ell$ coefficient on every such
+line.  Scalarity and containment in a fixed finite-dimensional linear
+space are linear conditions.  Each of their annihilators therefore
+gives a homogeneous polynomial in the $a_j$ that vanishes identically.
+Over the infinite characteristic-zero field all its coefficients vanish.
+This proves the commuting-parameter assertion.
+\end{proof}
+
+Proposition~\ref{prop:word-transfer} is the bridge from responses to
+free-series equations.  It controls a first unsolved coefficient without
+assuming that the perturbation is commutative or that the auxiliary
+matrix has a separated spectrum before regularization.  We next use its
+commuting-parameter consequence to show that linked responses arise from
+actual formal relations between eigenvalues.

preprints/A-Counterexample-to-Kuroshs-Division-Ring-Problem-September-23-2026/paper.pdf

@@ -0,0 +1,3 @@
+version https://git-lfs.github.com/spec/v1
+oid sha256:9195b0a83fc633704a735973a068257d424a46196a74cad21176e49348147aec
+size 541378

preprints/A-Counterexample-to-Rysers-Covering-Conjecture-September-23-2026/README.md

@@ -0,0 +1,18 @@
+# [A counterexample to Ryser's covering conjecture](paper.pdf)
+
+**Author:** OpenAI
+
+**Date:** September 23, 2026
+
+## Citation
+
+```bibtex
+@misc{OAI:A-Counterexample-to-Rysers-Covering-Conjecture-September-23-2026,
+  author = {{OpenAI}},
+  title = {{A counterexample to Ryser's covering conjecture}},
+  howpublished = {OpenAI Math Release preprint
+                  \href{https://github.com/openai/math/blob/main/preprints/A-Counterexample-to-Rysers-Covering-Conjecture-September-23-2026/paper.pdf}{OAI:A-Counterexample-to-Rysers-Covering-Conjecture-September-23-2026}},
+  year = {2026}
+}
+```
+

preprints/A-Counterexample-to-Rysers-Covering-Conjecture-September-23-2026/build/construction.tex

@@ -0,0 +1,857 @@
+\Needspace{14\baselineskip}
+\section{A configuration with two coordinatizations}
+\label{sec:gadget}
+
+We now use the deletion lemma to construct the exceptional configuration.
+We need two properties of the same set: many points on each undeleted
+line, and rigidity of its small line covers.  A change of coordinates
+lets us obtain both.  In the original coordinates we delete one line
+in every finite nonzero direction.  After the change of coordinates,
+the deleted sets become curves, so the line-survival conclusion of
+Lemma~\ref{lem:linecover} applies to every line.  It will force every
+small mixed cover by lines and individual points to be a parallel class.
+
+\begin{lemma}
+\label{lem:gadget}
+For every sufficiently large prime \(s\equiv2\pmod3\), there is
+a set \(Y\subseteq\F_s^2\) with the following properties.  For each
+\(d\in\F_s^*\), there is an affine line \(\ell_d\) of slope \(d\),
+and
+\[
+ Y=\F_s^2\setminus\bigcup_{d\in\F_s^*}\ell_d.
+\]
+For each \(d\in\F_s^*\), every line of slope \(d\) other than
+\(\ell_d\) contains at least five points of \(Y\).  The map
+\[
+ f:\F_s^2\longrightarrow\F_s^2,
+ \qquad f(x,y)=(x^3,y),
+\]
+is a bijection, and every affine line contains at least \(s/8\) points
+of \(f(Y)\).  Moreover, if \(a\) affine lines and \(b\) individual
+points cover \(f(Y)\), and
+\begin{equation}
+\label{con-weighted-budget}
+ a+\frac b2\le s,
+\end{equation}
+then \(a=s\), \(b=0\), and the lines form a full parallel class.
+Consequently every such mixed cover has weighted cost at least \(s\),
+and no cover of \(f(Y)\) by at most \(s\) lines contains two
+nonparallel lines.
+\end{lemma}
+
+\begin{proof}
+Let \(s\) tend to infinity through primes congruent to \(2\pmod3\).
+Since
+\(\gcd(3,s-1)=1\), cubing permutes \(\F_s^*\) and fixes zero.
+Thus \(f\) is bijective.  It maps horizontal lines to horizontal
+lines and vertical lines to vertical lines.  We first prove the two
+survival properties simultaneously with high probability, then use
+them to classify mixed covers satisfying~\eqref{con-weighted-budget}.
+
+\paragraph{Survivors in the original coordinates.}
+Choose the lines \(\ell_d\), independently for the \(s-1\) values
+of \(d\), uniformly within their parallel classes.  Fix a line
+\(\ell\) of finite nonzero slope \(d\), and temporarily ignore the
+choice \(\ell_d\).  Each of the other \(s-2\) choices deletes a
+uniformly distributed point of \(\ell\).  Let \(Z_\ell\) count
+the surviving points on \(\ell\).  Its expectation is
+\[
+ \mu:=\E Z_\ell=s(1-1/s)^{s-2}.
+\]
+Changing one of these choices changes the survivor count by at most
+two.  Apply the independent-trial jump bound and bounded-drift
+concentration estimate of
+\cite[Lemmas~A.2--A.3]{OpenAI2026Balanced} to the exposure martingale,
+with \(T=s-2\), \(L=2\), and \(u=\mu/2\).  The martingale has
+zero drift, and hence
+\[
+ \Pr(Z_\ell<\mu/2)
+ \le 2\exp\left(-\frac{\mu^2}{128(s-2)}\right)
+ =\exp(-\Omega(s)).
+\]
+There are fewer than \(s^2\) lines under consideration.  A union
+bound, together with \(\mu/2\ge5\) for sufficiently large \(s\),
+shows that all these counts are at least five with probability tending
+to one.  If \(\ell\ne\ell_d\), reinstating the deletion by
+\(\ell_d\) changes no point on \(\ell\), since the two lines are
+parallel.  Thus every required line in the original coordinates has
+at least five points of \(Y\).
+
+\paragraph{Survivors and almost-pencil covers in the new coordinates.}
+In the new coordinates, the possible deletion shapes are
+\[
+ F_{d,z}=\{(x^3,dx+z):x\in\F_s\},
+ \qquad d\in\F_s^*,\quad z\in\F_s.
+\]
+There are \(s(s-1)\) such shapes, each of size \(s\), and each of
+the \(s-1\) independent trials chooses one of them.  Every point
+is deleted by a specified trial with probability \(1/s\), so its
+overall survival probability is the uniform value
+\[
+ \delta_s=(1-1/s)^{s-1}\ge\frac14
+\]
+for all sufficiently large \(s\).
+
+Each shape meets every affine line in at most three points.  For a
+vertical line this follows from bijectivity of cubing; for a
+horizontal line it follows from \(d\ne0\).  For a line of finite
+nonzero slope \(m\), the intersection is determined by
+\[
+ dx+z=mx^3+u,
+\]
+a nonzero polynomial equation of degree three.  In particular, no
+shape is itself a line when \(s>5\).  All the hypotheses of
+Lemma~\ref{lem:linecover} hold with field size \(s\), a fixed
+integer constant, for example \(D=3\), survival lower bound
+\(\delta=1/4\), and \(\rho_0=1/100\).  The exact-line
+occurrence probabilities are zero, and the subfield restriction is
+automatic for a prime field.  With probability tending to one, every
+line consequently contains at least \(s/8\) points of \(f(Y)\),
+and its almost-cover conclusion holds for \(f(Y)\) as well.  We now
+combine that conclusion with the lower bound on every line to replace
+an almost pencil by a complete one.
+
+\paragraph{From an almost pencil to a full parallel class.}
+Assume these events hold, and consider a cover satisfying
+\eqref{con-weighted-budget}.  Repeated resources can be removed;
+alternatively, one may keep their original counts, since removal
+only reduces the budget.  The line portion has at most \(s\)
+distinct lines and leaves at most \(b\le2s\) points of \(f(Y)\)
+uncovered.  Lemma~\ref{lem:linecover} supplies at least
+\((1-\rho_0)s\) distinct lines in this portion through a common
+projective point \(p\).  There are at most \(\rho_0s\) other
+lines, and
+\[
+ b\le2(s-a)\le2\rho_0s.
+\]
+If an affine line through \(p\) were absent from the cover, all the
+chosen lines through \(p\) would cover at most one of its affine
+points.  The other lines would cover at most \(\rho_0s\) further
+points, and the individual point resources at most \(2\rho_0s\).
+Thus at most \(1+3\rho_0s\) points of this missing line would be
+covered.  For sufficiently large \(s\), this contradicts
+\[
+ \frac s8>1+3\rho_0s.
+\]
+Every affine line through \(p\) must therefore be present.  A
+finite \(p\) requires \(s+1\) lines, which is impossible.  An
+infinite \(p\) requires all \(s\) lines in its parallel class,
+and then \eqref{con-weighted-budget} forces \(a=s\) and \(b=0\).
+This also rules out repetitions in a cover attaining the budget.
+
+The events used in the two coordinate systems hold simultaneously
+with probability tending to one.  Consequently every sufficiently
+large prime \(s\equiv2\pmod3\) admits a successful configuration
+\(Y\).  A mixed cover of cost less than
+\(s\) would satisfy \eqref{con-weighted-budget} but not its
+necessary equality conclusion.  The two stated consequences follow.
+\end{proof}
+
+There are arbitrarily large primes congruent to \(2\pmod3\).  Indeed,
+if the list were finite and its product were \(P\), the integer
+\(3P-1\equiv2\pmod3\) would have a prime factor congruent to
+\(2\pmod3\), and none of the primes dividing \(P\) could divide it.
+Fix any sufficiently large prime \(s\equiv2\pmod3\), and choose
+\(Y\) and the omitted lines from Lemma~\ref{lem:gadget}.
+These data remain fixed for the rest of the proof.
+
+\section{Ambient placement}
+\label{sec:placement}
+
+We will place copies of \(Y\) in a larger affine plane, assign each
+copy the directions in which its labels will change, and delete the
+ambient lines of those directions that meet it.  The placement must
+meet two requirements.  Joins between different copies must avoid
+their changed directions, while the undeleted points must retain the
+pencil and line-survival properties of Lemma~\ref{lem:linecover}.
+We first specify the copies and direction sets, then prove the
+deletion estimates for independent pools of translates, and finally
+select one compatible translate from each pool.
+
+\subsection{Fields, copies, and direction sets}
+
+Let \(n\ge3\) be odd, and put \(q=s^n\). We identify \(\F_s\)
+with the prime subfield of \(\F_q\). If \(E\) is a subfield, then
+\(|E|=s^a\), where \(a=[E:\F_s]\), and multiplication of vector-space
+bases gives
+\[
+ n=a[\F_q:E].
+\]
+For a proper subfield, the index \([\F_q:E]\) is an odd integer
+larger than one, so it is at least three. Consequently
+\[
+ |E|=s^a\le s^{n/3}=q^{1/3}
+ \qquad\text{for every proper subfield }E.
+\]
+Thus Theorem~\ref{thm:incidence} and
+Lemma~\ref{lem:linecover} apply along this sequence of ambient
+fields.  In this section all constants may depend on the already
+fixed \(s\) and \(Y\), but not on \(n\).
+
+The odd-degree condition has now supplied the required subfield bound.
+It remains to assign every ambient direction to at least one copy,
+with bounded overlap between assignments.  Identify the ambient
+directions with
+\(\PP^1(\F_q)=\F_q\cup\{\infty\}\), using slope.  Choose a
+set \(\Lambda\) of representatives of
+\(\F_q^*/\F_s^*\), with \(1\in\Lambda\).  There will be one
+configuration group for each \(\lambda\in\Lambda\), together
+with one further group denoted by \(*\).  Write
+\[
+ \mathcal I=\Lambda\sqcup\{*\},
+ \qquad N_q=|\mathcal I|=\frac{q-1}{s-1}+1.
+\]
+Each index \(i\in\mathcal I\) will specify a pool of translates
+of one configuration.  We call this pool a group and will choose one
+translate from each group.
+
+For \(\lambda\in\Lambda\), define
+\[
+ M_\lambda(x,y)=(x,\lambda y),
+ \qquad S_\lambda=\lambda\F_s^*,
+ \qquad A_\lambda=\{0,\infty\}.
+\]
+Choose distinct \(u,v\in\F_s^*\) and define
+\[
+ M_*=
+ \begin{pmatrix}1&1\\u&v\end{pmatrix},
+ \qquad
+ S_*=(\F_s\cup\{\infty\})\setminus\{u,v\},
+ \qquad A_*=\{u,v\}.
+\]
+All these linear maps are invertible.  For \(M_*\), the image of
+a finite slope \(t\) is
+\[
+ \frac{u+vt}{1+t},
+\]
+where a zero denominator gives the infinite direction.  Its images
+of the horizontal and vertical directions are \(u\) and \(v\).
+The induced projective bijection therefore maps the \(s-1\)
+finite nonzero subfield slopes precisely to \(S_*\).  The same
+statement for \(M_\lambda\) follows from the slope map
+\(t\mapsto\lambda t\).
+
+For each \(i\in\mathcal I\), put
+\[
+ W_i=M_i(\F_s^2).
+\]
+The directions of joins of distinct points of \(W_i\) form exactly
+\(S_i\cup A_i\).  Each set \(S_i\) has size \(s-1\), and
+\(A_i\) consists of the two images of the coordinate axes.  The
+cosets \(S_\lambda\) partition all nonzero finite ambient slopes.
+The extra set \(S_*\) contains \(0\) and \(\infty\).  Hence
+every ambient direction belongs to at least one and at most two
+sets \(S_i\).  More precisely, the directions belonging to two
+sets are
+\begin{equation}
+\label{con-double-directions}
+ S_1\cap S_* = \F_s^*\setminus\{u,v\},
+\end{equation}
+and the two groups involved are always \(1\) and \(*\).
+
+The sets \(S_i\) are the directions whose labels will change, and
+\(A_i\) are the images of the two axis directions preserved by \(f\).
+For compatibility between copies we use the
+slightly larger direction sets
+\[
+ D_i=
+ \begin{cases}
+ S_i\cup A_i,&i\in\{1,*\},\\
+ S_i,&i\notin\{1,*\}.
+ \end{cases}
+\]
+In particular,
+\begin{equation}
+\label{con-conflict-directions}
+ D_1=D_*=\PP^1(\F_s),
+\end{equation}
+whereas the remaining \(D_i\) are the nonidentity multiplicative
+cosets of \(\F_s^*\).  Every direction occurs in at most two
+sets \(D_i\).  Since \(|W_i|=s^2\), \(|D_i|\le s+1\), and
+\(N_q=O_s(q)\), a fixed constant depending only on \(s\)
+satisfies all size and multiplicity assumptions of
+Theorem~\ref{thm:selection}.  In a conflict-free selection, including
+the axes in both \(D_1\) and \(D_*\) ensures that an axis line
+meeting one of these two selected copies cannot also meet the other.
+This will enter the weighted cover
+argument for the two groups sharing a direction
+in~\eqref{con-double-directions}.
+
+\subsection{Independent candidate pools}
+
+Let \(K\) be a fixed pool size supplied by
+Theorem~\ref{thm:selection}.  For each
+\(i\in\mathcal I\), choose independent uniform shifts
+\(b_{i,1},\ldots,b_{i,K}\in\F_q^2\), independently between
+groups, and write
+\[
+ P_{i,k}=b_{i,k}+W_i,
+ \qquad Y_{i,k}=b_{i,k}+M_iY.
+\]
+Associate to this candidate the following set of ambient deletion
+lines:
+\[
+ \mathcal B_{i,k}
+ =\{\ell:\operatorname{dir}(\ell)\in S_i,
+                  \ \ell\cap Y_{i,k}\ne\varnothing\}.
+\]
+
+\begin{lemma}
+\label{con-deletion-count}
+For every candidate and every \(d\in S_i\), exactly \(s-1\)
+lines of \(\mathcal B_{i,k}\) have direction \(d\).  Each of
+these lines contains at least five points of \(Y_{i,k}\).
+Consequently \(|\mathcal B_{i,k}|=(s-1)^2\).
+\end{lemma}
+
+\begin{proof}
+Undo the translation and apply \(M_i^{-1}\).  An ambient line in
+the given direction meeting \(P_{i,k}\) becomes a line through a
+point of \(\F_s^2\) in a finite nonzero subfield direction.  Its
+intersection with \(\F_s^2\) is the corresponding subfield
+line.  The \(s\) lines in that subfield parallel class extend to
+distinct parallel ambient lines.  Exactly one is the omitted line
+in the definition of \(Y\); the other \(s-1\) each contain at
+least five points of \(Y\), by Lemma~\ref{lem:gadget}.  These are
+precisely the lines meeting the candidate's retained configuration.
+\end{proof}
+
+Each candidate therefore prescribes a bounded list of deletion lines.
+We next apply the deletion lemma to all these independent lists at
+once.  Besides the almost-pencil and line-survival conclusions, we
+need one bound on how many groups a line can meet along joins from an
+external point.  All three properties will survive the subsequent
+selection of one candidate per group.
+
+\begin{lemma}
+\label{con-pool-properties}
+There are constants \(\delta\in(0,1)\) and
+\(\rho=\delta/10\), independent of \(n\), with the following
+properties with probability tending to one as \(n\) tends to
+infinity through odd integers. Let
+\[
+ X_{\rm pool}
+ =\F_q^2\setminus
+   \bigcup_{i\in\mathcal I}\bigcup_{k=1}^K
+          \bigcup_{\ell\in\mathcal B_{i,k}}\ell.
+\]
+\begin{enumerate}
+\item Every cover of \(X_{\rm pool}\) by at most \(q\) affine
+lines contains at least \((1-\rho)q\) lines through a common
+projective point.
+\item For every affine line \(\ell\), if only exact occurrences
+of \(\ell\) are ignored in the pool deletions, at least
+\(\delta q/2\) points of \(\ell\) remain.
+\item For every finite point \(p\) and affine line \(\ell\)
+not through \(p\), at most five groups \(i\) have a candidate
+\(P_{i,k}\) containing a point \(y\in\ell\) whose join to
+\(p\) has direction in \(S_i\).
+\end{enumerate}
+\end{lemma}
+
+\begin{proof}
+We verify the hypotheses of Lemma~\ref{lem:linecover} for the full
+pool, which gives the first two assertions.  The third follows from
+a separate union bound over finite points and lines.
+
+Treat each candidate shift as one independent deletion trial, listing
+the lines of \(\mathcal B_{i,k}\).  There are \(KN_q=O_{s,K}(q)\)
+trials, each listing \(m=(s-1)^2\) lines.  We may take the shape
+family to be all \(q(q+1)\) ambient affine lines.  Each shape has
+size \(q\), and two distinct line shapes intersect in at most one
+point.  Thus the shape and trial bounds in
+Lemma~\ref{lem:linecover} hold with a sufficiently large fixed
+integer \(D\) depending only on \(s,K\).
+
+A candidate's deleting set is a uniform translate of a fixed union
+of \(m\) lines.  Its deletion probability at a fixed point is
+independent of the point and is at most \(m/q\).  Independence of
+the shifts makes the overall survival probability \(\delta_q\)
+uniform on \(\F_q^2\).  For sufficiently large \(q\),
+\(m/q\le1/2\) and \(N_q\le2q/(s-1)\).  Using
+\(\log(1-t)\ge-2t\) for \(0\le t\le1/2\), we obtain
+\[
+ \delta_q\ge (1-m/q)^{KN_q}
+ \ge \exp\bigl(-2KN_qm/q\bigr)
+ \ge \exp\bigl(-4K(s-1)\bigr).
+\]
+Fix
+\(\delta=\exp(-4K(s-1))\) and \(\rho=\delta/10\).
+
+For completeness, the exact-line hypothesis also has a uniform
+bound.  A fixed line \(\ell\) of direction \(d\) can occur only
+in candidates of groups with \(d\in S_i\), of which there are at
+most two.  In each such candidate, there are \(s-1\) translated
+lines in this direction.  Each fixed line has probability \(1/q\)
+of becoming \(\ell\) under the uniform shift.  Therefore the sum,
+over all trials, of the probabilities of listing \(\ell\) is at
+most
+\begin{equation}
+\label{con-exact-line-probability}
+ \frac{2K(s-1)}q.
+\end{equation}
+Increasing the fixed \(D\) if necessary verifies every hypothesis
+of Lemma~\ref{lem:linecover}.  Its conclusions give the first two
+properties.
+
+To prove the third, fix \(p\notin\ell\).  For one candidate in
+group \(i\), there are at most \(s-1\) possible points
+\(y\in\ell\) whose join to \(p\) has a direction in \(S_i\):
+each such direction contributes at most one intersection with
+\(\ell\).  A fixed point belongs to a uniform translate of
+\(W_i\) with probability \(s^2/q^2\).  The probability for this
+candidate is thus at most
+\[
+ \frac{(s-1)s^2}{q^2}.
+\]
+If six distinct groups meet the condition, choose one witnessing
+candidate from each.  There are at most \((KN_q)^6=O_{s,K}(q^6)\)
+choices, and their shifts are independent.  For the fixed pair
+\((p,\ell)\), the probability of six such groups is therefore
+\(O_{s,K}(q^{-6})\).  There are at most
+\(q^2q(q+1)=O(q^4)\) pairs \((p,\ell)\).  The union bound gives
+a total failure probability \(O_{s,K}(q^{-2})\), proving the claim.
+\end{proof}
+
+\subsection{Selection and transfer of the deletion estimates}
+
+The estimates just proved concern independent candidates.  We can now
+select compatible copies, using only inclusions to transfer the
+estimates to the selected deletion set.
+By Theorem~\ref{thm:selection}, with probability tending to one the
+same pool admits one candidate per group with no conflicts.
+Intersect this event with those of Lemma~\ref{con-pool-properties}.
+No independence between these events is needed.  For every
+sufficiently large odd integer \(n\), their intersection has positive
+probability, so fix a pool in the intersection and a conflict-free
+selection.  Denote the selected sets and shifts by
+\[
+ P_i=b_i+M_i(\F_s^2),\qquad Y_i=b_i+M_iY.
+\]
+Conflict avoidance gives the useful implication
+\begin{equation}
+\label{con-separation}
+ i\ne j,\quad \operatorname{dir}(\ell)\in D_i,\quad
+ \ell\cap P_i\ne\varnothing
+ \quad\Longrightarrow\quad \ell\cap P_j=\varnothing.
+\end{equation}
+In particular the \(P_i\) are pairwise disjoint: a common point
+would give a zero difference, which belongs to every conflict
+direction, and each \(D_i\) is nonempty.
+
+Let \(B\) be the union, as a set of lines, of the deletion lines
+of the selected candidates, and put
+\begin{equation}
+\label{con-selected-survivors}
+ X=\F_q^2\setminus\bigcup_{\ell\in B}\ell.
+\end{equation}
+Then \(X\supseteq X_{\rm pool}\).  Moreover, if \(\ell\notin B\),
+every selected deletion line belongs to the pool list after all
+exact occurrences of \(\ell\) have been omitted.  Hence \(X\)
+contains the survivor set of that modified pool process. This
+inclusion is deterministic: it holds even if \(\ell\) occurs in an
+unselected candidate and even though the selected shifts depend on the
+whole pool. In particular,
+\begin{equation}
+\label{con-selected-line-bound}
+ \ell\notin B
+ \quad\Longrightarrow\quad |X\cap\ell|\ge\frac{\delta q}{2}.
+\end{equation}
+The third property of Lemma~\ref{con-pool-properties} also remains
+available for the selection, since it was established for all
+candidates in the pool.
+
+This completes the placement: the copies are separated, covers of the
+ordinary survivors by at most \(q\) lines contain an almost pencil,
+every undeleted line contains at least \(\delta q/2\) survivors,
+and the bound of five groups remains valid.  The order of choices is
+\(s,Y\), then the selection constant and \(K\), then the deletion
+constants \(D,\delta,\rho\), and finally \(q=s^n\).
+Only after fixing these constants do we require \(q=s^n\) to be
+sufficiently large.  All further numerical inequalities below involve
+these fixed constants
+and can be imposed by enlarging this final threshold.  Thus the
+construction is available for every sufficiently large odd integer
+\(n\), with the same \(s,Y\); its pools and selected shifts may
+depend on \(n\). The constant used for the ambient deletion lemma
+may depend on \(K\), but it changes no hypothesis of the already
+applied selection theorem. A small value of \(\delta\) only increases
+the final lower threshold on \(q\).
+
+\section{The hypergraph and its covering number}
+\label{sec:hypergraph}
+
+The placement supplies the ordinary points \(X\) and the exceptional
+copies \(Y_i\).  We now turn them into edges.  Ordinary edges retain
+all their affine-line labels.  In an exceptional copy, the directions
+\(S_i\) instead use line labels read from \(f(Y)\); all remaining
+directions retain their ordinary labels.  We will first check that
+every pair of edges still meets, then use the rigidity of \(f(Y)\)
+to rule out covers of size \(q\).
+
+\subsection{Vertices, edges, and intersection}
+
+Set \(r=q+1\).  There is one vertex part \(V_d\) for each
+\(d\in\PP^1(\F_q)\).  For every ambient affine line \(\ell\)
+of direction \(d\), place an ordinary vertex \(o_\ell\) in
+\(V_d\).  If \(d\in S_i\), let \(t_i(d)\in\F_s^*\) be
+the unique subfield direction whose image under \(M_i\) is
+\(d\).  For each of the \(s\) subfield affine lines \(L\) of
+direction \(t_i(d)\), place a special vertex \(v_{i,d,L}\) in
+\(V_d\).  These subfield lines are read in the new coordinates
+used for \(f(Y)\).  All vertices are formal labels, tagged to be
+distinct between parts, between ordinary and special types, and
+between different groups.  In particular each \(V_d\) is finite,
+and the parts are pairwise disjoint.
+
+Write \(\ell_d(z)\) for the ambient line of direction \(d\)
+through \(z\), and \(L_t(w)\) for the subfield line of direction
+\(t\) through \(w\in\F_s^2\).  Each \(x\in X\) defines the
+ordinary edge
+\[
+ e_x=\{o_{\ell_d(x)}:d\in\PP^1(\F_q)\}.
+\]
+For each \(i\in\mathcal I\) and \(w\in Y\), put
+\(y_{i,w}=b_i+M_iw\) and define an exceptional edge by selecting,
+in part \(V_d\), the vertex
+\begin{equation}
+\label{con-exceptional-vertex}
+ \begin{cases}
+ v_{i,d,L_{t_i(d)}(f(w))},& d\in S_i,\\
+ o_{\ell_d(y_{i,w})},&d\notin S_i.
+ \end{cases}
+\end{equation}
+Denote this edge by \(e_{i,w}\), and let
+\[
+ H=\{e_x:x\in X\}\cup
+   \{e_{i,w}:i\in\mathcal I,\ w\in Y\}.
+\]
+Thus an exceptional edge changes labels in the assigned directions,
+without introducing any new part.  Every edge contains exactly one
+vertex from every part, so \(H\)
+is \((q+1)\)-partite and \((q+1)\)-uniform.  It is nonempty,
+because \(Y\ne\varnothing\).
+
+Figure~\ref{fig:two-coordinates} shows how to read an exceptional
+edge from its parameter \(w\).  The deletion lines were defined from
+its ordinary base point \(y_{i,w}\), while its special labels use
+\(f(w)\).  The two axis parallel classes are preserved by \(f\),
+which joins these descriptions in the intersection proof.
+
+\begin{figure}[htbp]
+\centering
+\input{figures/two-coordinates}
+\caption{The two label rules for the vertex of \(e_{i,w}\) in part
+\(V_d\).  The left branch uses the embedded point \(b_i+M_iw\);
+the right branch uses \(f(w)=(w_1^3,w_2)\) in the small plane.
+For each \(d\in A_i\), two parameters give the same ordinary label
+exactly when their images under \(f\) lie on the same corresponding
+horizontal or vertical line.
+The diagram records maps and label choices, rather than geometric
+positions in the finite plane.}
+\label{fig:two-coordinates}
+\end{figure}
+
+\begin{proposition}
+\label{con-intersection}
+The hypergraph \(H\) is intersecting.  Furthermore, all the edges
+in its displayed parametrization are distinct.
+\end{proposition}
+
+\begin{proof}
+There are four types of pairs, according to whether their edges are
+ordinary or exceptional and, in the latter case, whether their group
+indices agree.
+
+Two ordinary edges with distinct base points share the ordinary
+label of their join.  Next, an exceptional base point \(y\in Y_i\)
+lies on a deletion line in every direction of the nonempty set
+\(S_i\), so \(y\notin X\).  Its join with \(x\in X\) cannot
+have direction in \(S_i\): otherwise that join would be a selected
+deletion line containing \(x\).  Its ordinary label therefore
+belongs to both the exceptional edge and \(e_x\).
+
+For exceptional edges from distinct groups \(i,j\), the base
+points are distinct by \eqref{con-separation}.  That same separation
+condition ensures that their join has direction outside both
+\(D_i\) and \(D_j\), and hence outside \(S_i\cup S_j\).
+The two edges share its ordinary label.
+
+Finally, consider \(e_{i,w}\) and \(e_{i,w'}\) with \(w\ne w'\).
+The transformed points \(f(w),f(w')\) are distinct.  If their
+join has finite nonzero slope \(t\), the exceptional edges share
+the special label of this subfield line in the part of direction
+\(M_i(t)\).  If their join is horizontal or vertical, the relevant
+original coordinates are equal as well, because \(f\) preserves
+these two parallel classes bijectively.  The original base points
+then share an ordinary axis-direction label, in a direction of
+\(A_i\).  This completes the intersection proof.
+
+We also check that the parameters count distinct edges, as required
+when we count exceptional edges below.  Two ordinary labels in
+nonparallel directions determine the ordinary base point, so ordinary
+edges are distinct.
+An exceptional edge has special labels and cannot equal an ordinary
+edge.  Within one group, the special labels in any two distinct
+directions of \(S_i\) determine \(f(w)\), since the corresponding
+subfield lines are nonparallel.  As \(|S_i|=s-1\ge2\), equal
+exceptional edges within a group would imply \(w=w'\).  Finally,
+any special label tagged by group \(i\) is absent from every edge
+of a different group.  Exceptional edges in distinct groups are
+therefore distinct too.
+\end{proof}
+
+\Needspace{14\baselineskip}
+\subsection{Covering vertices in the small-plane coordinates}
+
+We identify the positions of the exceptional edges of group \(i\)
+with \(f(Y)\), via \(e_{i,w}\leftrightarrow f(w)\).  The action
+of a covering vertex means the subset of these positions whose edges
+contain that vertex.  The next lemma translates every such action
+into one of the two resources in Lemma~\ref{lem:gadget}: a subfield
+line or an individual point.
+
+\begin{lemma}
+\label{con-label-action}
+Fix a group \(i\).
+\begin{enumerate}
+\item A special vertex of group \(i\) covers the positions on one
+subfield line.  A special vertex of another group covers none.
+\item An ordinary vertex of direction in \(S_i\) covers none.
+An ordinary vertex of direction in \(A_i\), if it covers any
+positions, covers them on a single horizontal or vertical line in
+the \(f\) coordinates.
+\item An ordinary vertex of direction outside \(S_i\cup A_i\)
+covers at most one position.
+\end{enumerate}
+In particular, the positions covered by any one vertex are contained
+in a subfield affine line.
+\end{lemma}
+
+\begin{proof}
+The assertions for special vertices and altered ordinary directions
+follow directly from \eqref{con-exceptional-vertex}.  For an axis
+direction, undoing \(b_i,M_i\) fixes one original coordinate, so
+applying \(f\) gives the corresponding horizontal or vertical
+line.  A line in a direction outside \(S_i\cup A_i\) cannot
+contain two points of \(P_i\), since the join directions of that
+embedded subplane are exactly \(S_i\cup A_i\).  It therefore
+covers at most one exceptional position.  A singleton can be
+included in an arbitrary subfield line through it, and an empty
+set requires no covering resource.
+\end{proof}
+
+\subsection{The final cover argument}
+
+\begin{proof}[Completion of the proof of Theorem~\ref{thm:main}]
+We prove that \(H\) has no vertex cover of size at most \(q\).
+Suppose to the contrary that \(C\) is such a cover.  Its ordinary
+vertices, regarded as affine lines, must cover \(X\): special
+vertices belong to no ordinary edge.  These lines also cover
+\(X_{\rm pool}\).  Lemma~\ref{con-pool-properties} therefore
+gives a projective point \(p\) such that at least
+\((1-\rho)q\) ordinary vertices of \(C\) are lines through
+\(p\).  The line-survival bound will force every undeleted member
+of this pencil into \(C\).  We then count the vertices left for
+exceptional edges, separating centers at infinity, centers inside one
+of the sets \(P_i\), and finite centers outside them all.
+
+\paragraph{Every undeleted line in this pencil is forced.}
+We claim that
+\begin{equation}
+\label{con-forced-pencil}
+ p\in\ell,\quad\ell\notin B
+ \quad\Longrightarrow\quad o_\ell\in C,
+\end{equation}
+where incidence with an infinite \(p\) means that \(\ell\) has
+the corresponding direction.  If \(o_\ell\) were missing, all
+other lines of \(C\) through \(p\) would cover at most one
+affine point of \(\ell\).  There are at most \(\rho q\)
+vertices outside the large pencil subcollection.  Each of their
+ordinary lines covers at most one point of \(\ell\), and their
+special vertices cover no ordinary edge.  Consequently
+\[
+ |X\cap\ell|\le1+\rho q.
+\]
+This contradicts \eqref{con-selected-line-bound}, since
+\(\delta q/2>1+\rho q\) for sufficiently large \(q\) and
+\(\rho=\delta/10\).  At infinity the pencil lines are parallel,
+so the possible contribution of one central point is absent; the
+same upper bound still applies.
+
+Let \(F_p\) be the set of ordinary vertices forced by
+\eqref{con-forced-pencil}, and call the vertices of \(C\setminus F_p\)
+the additional vertices.  The deleted pencil lines determine how
+many additional vertices the assumed budget \(|C|\le q\) permits.
+
+\paragraph{An infinite center.}
+Suppose \(p\) is the point at infinity in direction \(d\).  Put
+\[
+ I_d=\{i\in\mathcal I:d\in S_i\},
+ \qquad z=|I_d|\in\{1,2\}.
+\]
+Only these groups contribute deletion lines in direction \(d\),
+and each contributes \(s-1\), by
+Lemma~\ref{con-deletion-count}.  At most \(z(s-1)\) of the
+\(q\) lines in this parallel class belong to \(B\).  Thus
+\[
+ |C\setminus F_p|\le z(s-1).
+\]
+None of the forced vertices covers any exceptional edge of a group
+in \(I_d\), because those groups use special vertices in
+direction \(d\).
+
+If \(z=1\), all positions of this group's \(f(Y)\) must therefore
+be covered by at most \(s-1\) additional vertices.  By
+Lemma~\ref{con-label-action}, their effects can be included in at
+most \(s-1\) subfield lines.  This contradicts
+Lemma~\ref{lem:gadget}.
+
+If \(z=2\), the two groups are \(1,*\), by
+\eqref{con-double-directions}.  An additional vertex can now meet
+both groups, so we use the weighted line-and-point budget instead of
+counting one line per vertex.  For each of these groups let \(a_i\)
+count the additional vertices that cover at least one of its
+positions and are either special vertices of this group or ordinary
+vertices in its axis directions \(A_i\).  Let \(b_i'\) count the
+other additional vertices covering a position of the group.  By
+Lemma~\ref{con-label-action}, the latter vertices each cover only
+a singleton.  The former supply subfield lines.  The weighted
+conclusion of Lemma~\ref{lem:gadget} gives
+\begin{equation}
+\label{con-two-group-lower}
+ a_i+\frac{b_i'}2\ge s
+ \qquad (i=1,*).
+\end{equation}
+
+Each additional vertex contributes at most one to the sum of the
+two left sides.  A special vertex is confined to its tagged group.
+An ordinary axis-direction vertex counted in \(a_i\) is a line
+meeting \(P_i\) with direction in \(A_i\subseteq D_i\); by
+\eqref{con-separation} it cannot meet the other group's \(P_j\).
+It therefore contributes nothing to that other group, even as a
+singleton.  Any vertex of neither exclusive type contributes at
+most \(1/2\) to each group.  This accounts for every type of
+additional vertex, including those covering neither group.
+Summing \eqref{con-two-group-lower}, we obtain
+\[
+ 2s\le a_1+\frac{b_1'}2+a_*+\frac{b_*'}2
+ \le |C\setminus F_p|\le2(s-1),
+\]
+a contradiction.
+
+\paragraph{A finite center inside an embedded configuration.}
+Now let \(p\) be finite, and write
+\[
+ h_p=|\{\ell\in B:p\in\ell\}|.
+\]
+There are \(q+1\) affine lines through \(p\), so
+\begin{equation}
+\label{con-finite-budget}
+ |F_p|=q+1-h_p,
+ \qquad |C\setminus F_p|\le h_p-1.
+\end{equation}
+The loss of one in the second inequality comes from the \(q+1\)
+members of a finite pencil and the assumed cover size \(q\).
+If \(h_p=0\), the forced vertices already exceed the permitted
+size of \(C\).  We may therefore assume \(h_p\ge1\) in both
+finite-center cases.
+
+Suppose \(p\in P_i\), for the necessarily unique such group.
+Every deletion line through \(p\) comes from this group: a line
+from another group would contradict \eqref{con-separation}, since
+its direction belongs to that other group's conflict set.  There
+is at most one such line in each direction of \(S_i\), and hence
+\begin{equation}
+\label{con-internal-pencil-size}
+ h_p\le s-1.
+\end{equation}
+Let \(w_p=M_i^{-1}(p-b_i)\in\F_s^2\), whether or not
+\(w_p\in Y\).  We account for the exceptional positions of this
+group covered by \(F_p\): they will all lie on two nonparallel
+lines in the new coordinates.  For a position with base point
+\(y\ne p\), its join with \(p\) has a direction in
+\(S_i\cup A_i\).  A direction in \(S_i\) cannot cover this
+edge by an ordinary vertex.  Any actual forced coverage must
+therefore lie along one of the two axis-direction lines through
+\(p\).  A position based at \(p\) itself, if present, lies on
+both these axes as well.
+
+Since \(f\) preserves the two coordinate parallel classes, all
+these covered positions are contained, in the new coordinates, in
+the horizontal and vertical lines through \(f(w_p)\).  The
+remaining positions are covered by the additional vertices; each
+can be included in one subfield line by
+Lemma~\ref{con-label-action}.  Equations
+\eqref{con-finite-budget} and \eqref{con-internal-pencil-size}
+would thus give a cover of \(f(Y)\) by at most
+\[
+ 2+(h_p-1)\le s
+\]
+lines containing the two nonparallel coordinate lines. These
+are legitimate covering resources whether or not their ordinary labels
+belong to \(F_p\): their union contains every position covered by
+\(F_p\). Removing any duplicate replacement lines preserves these
+two distinct lines and can only decrease the count.  This contradicts
+Lemma~\ref{lem:gadget}.
+
+\paragraph{A finite center outside all embedded configurations.}
+It remains to consider \(p\notin\bigcup_{i\in\mathcal I}P_i\).
+Here we use the five points on each deletion line, rather than the
+small-plane cover obstruction.  We will find \(5h_p\) exceptional
+edges missed by \(F_p\), while each additional vertex covers at
+most five of them.
+
+Each group contributes at most one deletion line through \(p\).
+Indeed, two distinct such lines cannot be parallel.  After undoing
+\(b_i,M_i\), their subfield affine equations have distinct
+subfield directions.  Solving these two equations uses a
+nonsingular linear system over \(\F_s\), so their intersection
+lies in \(\F_s^2\).  This would place \(p\) in \(P_i\), a
+contradiction.  Also, deletion lines from different groups cannot
+coincide, by \eqref{con-separation}.
+
+The \(h_p\) deletion lines through \(p\) therefore come from
+\(h_p\) distinct groups.  From each such group \(i\), choose five
+distinct points of \(Y_i\) on its contributing line, possible by
+Lemma~\ref{con-deletion-count}.  These give \(5h_p\) distinct
+exceptional edges, by Proposition~\ref{con-intersection}.  None
+is covered by a forced ordinary vertex: its unique join line to
+\(p\) is precisely its contributing deletion line, whose
+direction is altered for that group.
+
+Any one additional vertex covers at most five of these selected
+edges.  For a special vertex this follows because it belongs to
+only one group.  An ordinary line through \(p\) covers none,
+by the same unique-join and altered-direction argument.  Finally,
+let an ordinary vertex \(o_\ell\), with \(\ell\) not through
+\(p\), cover some selected edges.  The line \(\ell\) meets each
+contributing line in at most one point, so \(o_\ell\) covers at most
+one selected edge from each contributing group.  Every such group
+has a selected point \(y\in Y_i\cap\ell\) whose join to \(p\)
+has direction in \(S_i\).  The third part of
+Lemma~\ref{con-pool-properties}, which applies to all candidates
+and hence to the selection, bounds the number of these groups by
+five.  This proves the asserted bound for every additional vertex.
+
+By \eqref{con-finite-budget}, at most \(h_p-1\) additional
+vertices are available.  They can cover at most \(5(h_p-1)\) of
+the \(5h_p\) selected edges, a final contradiction.
+
+We have ruled out every possible projective center \(p\), and
+hence every vertex cover of size at most \(q\).  Therefore
+\(\tau(H)\ge q+1\).  Conversely, any edge of the nonempty
+intersecting hypergraph \(H\) is itself a vertex cover and has
+size \(q+1\).  It follows that
+\[
+ \tau(H)=q+1.
+\]
+The parameter choices in Section~\ref{sec:placement} give this
+hypergraph for every sufficiently large odd integer \(n\), with
+\(q=s^n\), after fixing any sufficiently large prime
+\(s\equiv2\pmod3\) and a successful \(Y\). Choose one such \(Y\)
+for each eligible \(s\), and let \(n_0(s)\ge3\) exceed all the
+resulting degree thresholds. This proves Theorem~\ref{thm:main}.
+Since there are arbitrarily large odd integers, it also gives infinitely
+many ranks \(r=q+1\).
+\end{proof}

preprints/A-Counterexample-to-Rysers-Covering-Conjecture-September-23-2026/build/figures/two-coordinates.tex

@@ -0,0 +1,19 @@
+\begin{tikzpicture}[
+  >=Latex,
+  point/.style={draw,rounded corners=2pt,align=center,inner sep=7pt},
+  labelbox/.style={draw,align=center,text width=5.1cm,inner sep=7pt},
+  every node/.style={font=\small}]
+\node[point] (w) at (0,0) {$w=(w_1,w_2)\in Y$};
+\node[point] (old) at (-3.5,-1.55)
+  {Ambient coordinates\\[3pt]$y_{i,w}=b_i+M_iw\in Y_i$};
+\node[point] (new) at (3.5,-1.55)
+  {Small-plane coordinates\\[3pt]$f(w)=(w_1^3,w_2)\in f(Y)$};
+\node[labelbox] (ordinary) at (-3.5,-3.65)
+  {Ordinary vertex in $V_d$\\[3pt]$o_{\ell_d(y_{i,w})}$};
+\node[labelbox] (special) at (3.5,-3.65)
+  {Special vertex in $V_d$\\[3pt]$v_{i,d,L_{t_i(d)}(f(w))}$};
+\draw[->] (w) -- node[above left,inner sep=2pt,xshift=-4pt,yshift=-2pt] {embed} (old);
+\draw[->] (w) -- node[above right,inner sep=2pt,xshift=4pt,yshift=-2pt] {apply $f$} (new);
+\draw[->] (old) -- node[left] {$d\notin S_i$} (ordinary);
+\draw[->] (new) -- node[right] {$d\in S_i$} (special);
+\end{tikzpicture}

preprints/A-Counterexample-to-Rysers-Covering-Conjecture-September-23-2026/build/incidence.tex

@@ -0,0 +1,707 @@
+\section{A uniform incidence bound}\label{sec:incidence}
+
+For a set $P$ of points and a set $\mathcal L$ of lines in
+$\PP^2(\F_q)$, write
+\begin{equation}\label{eq:inc-definition}
+ I(P,\mathcal L)
+ =\bigl|\{(p,\ell)\in P\times\mathcal L:p\in\ell\}\bigr|.
+\end{equation}
+Our aim is a power saving with constants independent of the field and
+its characteristic. In Section~\ref{sec:linecover}, this estimate will
+control intersections of a proposed line cover when no projective point
+lies on many of its lines. This is the incidence input to the conclusion
+that a small cover must contain almost a full pencil.
+
+\begin{theorem}[Uniform incidence saving]\label{thm:incidence}
+There exist absolute constants $\kappa\in(0,1/10)$ and $Q$ with the following
+property. Suppose that $q\ge Q$ is a prime power and every proper subfield
+of $\F_q$ has order at most $q^{1/3}$. If $P$ and $\mathcal L$ are sets of
+projective points and projective lines over $\F_q$ satisfying
+\begin{equation}\label{eq:inc-theorem-sizes}
+ |P|,|\mathcal L|\le q^{1+\kappa},
+\end{equation}
+then
+\begin{equation}\label{eq:inc-theorem-bound}
+ I(P,\mathcal L)\le q^{3/2-\kappa}.
+\end{equation}
+\end{theorem}
+
+The proof follows the incidence and sum--product strategy of
+Bourgain, Katz, and Tao~\cite[Sections 3--4 and 6]{BKT2004}. Its geometric
+step turns an incidence configuration with too many incidences into a
+dense set of pairs whose differences and quotients range over small
+sets. A graph argument then extracts one subset with both a small
+difference set and a small quotient set. Finally, scalar coverings show
+that such a subset cannot have size near the square root of the field
+order under the proper-subfield hypothesis.
+
+We develop the two algebraic ingredients first, then return to the
+incidence configuration. The proofs include the simultaneous extraction
+of additive and multiplicative control and the bounds on scalar
+expressions that make the argument uniform over fields of nonprime
+order.
+
+\subsection{Finite-set estimates}
+
+The first task is to pass from control of differences and quotients
+along the edges of a dense bipartite graph to control on all pairs in
+one subset. We begin with the sumset estimates needed to use the
+resulting additive control.
+
+For subsets of an abelian group, $X+Y$ and $X-Y$ denote the usual sumset
+and difference set. For a nonnegative integer $j$, the notation $jY$
+in this subsection means a sum of $j$ copies of $Y$, with $0Y=\{0\}$.
+Later, a field element multiplying a set will always mean field scaling;
+we will specify explicitly when an integer denotes an iterated sumset.
+
+The first two estimates below are Ruzsa's triangle inequality and
+covering lemma; the maximal-translates proof of the latter appears in
+\cite[p.~324]{Ruzsa1999}. The multiple-sum bound is a Pl\"unnecke--Ruzsa
+estimate, for which we use Petridis's minimizing-subset
+proof~\cite[Proposition 2.1 and Section 3]{Petridis2012}.
+
+\begin{lemma}[Triangle, covering, and multiple-sum estimates]
+\label{inc-sumset-estimates}
+Let the sets below be finite subsets of an abelian group.
+\begin{enumerate}
+\item If $Z$ is nonempty, then
+\begin{equation}\label{eq:inc-triangle}
+ |X-Y|\,|Z|\le |X-Z|\,|Z-Y|.
+\end{equation}
+\item If $Y$ is nonempty and $|X-Y|\le M|Y|$, then $X$ is contained in
+the union of at most $M$ translates of $Y-Y$.
+\item If $Y$ is nonempty and $|Y-Y|\le M|Y|$, then, for all nonnegative
+integers $u,v$,
+\begin{equation}\label{eq:inc-multiple-sums}
+ |uY-vY|\le M^{u+v}|Y|.
+\end{equation}
+\end{enumerate}
+\end{lemma}
+
+\begin{proof}
+For the triangle inequality, choose one representation $d=x_d-y_d$
+for each $d\in X-Y$. The map
+\[
+ (d,z)\longmapsto (x_d-z,z-y_d)
+\]
+from $(X-Y)\times Z$ into $(X-Z)\times(Z-Y)$ is injective: summing the
+two coordinates recovers $d$, after which the chosen representation
+recovers $z$. This proves~\eqref{eq:inc-triangle}.
+
+For the covering statement, take a maximal set $T\subseteq X$ for
+which the translates $t-Y$, $t\in T$, are pairwise disjoint. Their union
+is contained in $X-Y$, so $|T|\le M$. Maximality implies that for every
+$x\in X$ some $x-Y$ intersects $t-Y$ with $t\in T$, whence
+$x\in t+(Y-Y)$.
+
+For the multiple-sum bound, the minimizing-subset argument of
+Petridis~\cite{Petridis2012} lets us iterate a one-step growth estimate.
+Put $Q_0=-Y$ and choose a nonempty
+$W\subseteq Y$ minimizing
+\[
+ \mu=\frac{|W+Q_0|}{|W|}.
+\]
+Then $\mu\le M$, and every $W'\subseteq W$, including the empty set,
+satisfies $|W'+Q_0|\ge\mu|W'|$. We claim that for every finite set $E_0$,
+\begin{equation}\label{eq:inc-petridis-growth}
+ |W+Q_0+E_0|\le\mu|W+E_0|.
+\end{equation}
+The assertion is immediate if $E_0$ is empty. Otherwise order its
+elements as $e_1,\ldots,e_k$. When the translate $W+e_i$ is added to
+the preceding translates, let
+\[
+ W_i'=\{w\in W:w+e_i\in\bigcup_{j<i}(W+e_j)\}.
+\]
+It contributes exactly $|W|-|W_i'|$ new elements. Moreover,
+$W_i'+Q_0+e_i$ is contained in the union of the preceding enlarged
+translates $W+Q_0+e_j$. Thus the contribution of $W+Q_0+e_i$ is at most
+\[
+ |W+Q_0|-|W_i'+Q_0|
+ \le\mu\bigl(|W|-|W_i'|\bigr).
+\]
+Summing proves~\eqref{eq:inc-petridis-growth}. Iterating this estimate
+gives $|W+jQ_0|\le\mu^j|W|$ for every nonnegative integer $j$.
+We now remove the auxiliary set $W$ by applying
+\eqref{eq:inc-triangle} to $uQ_0,vQ_0$ with intermediate set $-W$:
+\[
+ |uQ_0-vQ_0|\,|W|
+ \le |W+uQ_0|\,|W+vQ_0|
+ \le\mu^{u+v}|W|^2.
+\]
+Negation preserves cardinality, so $|uQ_0-vQ_0|=|uY-vY|$. Since
+$|W|\le|Y|$ and $\mu\le M$, this is~\eqref{eq:inc-multiple-sums}.
+\end{proof}
+
+We next carry out the extraction from a dense graph. The argument uses
+the common-neighborhood method in Gowers's quantitative refinement
+\cite[Section 4, Lemma 11 and Proposition 12]{Gowers1998} of the
+Balog--Szemer\'edi theorem~\cite{BalogSzemeredi1994}; see also
+\cite[Section 5, Lemmas 5.1--5.2]{FoxSudakov2011}. The point is to use the same walks
+for both additive and multiplicative edge labels. We can therefore
+obtain both bounds on one subset, which is what the subsequent
+subfield argument requires. In the geometric application, the edge
+labels will come from the four pencils through four chosen points.
+
+\begin{lemma}[Simultaneous difference and quotient extraction]
+\label{inc-simultaneous-extraction}
+Let $F$ be a field, let $A,B\subseteq F^*$ be finite, and let
+$G\subseteq A\times B$. Define
+\[
+ D=\{a-b:(a,b)\in G\},\qquad
+ J=\{a/b:(a,b)\in G\}.
+\]
+Suppose that, for real numbers $n>0$ and $V\ge2$,
+\begin{equation}\label{eq:inc-graph-hypotheses}
+ |A|,|B|,|D|,|J|\le Vn,
+ \qquad |G|\ge n^2/V.
+\end{equation}
+Then there is $S\subseteq A$ such that
+\begin{equation}\label{eq:inc-extracted-size}
+ \frac{3n}{8V^2}\le |S|\le Vn
+\end{equation}
+and
+\begin{equation}\label{eq:inc-extracted-expansion}
+ |S-S|,|S/S|\le 2^{20}V^{22}n.
+\end{equation}
+Here $S/S=\{a/a':a,a'\in S\}$.
+\end{lemma}
+
+\begin{proof}
+We first find a large set of vertices such that every ordered pair
+is joined by many walks of length four. Counting the possible edge
+labels of these walks will then bound both differences and quotients
+of the endpoints.
+
+Regard $G$ as a bipartite graph, and put
+$\tau=(512V^8)^{-1}$. Call an ordered pair $(a,a_2)\in A^2$ bad if
+it has fewer than $\tau n$ common neighbors in $B$. To find a
+neighborhood with few bad pairs, first note that vertices of $B$
+with fewer than $n/(2V^2)$ neighbors account for at most
+$|B|n/(2V^2)\le n^2/(2V)$ edges. Since every vertex of $B$ has at
+most $Vn$ neighbors, the set of vertices with at least $n/(2V^2)$
+neighbors has size at least $n/(2V^2)$.
+
+Summed over all $b\in B$, the number of bad ordered pairs in the
+neighborhood of $b$ is at most
+\[
+ |A|^2\tau n\le V^2\tau n^3.
+\]
+Averaging over the large neighborhoods just identified, one obtains
+a neighborhood $A_0$ of size at least $n/(2V^2)$ in which the number
+of bad ordered pairs is at most
+\[
+ 2V^4\tau n^2=\frac{n^2}{256V^4}
+ \le \frac{|A_0|^2}{64}
+ \le \frac{|A_0|^2}{16}.
+\]
+Let $S$ consist of those $a\in A_0$ having at most $|A_0|/4$ bad
+partners in $A_0$. At most $|A_0|/4$ elements are discarded, so
+$|S|\ge3|A_0|/4$, proving~\eqref{eq:inc-extracted-size}.
+
+The choice of $S$ ensures that, for every $a,a'\in S$, at least
+$|A_0|/2$ elements $a_2\in A_0$ form nonbad pairs with both endpoints.
+Each such $a_2$ gives at least $(\tau n)^2$
+ordered walks
+\[
+ a,b_1,a_2,b_2,a'
+\]
+whose four edges belong to $G$. Repeated vertices are allowed.
+Consequently each ordered endpoint pair in $S^2$ has at least
+\begin{equation}\label{eq:inc-walk-count}
+ \frac{|A_0|}{2}(\tau n)^2
+ \ge\frac{n^3}{2^{20}V^{18}}
+\end{equation}
+such walks. This lower bound now converts the limited number of edge
+labels into a bound on endpoint differences.
+
+For one of these walks, write its edge differences, always taking
+the $A$ coordinate minus the $B$ coordinate, as
+\[
+ d_1=a-b_1,\quad d_2=a_2-b_1,\quad
+ d_3=a_2-b_2,\quad d_4=a'-b_2.
+\]
+They satisfy $a-a'=d_1-d_2+d_3-d_4$. For a fixed endpoint pair,
+the four differences determine the walk uniquely: starting from $a$,
+one recovers successively $b_1,a_2,b_2,a'$. Choose one ordered
+endpoint pair for each element of $S-S$. The resulting collections
+of difference quadruples are disjoint for different differences,
+because their alternating sums differ. Counting in $D^4$ and using
+\eqref{eq:inc-walk-count} gives
+\[
+ |S-S|\le
+ \frac{|D|^4}{n^3/(2^{20}V^{18})}
+ \le 2^{20}V^{22}n.
+\]
+
+The same walks give the quotient bound on the same set $S$. Put
+\[
+ j_1=a/b_1,\quad j_2=a_2/b_1,\quad
+ j_3=a_2/b_2,\quad j_4=a'/b_2.
+\]
+All are nonzero, and $a/a'=j_1j_3/(j_2j_4)$. From a fixed initial
+$a$ and these four quotients one recovers
+$b_1=a/j_1$, $a_2=j_2b_1$, $b_2=a_2/j_3$, and $a'=j_4b_2$.
+Choosing one endpoint pair for each element of $S/S$ and counting
+in $J^4$ therefore proves the other bound in
+\eqref{eq:inc-extracted-expansion}.
+\end{proof}
+
+\subsection{Bounded scalar expressions and subfields}
+
+We now turn to the algebraic obstruction: a set near the square-root
+field scale cannot have both the small difference set and the small
+quotient set supplied by the extraction lemma. The first step is to
+bound sums involving scalars formed using a bounded number of field
+operations. We use the scalar-covering method of Bourgain, Katz, and
+Tao~\cite[Lemma 3.2 and Proposition 3.3]{BKT2004}, keeping the dependence
+on the number of operations explicit.
+
+An expression in a set $E\subseteq F$ is built from members of
+$E\cup\{0,1\}$ using binary addition, subtraction, and multiplication,
+and unary negation and inversion. Each of these operations counts
+as one operation. Inversion is allowed only when its argument is
+nonzero.
+
+\begin{lemma}[Uniform scalar coverings]\label{inc-scalar-coverings}
+Let $B$ be a nonempty finite subset of a field $F$, let $E\subseteq B$,
+and let $K\ge2$. Suppose that
+\begin{equation}\label{eq:inc-scalar-hypotheses}
+ |B-B|\le K|B|,
+ \qquad |B-eB|\le 2K^3|B|\quad(e\in E).
+\end{equation}
+Define
+\begin{equation}\label{eq:inc-scalar-exponents}
+ C_0=4,\qquad C_{m+1}=3C_m+5,
+ \qquad\text{so that }C_m=\frac{13\cdot3^m-5}{2}.
+\end{equation}
+If $\lambda\in F$ has an expression in $E$ using at most $m$
+operations, then $\lambda B$ is contained in at most $K^{C_m}$
+translates of $B-B$. In particular,
+\begin{equation}\label{eq:inc-scalar-sum-bound}
+ |E+\lambda E|\le |B+\lambda B|
+ \le K^{C_m+3}|B|.
+\end{equation}
+These bounds are uniform over all such expressions and all choices
+of their entries from $E$.
+\end{lemma}
+
+\begin{proof}
+Put $H=B-B$, so $H=-H$ and $0\in H$. In the remainder of this proof,
+integer multiples of $B$ or $H$ inside sumsets denote iterated sums.
+We prove the covering assertion by induction on expression length.
+To combine coverings at an induction step, we will need a fixed
+covering of $2H$ by translates of $H$. By
+Lemma~\ref{inc-sumset-estimates},
+\[
+ |2H-B|=|2B-3B|\le K^5|B|.
+\]
+The covering statement of that lemma consequently supplies a set
+$T_*\subseteq F$ with $|T_*|\le K^5$ such that
+\begin{equation}\label{eq:inc-two-differences-cover}
+ 2H\subseteq T_*+H.
+\end{equation}
+
+Every $e\in E$ has a covering $eB\subseteq T_e+H$ with
+$|T_e|\le2K^3\le K^4$, by
+\eqref{eq:inc-scalar-hypotheses} and the covering lemma.
+The scalars $0$ and $1$ also have coverings by one translate of $H$:
+use $\{0\}\subseteq H$ and $B\subseteq b_0+H$ for any $b_0\in B$.
+This proves the case $m=0$.
+
+We now check how the number of translates changes under each allowed
+operation. Suppose that
+\[
+ \lambda B\subseteq T+H,\qquad
+ \mu B\subseteq U+H,\qquad
+ |T|\le K^a,\quad |U|\le K^b.
+\]
+Negation replaces $T$ by $-T$. For addition and subtraction,
+\[
+ (\lambda\pm\mu)B
+ \subseteq \lambda B\pm\mu B
+ \subseteq T\pm U+2H
+ \subseteq T\pm U+T_*+H,
+\]
+so the number of translates is at most $K^{a+b+5}$. For a product,
+\[
+ \lambda\mu B
+ \subseteq \lambda U+\lambda H
+ =\lambda U+(\lambda B-\lambda B)
+ \subseteq\lambda U+T-T+T_*+H.
+\]
+This uses at most $K^{b+2a+5}$ translates and is valid even if
+$\lambda=0$. Finally, if $\lambda\ne0$, then
+\[
+ |\lambda B-B|
+ \le |T|\,|H-B|
+ =|T|\,|B-2B|
+ \le K^{a+3}|B|.
+\]
+Scaling by $\lambda^{-1}$ and then negating gives
+$|\lambda^{-1}B-B|\le K^{a+3}|B|$. The covering lemma now covers
+$\lambda^{-1}B$ by at most $K^{a+3}$ translates of $H$.
+
+These bounds give the claimed uniform induction: in an expression
+with at most $m+1$ operations, each immediate subexpression has at
+most $m$ operations, so every resulting covering uses at most
+$K^{3C_m+5}$ translates. This proves the covering assertion
+with~\eqref{eq:inc-scalar-exponents}.
+
+To deduce the sumset bound from the covering, let
+$\lambda B\subseteq T+H$ be the covering just obtained. Then
+\[
+ |B+\lambda B|
+ \le |T|\,|B+H|
+ =|T|\,|2B-B|
+ \le K^{C_m+3}|B|,
+\]
+again by Lemma~\ref{inc-sumset-estimates}. Since $E\subseteq B$,
+this also bounds $|E+\lambda E|$.
+\end{proof}
+
+When working along a sequence $q\to\infty$, a positive quantity
+$K=q^{o(1)}$ means that $\log K/\log q\to0$. Likewise,
+$|S|=q^{1/2+o(1)}$ means that $\log|S|/\log q\to1/2$.
+In particular, every fixed power of a subpower factor is still a
+subpower factor. This is why a uniform bound for expressions of
+fixed length suffices for the next lemma.
+
+\begin{lemma}[Exclusion of an intermediate approximate subfield]
+\label{inc-no-approximate-subfield}
+There is no sequence of prime powers $q\to\infty$, each satisfying
+the proper-subfield hypothesis of Theorem~\ref{thm:incidence}, together
+with sets $S\subseteq\F_q^*$ and numbers $K\ge2$ such that
+\begin{equation}\label{eq:inc-approximate-subfield-hypotheses}
+ |S|=q^{1/2+o(1)},\qquad K=q^{o(1)},\qquad
+ |S-S|,|S/S|\le K|S|.
+\end{equation}
+\end{lemma}
+
+\begin{proof}
+Suppose that such a sequence exists. We first use multiplicative
+overlaps to obtain sets satisfying the scalar-covering hypotheses.
+Ratios of differences in a large subset will then form a subfield.
+The proper-subfield hypothesis will force that subfield to be the
+whole field, where averaging will contradict the scalar-covering
+bound.
+
+\paragraph{Extracting scalar-covering hypotheses.}
+Put $s=|S|$. The $s$ sets
+$S/a$, $a\in S$, each have size $s$, and their union is contained
+in $S/S$. Cauchy--Schwarz gives
+\begin{equation}\label{eq:inc-multiplicative-overlaps}
+ \sum_{a,b\in S}|S/a\cap S/b|
+ \ge\frac{s^4}{|S/S|}\ge\frac{s^3}{K}.
+\end{equation}
+Thus some $b\in S$ satisfies
+$\sum_{a\in S}|S/a\cap S/b|\ge s^2/K$.
+Since every summand is at most $s$, at least $s/(2K)$ values
+$a\in S$ have $|S/a\cap S/b|\ge s/(2K)$: otherwise the sum of
+the large summands would be less than $s^2/(2K)$ and that of all
+the other summands at most $s^2/(2K)$.
+
+Let $E$ be the set of $e=a/b$ for these values of $a$, and put
+$B=S/b$. Then
+\begin{equation}\label{eq:inc-overlap-subset}
+ E\subseteq B,\qquad |B|=s,\qquad
+ |E|\ge\frac{s}{2K},\qquad
+ |S\cap eS|\ge\frac{s}{2K}\quad(e\in E).
+\end{equation}
+The last inequality follows by multiplying $S/a\cap S/b$ by $a$.
+For $Z=S\cap eS$, the triangle inequality and the containments
+$Z\subseteq S,eS$ imply
+\[
+ |S-eS|
+ \le\frac{|S-Z|\,|Z-eS|}{|Z|}
+ \le\frac{(Ks)^2}{s/(2K)}=2K^3s.
+\]
+After scaling by $1/b$, we have
+\[
+ |B-B|\le Ks,\qquad |B-eB|\le2K^3s\quad(e\in E).
+\]
+Thus Lemma~\ref{inc-scalar-coverings} applies to $B$ and $E$.
+We have obtained uniform control for every scalar expressible using
+a bounded number of operations on elements of $E$.
+
+\paragraph{Ratios of differences form a subfield.}
+Write $t=|E|$. By~\eqref{eq:inc-overlap-subset} and
+\eqref{eq:inc-approximate-subfield-hypotheses},
+$t=q^{1/2+o(1)}$. In particular, $t\ge2$ for all sufficiently
+large fields in the sequence. Define
+\begin{equation}\label{eq:inc-ratio-set}
+ R_0=\left\{\frac{x-x'}{y-y'}:
+ x,x',y,y'\in E,\ y\ne y'\right\}.
+\end{equation}
+To prove closure under addition and multiplication, we apply the
+scalar-covering bound to the result of either operation.
+Every member of $R_0$ has an expression in $E$ with at most four
+operations. Consequently, for $r,r'\in R_0$, each of $r+r'$ and
+$rr'$ has an expression with at most nine operations. For either
+candidate $\lambda$, Lemma~\ref{inc-scalar-coverings} gives
+\[
+ |E+\lambda E|\le K^{C_9+3}s<t^2
+\]
+for all sufficiently large $q$, uniformly over $r,r'$.
+Indeed, the ratio of this upper bound to $t^2$ is at most
+$4K^{C_9+5}/s$, which tends to zero. Therefore the map
+$(x,y)\mapsto x+\lambda y$ from $E^2$ is not injective. A collision
+between distinct pairs satisfies $y\ne y'$, since otherwise also
+$x=x'$, and gives
+\[
+ \lambda=\frac{x'-x}{y-y'}\in R_0.
+\]
+It follows that $R_0$ is closed under addition and multiplication.
+It contains $0$ and $1$ and, directly from its definition, is closed
+under negation and reciprocals of nonzero elements. Thus $R_0$ is a
+subfield of $\F_q$.
+
+The size of $E$ now prevents this subfield from being proper.
+Choose distinct $e_0,e_1\in E$. The map
+\[
+ e\longmapsto\frac{e-e_0}{e_1-e_0}
+\]
+injects $E$ into $R_0$, so $|R_0|\ge t>q^{1/3}$ for large $q$.
+The proper-subfield hypothesis forces $R_0=\F_q$.
+
+\paragraph{A scalar with a large sumset.}
+Every field element consequently has a four-operation representation
+using elements of $E$. The scalar-covering estimate therefore bounds
+$|E+\lambda E|$ for every scalar. To reach a contradiction, we average
+over the field and find a scalar for which this sumset is too large.
+
+For $\lambda\in\F_q$, let $N_\lambda$ be the number of ordered
+quadruples $(x,y,x',y')\in E^4$ satisfying
+$x+\lambda y=x'+\lambda y'$. There are $t^2$ identical pairs,
+which give a collision for every $\lambda$. If $y\ne y'$, exactly
+one value of $\lambda$ gives a collision; if $y=y'$ but $x\ne x'$,
+none does. Hence
+\begin{equation}\label{eq:inc-average-collisions}
+ \frac1q\sum_{\lambda\in\F_q}N_\lambda
+ =t^2+\frac{t^4-t^3}{q}
+ \le t^2+\frac{t^4}{q}.
+\end{equation}
+For some $\lambda$, Cauchy--Schwarz applied to the fibers of
+$(x,y)\mapsto x+\lambda y$ therefore yields
+\begin{equation}\label{eq:inc-large-dilate-sum}
+ |E+\lambda E|
+ \ge\frac{t^4}{t^2+t^4/q}
+ \ge\frac12\min(t^2,q)
+ =q^{1-o(1)}.
+\end{equation}
+But $R_0=\F_q$, so this particular $\lambda$ itself has a
+four-operation representation in~\eqref{eq:inc-ratio-set}.
+Lemma~\ref{inc-scalar-coverings} bounds its sumset by
+$K^{C_4+3}s=q^{1/2+o(1)}$, contradicting
+\eqref{eq:inc-large-dilate-sum}. The expression length here remains
+bounded; no iteration of field generation is being used.
+\end{proof}
+
+\Needspace{10\baselineskip}
+\subsection{Proof of the incidence theorem}
+
+The algebraic ingredients are now in place. It remains to obtain the
+hypotheses of Lemma~\ref{inc-simultaneous-extraction} from a configuration
+with too many incidences. We use the projective-coordinate reduction
+in the incidence strategy of Bourgain, Katz, and
+Tao~\cite[Section 6]{BKT2004}: after controlling degrees, we choose four
+points whose pencils encode the two coordinates, their difference,
+and their quotient.
+
+\begin{proof}[Proof of Theorem~\ref{thm:incidence}]
+We argue by contradiction, allowing all quantitative losses to be
+subpower factors along a sequence of fields. Suppose the theorem is
+false. For a sequence of positive exponents
+$\kappa_j\to0$ in $(0,1/10)$, choose admissible counterexamples with
+$q_j\to\infty$. Such choices are possible because the failure of
+the theorem says that for each fixed exponent there are arbitrarily
+large counterexamples. Along this sequence, put
+$U=\max(2,q_j^{\kappa_j})$ and suppress the index $j$. We have
+\begin{equation}\label{eq:inc-counterexample-sequence}
+ |P|,|\mathcal L|\le Uq,\qquad
+ I(P,\mathcal L)\ge\frac{q^{3/2}}{U},\qquad
+ U\ge2,\qquad U=q^{o(1)}.
+\end{equation}
+All bounds below use absolute constants. Every assertion made for
+sufficiently large $q$ is along this sequence. Our goal is to construct
+from these counterexamples sets of the size and small-expansion form
+excluded by Lemma~\ref{inc-no-approximate-subfield}.
+
+\paragraph{Pruning the incidence graph.}
+We first retain many incidences while putting both an upper and a
+lower bound on every remaining degree. Consider the bipartite graph
+on $P$ and $\mathcal L$ whose edges are incidences. If $d(v)$ denotes a degree on either side, then counting
+ordered pairs on the opposite side and using the unique join of two
+distinct projective points, or the unique intersection of two
+distinct projective lines, gives
+\begin{equation}\label{eq:inc-degree-squares}
+ \sum_v d(v)^2\le2U^2q^2
+ \quad\text{on each side}.
+\end{equation}
+Here $\sum_vd(v)(d(v)-1)\le U^2q^2$, and
+$\sum_vd(v)=I(P,\mathcal L)\le U^2q^2$.
+
+Set
+\begin{equation}\label{eq:inc-degree-cutoffs}
+ H=16U^3q^{1/2},\qquad d=\frac{q^{1/2}}{8U^2}.
+\end{equation}
+Delete every vertex whose initial degree exceeds $H$. By
+\eqref{eq:inc-degree-squares}, the total number of incidences lost,
+counting both sides, is at most
+\[
+ \frac{4U^2q^2}{H}=\frac{q^{3/2}}{4U}.
+\]
+Next, iteratively delete vertices whose current degree is less than
+$d$. Each of the at most $2Uq$ original vertices is deleted at most
+once, so this stage loses at most
+$2Uq\,d=q^{3/2}/(4U)$ incidences. Denote the surviving sets again by
+$P,\mathcal L$. They satisfy
+\begin{equation}\label{eq:inc-pruned-bounds}
+ I(P,\mathcal L)\ge\frac{q^{3/2}}{2U},\qquad
+ \frac{q}{32U^4}\le |P|,|\mathcal L|\le Uq,
+ \qquad d\le d(v)\le H.
+\end{equation}
+The size lower bounds follow by dividing the surviving incidence
+count by the maximum degree $H$.
+
+The pruning also controls the intersection of $P$ with
+\emph{every} projective line, including lines not in $\mathcal L$.
+We will need this stronger form of control when discarding points
+on a few lines chosen later. Fix a projective line $\ell$. Each point
+of $P\cap\ell$ is incident with at least $d-1$ lines of $\mathcal L$ different from $\ell$.
+These lines are distinct for different points of $P\cap\ell$.
+Since $d\to\infty$, for large $q$ we obtain
+\begin{equation}\label{eq:inc-all-line-cap}
+ |P\cap\ell|
+ \le\frac{|\mathcal L|}{d-1}
+ \le16U^3q^{1/2}=H
+ \qquad\text{for every projective line }\ell.
+\end{equation}
+
+\paragraph{Four anchors with many common witnesses.}
+We seek four points, three collinear and the fourth off their line,
+together with many points joined to each of them by lines of
+$\mathcal L$. We obtain the four points by first counting such
+configurations around each possible common point and then averaging.
+
+For $x\in P$, let $N(x)$ consist of the points in $P\setminus\{x\}$
+joined to $x$ by a line of $\mathcal L$. Distinct lines through $x$
+have disjoint point sets away from $x$, so
+\begin{equation}\label{eq:inc-neighborhood-size}
+ |N(x)|\ge d(d-1)\ge\frac{q}{128U^4}
+\end{equation}
+for large $q$. Put $t_\ell=|N(x)\cap\ell|$. Summing degrees of the
+points of $N(x)$ gives
+\[
+ \sum_{\ell\in\mathcal L}t_\ell
+ \ge d|N(x)|\ge\frac{q^{3/2}}{1024U^6}.
+\]
+Lines with $t_\ell\le2$ contribute at most $2Uq$. Because $U=q^{o(1)}$,
+for sufficiently large $q$ this is at most half of the displayed
+lower bound. Consequently
+\[
+ \sum_{\ell:t_\ell\ge3}t_\ell
+ \ge\frac{q^{3/2}}{2048U^6}.
+\]
+For every integer $t\ge3$, one has
+$t(t-1)(t-2)\ge(2/9)t^3$. H\"older's inequality therefore shows
+that the number of ordered triples of distinct points of $N(x)$
+lying on a line of $\mathcal L$ is at least
+\begin{equation}\label{eq:inc-collinear-triples}
+ \begin{split}
+ \sum_{\ell\in\mathcal L}t_\ell(t_\ell-1)(t_\ell-2)
+ &\ge\frac{2}{9|\mathcal L|^2}
+       \left(\sum_{\ell:t_\ell\ge3}t_\ell\right)^3\\
+ &\ge\frac{q^{5/2}}{9\cdot2^{32}U^{20}}.
+ \end{split}
+\end{equation}
+By~\eqref{eq:inc-all-line-cap} and
+\eqref{eq:inc-neighborhood-size}, each such triple admits at least
+\[
+ \frac{q}{128U^4}-H\ge\frac{q}{256U^4}
+\]
+choices of a fourth point in $N(x)$ outside its line, once $q$ is
+large enough. Summing over $x$ and using
+\eqref{eq:inc-pruned-bounds}, we obtain at least
+\begin{equation}\label{eq:inc-witness-quadruple-count}
+ \frac{q^{9/2}}{9\cdot2^{45}U^{28}}
+\end{equation}
+ordered choices $(x,p_1,p_2,p_3,p_4)$ with $p_1,p_2,p_3$ distinct
+and collinear on a line of $\mathcal L$, with $p_4$ outside that
+line, and with all four $p_i$ in $N(x)$.
+
+We now reverse the count to fix the four anchor points while retaining
+many choices of the common point $x$. The number of possible ordered
+quadruples $(p_1,p_2,p_3,p_4)$ in this count is at most
+\begin{equation}\label{eq:inc-quadruple-upper-bound}
+ |\mathcal L|H^3|P|
+ \le2^{12}U^{11}q^{7/2}.
+\end{equation}
+Some quadruple thus has at least
+\begin{equation}\label{eq:inc-common-witnesses}
+ c_0\frac{q}{U^{39}}
+ \quad\text{witnesses }x,\qquad
+ c_0=\frac1{9\cdot2^{57}}.
+\end{equation}
+Discard the witnesses on the common line of $p_1,p_2,p_3$ and on
+the joins of $p_1,p_4$ and of $p_2,p_4$. The cap
+\eqref{eq:inc-all-line-cap} discards at most $3H$ witnesses.
+Since $U=q^{o(1)}$, this is eventually less than
+$c_0q/(2U^{39})$. At least that many witnesses remain. Their joins
+to the four anchors will supply the graph and the four small sets
+required by Lemma~\ref{inc-simultaneous-extraction}.
+
+\paragraph{The four pencils as coordinates.}
+Choose projective coordinates in which
+\[
+ p_1=(1:0:0),\qquad p_2=(0:1:0),\qquad
+ p_3=(1:1:0),\qquad p_4=(0:0:1).
+\]
+This is possible over $\F_q$: representatives of $p_1,p_2$ span
+their line, the representative of the distinct third point has
+both coefficients nonzero and permits rescaling of the first two,
+and a representative of $p_4$ completes a basis. The discarded
+lines are precisely the coordinate line at infinity and the two
+coordinate axes through $p_4$. Hence every remaining witness has
+coordinates $(a:b:1)$ with $a,b\ne0$.
+
+Let $G$ be the set of these pairs $(a,b)$, and let $A,B$ be its
+coordinate projections. Define $D$ and $J$ from $G$ as in
+Lemma~\ref{inc-simultaneous-extraction}. The chosen coordinates make
+the roles of the four pencils explicit: for a witness $(a:b:1)$,
+the joins to $p_2,p_1,p_3,p_4$, respectively, have equations
+\[
+ X=aZ,\qquad Y=bZ,\qquad
+ X-Y=(a-b)Z,\qquad X=(a/b)Y.
+\]
+Each join is a member of $\mathcal L$, by the witness property.
+Each anchor belongs to $P$ and has degree at most $H$.
+Distinct values of the corresponding parameter give distinct
+lines in its pencil. Thus
+\[
+ |A|,|B|,|D|,|J|\le H,
+ \qquad |G|\ge\frac{c_0q}{2U^{39}}.
+\]
+With
+\begin{equation}\label{eq:inc-graph-parameters}
+ n=q^{1/2},\qquad V=\frac{2}{c_0}U^{39},
+\end{equation}
+we have $V\ge2$, $V=q^{o(1)}$, and all the hypotheses
+\eqref{eq:inc-graph-hypotheses} hold.
+
+The incidence configuration now meets the extraction hypotheses.
+Lemma~\ref{inc-simultaneous-extraction} gives
+$S\subseteq\F_q^*$ satisfying
+\[
+ \frac{3q^{1/2}}{8V^2}\le |S|\le Vq^{1/2},\qquad
+ |S-S|,|S/S|\le2^{20}V^{22}q^{1/2}.
+\]
+In particular $|S|=q^{1/2+o(1)}$, and, on setting
+$K=2^{23}V^{24}$, we have $K\ge2$, $K=q^{o(1)}$, and
+$|S-S|,|S/S|\le K|S|$. This contradicts
+Lemma~\ref{inc-no-approximate-subfield}.
+
+We have excluded every sequence of counterexamples of the form
+\eqref{eq:inc-counterexample-sequence}. The negation of the theorem
+would have produced such a sequence, so some absolute
+$\kappa\in(0,1/10)$ and $Q$ satisfy the stated conclusion.
+\end{proof}

preprints/A-Counterexample-to-Rysers-Covering-Conjecture-September-23-2026/build/linecover.tex

@@ -0,0 +1,447 @@
+\section{Line covers after random deletions}\label{sec:linecover}
+
+We regard the affine plane $\F_q^2$ as a subset of its projective
+completion $\PP^2(\F_q)$. Thus a pencil with centre at infinity is a
+parallel class of affine lines. The purpose of this section is to show
+that a set of at most $q$ lines covering almost all survivors of a
+suitable random deletion process must contain almost a full pencil.
+The deletions within a single trial may be dependent; independence is
+required only between trials.
+
+Classical covering results for finite planes are closely related by
+duality to blocking sets. Jamison and Brouwer--Schrijver established the
+affine blocking-set lower bound $2q-1$~\cite{Jamison1977,BrouwerSchrijver1978},
+and Blokhuis, Brouwer, and Sz\H{o}nyi studied partial projective covers
+and bounds depending on their uncovered points
+\cite[Section 1.1, Proposition 1.5]{BBS2010}. Here the covered set instead consists of the
+positive-density survivors of structured random deletions. We prove the
+pencil statement needed in this setting directly.
+
+There are two different concentration tasks. The total number of
+surviving points and the number on a fixed line have bounded changes
+under one trial; for these we use the elementary estimates
+in~\cite[Lemmas~A.2--A.3]{OpenAI2026Balanced}. The more delicate task
+is to retain many intersections of a short list of lines, simultaneously
+for every such list. A deletion trial rarely hits this small set, and
+we must use that rarity in the variance estimate. We prove the needed
+Bernstein-type martingale bound here, following the method of
+Freedman~\cite[Theorem (1.6) and Proposition (2.1)]{Freedman1975}.
+
+\begin{lemma}[Martingale concentration]\label{lem:martingale}
+Let $Z_i$, $1\leq i\leq n$, be martingale differences for a filtration
+$(\mathcal A_i)_{i=0}^n$. Suppose that $M,V>0$ and deterministic numbers
+$v_i\geq0$ satisfy, almost surely,
+\[
+ |Z_i|\leq M,\qquad
+ \E(Z_i^2\mid\mathcal A_{i-1})\leq v_i,\qquad
+ \sum_{i=1}^n v_i\leq V.
+\]
+Then, for every $u>0$,
+\begin{equation}\label{eq:lc-martingale}
+ \Pr\left(\left|\sum_{i=1}^n Z_i\right|\geq u\right)
+ \leq 2\exp\left[-\frac14
+   \min\left\{\frac{u^2}{V},\frac{u}{M}\right\}\right].
+\end{equation}
+\end{lemma}
+
+\begin{proof}
+For $|z|\leq1$ we have $e^z\leq1+z+z^2$. Consequently, if
+$|\theta|\leq M^{-1}$, the conditional mean-zero property gives
+\[
+ \E(e^{\theta Z_i}\mid\mathcal A_{i-1})
+ \leq 1+\theta^2v_i\leq e^{\theta^2v_i}.
+\]
+Iteration and Markov's inequality bound the upper tail by
+$\exp(\theta^2V-\theta u)$ for $0<\theta\leq M^{-1}$.
+Take $\theta=\min\{u/(2V),1/M\}$. If $u\leq2V/M$, the exponent is
+$-u^2/(4V)$; otherwise it is at most $-u/(2M)$.
+Applying the same argument to $-Z_i$ proves
+\eqref{eq:lc-martingale}.
+
+\end{proof}
+
+\begin{lemma}[Line covers after random deletions]\label{lem:linecover}
+Let $q\to\infty$ through prime powers for which every proper subfield
+of $\F_q$ has order at most $q^{1/3}$. Fix a positive integer $D$ and
+constants $\delta>0$ and $0<\rho<1$. For each $q$, let $\mathcal F$ be a
+family of at most $Dq^D$ subsets of $\F_q^2$, called shapes, such that
+\begin{enumerate}
+ \item every $F\in\mathcal F$ has at most $Dq$ points;
+ \item for every $F\in\mathcal F$ and affine line $\ell$, either
+       $F=\ell$ or $|F\cap\ell|\leq D$.
+\end{enumerate}
+There are at most $Dq$ independent trials, each producing a random list
+of at most $D$ shapes from $\mathcal F$. Delete every point on any listed
+shape, and denote the surviving set by $X$. Assume that
+\begin{enumerate}
+ \item for each point $x$ and each trial, the probability that this
+       trial deletes $x$ is at most $D/q$;
+ \item $\Pr(x\in X)=\delta_q\geq\delta$ for every $x\in\F_q^2$;
+ \item for each affine line $\ell$, the sum over trials of the
+       probabilities that the trial lists the exact shape $\ell$ is
+       at most $D/q$.
+\end{enumerate}
+Write $X^\ell$ for the surviving set when all exact occurrences of the
+shape $\ell$ are ignored, with every other listed shape still deleted.
+With probability tending to one, both of the following statements
+hold:
+\begin{enumerate}
+ \item $|X^\ell\cap\ell|\geq\delta q/2$ for every affine line $\ell$;
+ \item every set of at most $q$ affine lines covering all but at most
+       $2q$ points of $X$ contains at least $(1-\rho)q$ lines through a
+       common projective point.
+\end{enumerate}
+\end{lemma}
+
+\begin{proof}
+We first establish a simultaneous lower bound on surviving
+cross-intersections for every admissible pair of short line lists,
+ignoring any deletion shape equal to one of those lines. After fixing
+such a realization, we show that any cover lacking an almost full
+pencil contains an admissible pair with too few surviving
+cross-intersections. This second step samples only within the fixed
+cover, with separate arguments according to its maximum concurrency.
+
+Fix the absolute constant $\kappa\in(0,1/10)$ of
+Theorem~\ref{thm:incidence}, and put $\epsilon=\kappa/10$.
+All constants below may depend on the fixed parameters, but not on
+$q$, the particular deletion distributions, or a subsequently chosen
+cover.
+
+\paragraph{Uniform point and line counts.}
+Expose the independent trials one at a time. Changing one trial
+changes $|X|$ by at most $2D^2q$, because the old and new lists contain
+at most $2D$ shapes in total and each has at most $Dq$ points.
+The exposure observation in~\cite[Lemma~A.3]{OpenAI2026Balanced}
+therefore bounds the absolute martingale increment by $L=2D^2q$.
+There are $T\le Dq$ trials, and $\E|X|=\delta_q q^2$. The fixed-time
+concentration estimate in~\cite[Lemma~A.2]{OpenAI2026Balanced}, with
+$u=q^{3/2+\epsilon}$, gives
+\[
+ \Pr\bigl(\bigl||X|-\delta_q q^2\bigr|\ge q^{3/2+\epsilon}\bigr)
+ \le2\exp\left(-\frac{q^{2\epsilon}}{32D^5}\right).
+\]
+If there are no trials, the count is deterministic and this bound
+holds as well. Equivalently, we may add a dummy trial of zero change.
+
+For the line count $|X^\ell\cap\ell|$, the rule ignoring exact copies
+of $\ell$ is applied separately to each trial. Changing one trial
+then changes the count by at most $2D^2$: each of its at most $2D$
+old or new shapes meets $\ell$ in at most $D$ points. With
+$L=2D^2$ and $u=q^{1/2+\epsilon}$, the same two elementary estimates
+and $T\le Dq$ give
+\[
+ \Pr\bigl(\bigl||X^\ell\cap\ell|-\E|X^\ell\cap\ell|\bigr|
+              \ge q^{1/2+\epsilon}\bigr)
+ \le2\exp\left(-\frac{q^{2\epsilon}}{32D^5}\right).
+\]
+
+Moreover, a point restored by ignoring $\ell$ can be restored only if
+some trial lists $\ell$ exactly. The probability of this event is at
+most $D/q$, by the third deletion hypothesis. Therefore
+\[
+ \delta_q q\leq \E|X^\ell\cap\ell|\leq\delta_q q+D.
+\]
+There are $q(q+1)$ affine lines. The superpolynomial tails and a union
+bound consequently show that, with probability tending to one,
+\begin{equation}\label{eq:lc-counts}
+ \begin{split}
+ |X|&=\delta_q q^2+O_D(q^{3/2+\epsilon}),\\
+ |X^\ell\cap\ell|&=\delta_q q+O_D(q^{1/2+\epsilon})
+       \qquad\text{for every affine line }\ell.
+ \end{split}
+\end{equation}
+The lower line estimate proves the first conclusion of the lemma.
+For the pencil conclusion we will also need to bound overlaps in an
+alleged cover. The same estimates give the uniform upper bound
+\begin{equation}\label{eq:lc-line-upper}
+ |X\cap\ell|\leq\delta_q q+O_D(q^{1/2+\epsilon}),
+\end{equation}
+because $X\subseteq X^\ell$.
+
+\paragraph{A simultaneous event for small line samples.}
+The counts just proved control a single line but do not yet restrict
+how a cover's lines meet. We now obtain a stronger test on short lists
+of lines: provided their intersections are distinct and not concentrated
+on a deletion shape, many of those intersections survive. The list
+length will be logarithmic in $q$, so that the variance-sensitive tail
+can pay for a union bound over all lists.
+
+Set
+\[
+ h=10(D+1),\qquad H_0=2Dh,\qquad
+ t=\lceil A\log q\rceil,
+\]
+where the fixed constant $A$ will be chosen below. Consider any ordered
+tuple of $2t$ distinct affine lines, separated into two lists of length
+$t$. Call the tuple admissible if its cross-list intersections form a
+set $P_0$ of $t^2$ distinct finite points, and
+\begin{equation}\label{eq:lc-occupancy}
+ |F\cap P_0|\leq H_0
+ \quad\text{for every }F\in\mathcal F
+       \text{ which is not one of the tuple lines}.
+\end{equation}
+For this tuple, ignore deletions by every shape equal to a tuple line,
+and let $Y_0$ count the survivors in $P_0$ after the remaining deletions.
+We claim that, with probability tending to one,
+\begin{equation}\label{eq:lc-tuple-event}
+ Y_0\geq\frac{\delta t^2}{2}
+ \qquad\text{for every admissible tuple}.
+\end{equation}
+
+To prove the claim, fix a tuple in advance. Omitting shapes is a
+deterministic operation on each trial, so the trials remain
+independent. Omission can only restore points, and hence
+$\E Y_0\geq\delta t^2$. By \eqref{eq:lc-occupancy}, each trial's
+contribution to the deletion set within $P_0$ has size at most
+$C_0=DH_0$. The probability that this contribution is nonempty is at
+most $Dt^2/q$, by a union bound over $P_0$ and the per-point deletion
+hypothesis.
+
+For completeness, the rarity of a nonempty contribution also controls
+the conditional variance of the exposure martingale. Fix the past,
+and let $g_i(w)$ be the conditional expected final value of $Y_0$ when
+the current trial has outcome $w$. Let $g_i(\varnothing)$ denote the
+hypothetical value obtained by making this trial delete nothing on
+$P_0$. Independence leaves the law of the future trials unchanged, so
+\[
+ |g_i(w)-g_i(\varnothing)|\leq C_0,
+\]
+with equality to zero whenever the trial's contribution is empty.
+The current trial also retains its original law conditional on the
+past. Thus the martingale increment has absolute value at most $2C_0$, and
+its conditional variance is at most
+\[
+ \E\bigl((g_i(w)-g_i(\varnothing))^2\mid\text{past}\bigr)
+ \leq C_0^2\frac{Dt^2}{q}.
+\]
+The sum of conditional variances is at most $D^2C_0^2t^2$, since there
+are at most $Dq$ trials. Applying Lemma~\ref{lem:martingale} at deviation
+$\delta t^2/2$ yields
+\begin{equation}\label{eq:lc-tuple-tail}
+ \Pr\left(Y_0<\frac{\delta t^2}{2}\right)
+ \leq2e^{-c_0t^2},\qquad
+ c_0=\min\left\{\frac{\delta^2}{16D^2C_0^2},
+                       \frac{\delta}{16C_0}\right\}>0.
+\end{equation}
+In particular, $c_0$ is independent of $A$.
+
+There are at most $(q(q+1))^{2t}$ ordered tuples. The logarithm of the
+union bound in \eqref{eq:lc-tuple-tail} is
+\[
+ -c_0t^2+4t\log q+O(t/q)+O(1)
+ =\bigl(-c_0A^2+4A+o(1)\bigr)(\log q)^2.
+\]
+Choose, for example, any fixed $A>5/c_0$. This proves
+\eqref{eq:lc-tuple-event} simultaneously for all admissible tuples.
+
+\paragraph{Consequences of an alleged cover.}
+The random part of the argument is complete. We now keep the deletion
+outcome fixed and show deterministically that the simultaneous events
+exclude any cover without an almost full pencil. Random sampling below
+is only a way to find a short witness inside a fixed cover.
+
+Fix a realization satisfying both \eqref{eq:lc-counts} and
+\eqref{eq:lc-tuple-event}. Suppose that the second conclusion of the
+lemma fails. From an offending cover remove every line which actually
+occurred as a listed deletion shape. Such a line contains no point of
+$X$, so the remaining line set $L$ still covers all but at most $2q$
+points of $X$. Put $m=|L|$, and for any projective point $p$ write
+$n_p=|\{\ell\in L:p\in\ell\}|$. Define
+\[
+ E_L=\sum_{x\in X}(n_x-1)_+,
+ \qquad (a)_+=\max\{a,0\}.
+\]
+The count of covered survivors and \eqref{eq:lc-line-upper} give
+\[
+ m\bigl(\delta_q q+O_D(q^{1/2+\epsilon})\bigr)
+ \geq\delta_q q^2-O_D(q^{3/2+\epsilon})-2q.
+\]
+Since $\delta_q\geq\delta$ and $m\leq q$, this implies
+\begin{equation}\label{eq:lc-cover-counts}
+ m=q-O_{D,\delta}(q^{1/2+\epsilon}),\qquad
+ E_L=O_D(q^{3/2+\epsilon}).
+\end{equation}
+For the second estimate, subtract the number of covered survivors
+from $\sum_{\ell\in L}|X\cap\ell|$ and use $m\leq q$.
+Finally, put
+\[
+ R=\max_{p\in\PP^2(\F_q)}n_p.
+\]
+The failure of the pencil conclusion implies $R\leq(1-\rho)q$.
+We will find an admissible tuple of lines from $L$ with fewer than
+$\delta t^2/2$ survivors in its cross-intersection set.
+
+\paragraph{Low concurrency: $R\leq q^{1/2+\kappa/4}$.}
+Let $T=q^{1/2-\kappa/3}$. The number of ordered pairs of distinct lines
+of $L$ meeting at a point $x\in X$ is $\sum_{x\in X}n_x(n_x-1)$.
+Points with $n_x<T$ contribute at most
+\begin{equation}\label{eq:lc-low-light}
+ T E_L=O(q^{2-7\kappa/30}).
+\end{equation}
+The set $P=\{x\in X:n_x\geq T\}$ has size
+\[
+ |P|\leq\frac{E_L}{T-1}
+      =O(q^{1+13\kappa/30})\leq q^{1+\kappa}
+\]
+for sufficiently large $q$. Theorem~\ref{thm:incidence}, applied to
+$P$ and $L$, therefore gives $\sum_{x\in P}n_x\leq q^{3/2-\kappa}$.
+The contribution of these points to the ordered-pair count is at most
+\begin{equation}\label{eq:lc-low-heavy}
+ R\sum_{x\in P}n_x\leq q^{2-3\kappa/4}.
+\end{equation}
+Together, \eqref{eq:lc-low-light} and \eqref{eq:lc-low-heavy} show that
+the fraction of ordered distinct line pairs meeting in $X$ is
+\begin{equation}\label{eq:lc-low-pair-fraction}
+ O\bigl(q^{-7\kappa/30}+q^{-3\kappa/4}\bigr)=o(1).
+\end{equation}
+
+Sample an ordered tuple of $2t$ distinct lines uniformly from $L$.
+The expected number of its cross pairs meeting in $X$ is $t^2$ times
+the fraction in \eqref{eq:lc-low-pair-fraction}. Markov's inequality
+shows that this number is less than $\delta t^2/2$ with probability
+tending to one.
+
+We next check the two admissibility conditions. A given sampled pair
+is parallel with probability at most $R/(m-1)$. Once two distinct
+lines have been sampled, at most $R$ population lines pass through
+their projective intersection, so the probability of concurrency for
+any specified three sampled positions is at most $R/(m-2)$. The
+probability of any parallel pair or concurrent triple in the tuple
+is consequently
+\begin{equation}\label{eq:lc-low-degeneracy}
+ O\left(\frac{(t^2+t^3)R}{m}\right)
+ =O\bigl((\log q)^3q^{-1/2+\kappa/4}\bigr)=o(1).
+\end{equation}
+If neither event occurs, every cross intersection is finite and the
+cross intersections are distinct: equality of two different cross
+intersections would make at least three of the tuple lines concurrent.
+
+To verify \eqref{eq:lc-occupancy}, fix $F\in\mathcal F$. Consider the
+event that $F$ is not a tuple line but contains more than $H_0$ distinct
+finite cross intersections. Each tuple line then contains at most
+$D$ points of $F$. Greedily choose a cross intersection on $F$ and its
+two representing positions, and discard all cross intersections on
+either of those lines. Each step discards at most $2D$ points. Since
+$H_0=2Dh$, this produces $h$ cross pairs whose $2h$ positions are
+distinct.
+
+Among the population $L$, there are at most $mDR$ ordered distinct
+line pairs, neither line equal to $F$, whose intersection lies in
+$F$. Indeed, the first line meets $F$ in at most $D$ points, and at
+each such point there are at most $R$ choices of the second line.
+There are at most $t^{2h}$ ordered choices of the $h$ cross pairs of
+positions. For each choice, the sampled lines at those $2h$ distinct
+positions are uniform among $(m)_{2h}$ assignments, where
+$(z)_a=z(z-1)\cdots(z-a+1)$. The probability of the event for this
+fixed shape is therefore at most
+\begin{equation}\label{eq:lc-low-shape}
+ \frac{t^{2h}(mDR)^h}{(m)_{2h}}
+ =O_{D,h}\left(\left(\frac{t^2R}{m}\right)^h\right).
+\end{equation}
+The numerator may count assignments with repeated population lines;
+this only enlarges the upper bound. Summing
+\eqref{eq:lc-low-shape} over all shapes gives
+\[
+ O\bigl((\log q)^{2h}q^{D-h(1/2-\kappa/4)}\bigr)=o(1),
+\]
+because $\kappa<1/10$ and $h=10(D+1)$ imply
+\[
+ D-h(1/2-\kappa/4)
+ <D-\frac{19}{4}(D+1)<0.
+\]
+Thus, with probability tending to one, the sampled tuple is admissible
+and has fewer than $\delta t^2/2$ cross pairs meeting in $X$.
+
+\paragraph{High concurrency: $R>q^{1/2+\kappa/4}$.}
+Choose a projective point $p$ with $n_p=R$, and take all $R$ lines of
+$L$ through $p$ as the pencil. Its complement has size
+\[
+ N=m-R\geq\rho q/2
+\]
+for sufficiently large $q$, by \eqref{eq:lc-cover-counts}; no
+complement line passes through $p$. Sample the first list of $t$ lines
+uniformly without replacement from the pencil, and independently
+sample the second list in the same way from the complement.
+
+Any cross intersection lies off $p$. At a cross intersection
+$x\in X$, exactly one pencil line passes through $x$, and the number
+of cross pairs meeting there is the number of complement lines
+through $x$, namely $n_x-1$. Thus at most $E_L$ of all $RN$ population
+cross pairs meet in $X$. Their fraction is at most
+\begin{equation}\label{eq:lc-high-pair-fraction}
+ \frac{E_L}{RN}
+ =O\bigl(q^{\epsilon-\kappa/4}\bigr)
+ =O(q^{-3\kappa/20})=o(1).
+\end{equation}
+As before, expectation and Markov's inequality show that fewer than
+$\delta t^2/2$ sampled cross pairs meet in $X$, with probability
+tending to one.
+
+For a fixed complement line, at most one pencil line meets it at
+infinity when $p$ is finite. When $p$ is at infinity, the pencil lines
+have direction $p$ and every complement line has a different
+direction, so all cross intersections are finite. The probability of
+any infinite cross intersection is therefore at most $t^2/R$.
+Two equal cross intersections must use the same pencil line, since
+distinct pencil lines meet only at $p$, which lies on no complement
+line. Given two complement lines, their projective intersection
+determines at most one affine pencil line, namely its join with $p$.
+If this join is the line at infinity, there is no affine candidate.
+Conditioning on the second list and taking a union bound over its
+pairs shows that the probability of a collision is $O(t^3/R)$.
+The total probability of a failure of finiteness or distinctness is
+therefore
+\begin{equation}\label{eq:lc-high-degeneracy}
+ O\left(\frac{t^2+t^3}{R}\right)
+ =O\bigl((\log q)^3q^{-1/2-\kappa/4}\bigr)=o(1).
+\end{equation}
+
+Fix $F\in\mathcal F$, and consider the event that $F$ is not a tuple
+line but contains more than $H_0$ distinct finite cross intersections.
+The same greedy argument gives $h$ cross pairs with disjoint
+positions. There are at most
+$DN$ population cross pairs, with neither line equal to $F$, whose
+intersection lies in $F$: a complement line distinct from $F$ meets
+$F$ in at most $D$ points, all off $p$, and each such point determines
+at most one pencil line. The two lists are sampled independently, so
+the probability of this occupancy violation for a fixed shape is at
+most
+\begin{equation}\label{eq:lc-high-shape}
+ \frac{t^{2h}(DN)^h}{(R)_h(N)_h}
+ =O_{D,h}\left(\left(\frac{t^2}{R}\right)^h\right).
+\end{equation}
+After summing over $\mathcal F$, the bound becomes
+\[
+ O\bigl((\log q)^{2h}q^{D-h(1/2+\kappa/4)}\bigr)=o(1),
+\]
+since
+\[
+ D-h(1/2+\kappa/4)<D-5(D+1)<0.
+\]
+The sampled tuple is consequently admissible and has fewer than
+$\delta t^2/2$ cross pairs meeting in $X$, with probability tending
+to one in this regime as well.
+
+\paragraph{The contradiction and the order of choices.}
+All sampling estimates above are uniform over line sets satisfying
+\eqref{eq:lc-cover-counts} and the appropriate bound on $R$. In either
+regime, for every sufficiently large $q$ there is therefore an
+admissible tuple from the fixed line set $L$ with
+\[
+ |X\cap P_0|<\delta t^2/2.
+\]
+No line of $L$ actually occurred as a listed deletion shape. Hence
+ignoring exact occurrences of its tuple lines changes none of the
+deletions in the fixed realization, and the tuple's count $Y_0$ is
+exactly $|X\cap P_0|$. This contradicts
+\eqref{eq:lc-tuple-event}.
+
+In particular, the cover and the tuple are allowed to depend on the
+realized deletions: the event \eqref{eq:lc-tuple-event} was established
+simultaneously for every admissible tuple before fixing the
+realization. Sampling from $L$ is used only to prove the existence of
+a tuple inside that fixed realization. No concentration estimate is
+conditioned on the choice of the cover. The contradiction proves the
+second conclusion, and finishes the proof.
+\end{proof}

preprints/A-Counterexample-to-Rysers-Covering-Conjecture-September-23-2026/build/paper.tex

@@ -0,0 +1,301 @@
+\documentclass[11pt]{article}
+\usepackage[T1]{fontenc}
+\usepackage[utf8]{inputenc}
+\usepackage{lmodern}
+\input{glyphtounicode}
+\input{glyphtounicode-cmex}
+\pdfgentounicode=1
+\usepackage{tikz}
+\usetikzlibrary{arrows.meta,positioning}
+\usepackage[margin=1in]{geometry}
+\usepackage{amsmath,amssymb,amsthm,mathtools}
+\usepackage{microtype}
+\usepackage{enumitem}
+\usepackage{booktabs,array}
+\usepackage{needspace}
+\usepackage{xcolor}
+\usepackage[hyphens]{url}
+\usepackage{hyperref}
+\hypersetup{colorlinks=true,linkcolor=blue!45!black,citecolor=blue!45!black,
+  urlcolor=blue!45!black,
+  pdftitle={A counterexample to Ryser's covering conjecture},pdfauthor={OpenAI}}
+\urlstyle{same}
+\setlength{\emergencystretch}{2em}
+\setlist{itemsep=3pt,topsep=5pt}
+\newtheorem{theorem}{Theorem}[section]
+\newtheorem{lemma}[theorem]{Lemma}
+\newtheorem{proposition}[theorem]{Proposition}
+\newtheorem{corollary}[theorem]{Corollary}
+\theoremstyle{remark}
+\newtheorem{remark}[theorem]{Remark}
+\newtheorem{definition}[theorem]{Definition}
+\newcommand{\F}{\mathbb F}
+\newcommand{\PP}{\mathbb P}
+\newcommand{\R}{\mathbb R}
+\newcommand{\N}{\mathbb N}
+\newcommand{\E}{\mathbb E}
+\DeclareMathOperator{\Var}{Var}
+\DeclareMathOperator{\Cov}{Cov}
+\DeclareMathOperator{\supp}{supp}
+\title{A counterexample to Ryser's covering conjecture}
+\author{OpenAI}
+\date{September 23, 2026}
+
+\begin{document}
+\maketitle
+\begin{abstract}
+For every sufficiently large prime $s\equiv2\pmod3$ and every sufficiently
+large odd integer $n$, with the threshold depending on $s$, we construct
+an intersecting $(s^n+1)$-partite $(s^n+1)$-uniform hypergraph with covering
+number $s^n+1$. This disproves Ryser's covering conjecture in its
+intersecting case.
+\end{abstract}
+
+\section{Introduction}
+
+A vertex cover of a finite hypergraph $H$ is a set of vertices meeting
+every edge. Its covering number $\tau(H)$ is the minimum size of such a
+set, and its matching number $\nu(H)$ is the maximum number of pairwise
+disjoint edges. For an integer $r\ge2$, an $r$-partite $r$-uniform
+hypergraph has disjoint vertex parts $V_1,\ldots,V_r$, with every edge
+containing exactly one vertex from each part. Ryser's covering
+conjecture asserts that every such hypergraph satisfies
+\begin{equation}\label{eq:ryser}
+ \tau(H)\le(r-1)\nu(H).
+\end{equation}
+A nonempty hypergraph is intersecting when every two distinct edges
+meet; equivalently, its matching number is one. In this case the
+conjecture asks for a cover of size at most $r-1$. The vertices of a
+cover may lie in several different parts, which is the central
+obstacle in constructing a counterexample.
+
+For $r=2$, inequality~\eqref{eq:ryser} is K\H{o}nig's matching theorem.
+An equivalent formulation appears in Henderson's 1971
+thesis~\cite{Henderson1971}; Best and Wanless discuss the history of its
+attribution to Ryser~\cite{BestWanless2018}. Aharoni proved the full
+tripartite case~\cite{Aharoni2001}. In the intersecting case, the results
+through rank four are due to Gy\'arf\'as, and the rank-five result to
+Tuza, as recorded by Kir\'aly and T\'othm\'er\'esz~\cite{KiralyTothmeresz2017};
+see also Tuza's early study of the covering conjecture~\cite{Tuza1983}.
+Haxell and Scott proved $\tau(H)\le(r-\epsilon)\nu(H)$ for some
+$\epsilon>0$ when $r=4$ or $5$, without assuming
+intersection~\cite{HaxellScott2012}. White proved the particular
+rank-four case $\nu(H)=2$, obtaining $\tau(H)\le6$~\cite{White2026}.\footnote{White
+reports assistance from GPT-5.6 Sol and Claude in the Methods section
+of his preprint.}
+Under the additional assumption that any two distinct edges meet in
+exactly one vertex, Franceti\'{c}, Herke, McKay, and Wanless established
+the intersecting case through rank nine~\cite{FranceticHerkeMcKayWanless2017}.
+Bishnoi, Das, Morris, and Szab\'o obtained bounds when every pair of
+edges has a larger prescribed intersection~\cite{BDMS2021}.
+
+Finite planes supply the classical equality examples. Delete a point
+and all lines through it from a projective plane of order $q$. Partition
+the remaining points according to the deleted line through them, and
+take the remaining lines as hyperedges. The result is intersecting,
+has $q+1$ parts and covering number $q$. Thus equality in the proposed
+bound occurs whenever $r-1$ is a prime power. Work of Mansour, Song, and
+Yuster~\cite{MansourSongYuster2009} and Aharoni, Bar\'{a}t, and
+Wanless~\cite{AharoniBaratWanless2016} studies the structure and sparsity
+of equality examples. Abu-Khazneh, Bar\'{a}t, Pokrovskiy, and
+Szab\'{o} exploit the rigidity of minimum covers in a truncated plane
+and add exceptional edges to obtain further equality
+examples~\cite{AbuKhaznehBaratPokrovskiySzabo2019}. In that construction,
+the number of parts and the covering number both increase by one.
+Haxell and Scott combine affine planes with intersecting configurations
+to obtain covering number at least $r-4$ for every sufficiently large
+$r$, improved to $r-3$ when $r$ is even~\cite{HaxellScott2017}.
+
+Our construction also starts from a finite plane and inserts exceptional
+edges, but keeps one part for each ambient direction. It changes the
+line labels within these parts to raise the covering number beyond
+$q$. The following theorem gives the precise field orders for which
+this can be done.
+
+\begin{theorem}\label{thm:main}
+There exists $s_0$ such that, for every prime $s\ge s_0$ with
+$s\equiv2\pmod3$, there is an integer $n_0(s)\ge3$ with the following
+property. For every odd integer $n\ge n_0(s)$, put $q=s^n$.
+There is a finite intersecting $(q+1)$-partite $(q+1)$-uniform
+hypergraph $H$ satisfying
+\[
+ \tau(H)=q+1.
+\]
+In particular, the intersecting case of~\eqref{eq:ryser} is false.
+\end{theorem}
+
+Every edge of a nonempty intersecting hypergraph is itself a cover,
+so the assertion $\tau(H)\le q+1$ is automatic. The proof must construct
+an intersecting hypergraph and exclude every cover of size at most $q$.
+We first fix any sufficiently large eligible prime $s$, construct a
+configuration in $\F_s^2$, and then take $n$ sufficiently large among
+the odd integers. All subsequent thresholds may depend on that fixed
+configuration. The result is asymptotic; the proof does not give a
+numerical threshold.
+
+A companion paper~\cite{OpenAI2026Balanced} proves a different theorem:
+for every sufficiently large prime $q$, it constructs such a hypergraph
+with exactly $q+1$ vertices in every part and with every vertex
+nonisolated. Its construction modifies line partitions directly in the
+prime plane. Here the field order is a proper odd power, and the proof
+uses a small-field configuration with a nonlinear change of
+coordinates. The two constructions give distinct ranges of ranks.
+The only results imported from the companion are its three elementary
+estimates in Appendix~A: the affine second-moment identity
+(Lemma~A.1), bounded-drift concentration (Lemma~A.2), and the exposure
+bound for independent trials (Lemma~A.3).
+
+Lov\'asz proposed the stronger assertion that deleting $r-1$ vertices
+can always reduce the matching number. Clow, Haxell, and Mohar
+disproved that assertion at $r=3$~\cite{CHM2025}; Abiad, Garbe, Povill,
+and Spiegel subsequently gave an infinite family at $r=3$ and
+examples at $r=4$~\cite{AbiadGarbePovillSpiegel2025}. These examples
+are consistent with~\eqref{eq:ryser}. They exclude a prescribed
+matching-reduction procedure, whereas Theorem~\ref{thm:main}
+violates the covering inequality itself.
+
+\subsection*{Monochromatic tree covers}
+
+An $r$-edge-coloring here assigns exactly one of $r$ colors to each edge.
+A monochromatic tree cover is a family of monochromatic trees whose vertex
+sets cover the graph; the trees may overlap, and singleton trees are allowed.
+Gy\'arf\'as's conjecture asks whether $r-1$ such trees suffice for every
+$r$-edge-colored complete graph; see \cite[Conjecture~1]{Milicevic2017}.
+This is the covering formulation recorded after Theorem~2 in
+\cite[Section~3]{ErdosGyarfasPyber1991}, not the distinct vertex-disjoint
+tree-partition conjecture numbered Conjecture~2 there.
+
+\begin{corollary}\label{cor:tree-cover}
+Let $s_0$ and $n_0(s)$ be as in Theorem~\ref{thm:main}. For every prime
+$s\ge s_0$ with $s\equiv2\pmod3$ and every odd integer $n\ge n_0(s)$,
+put $r=s^n+1$. There is a finite simple complete graph with an
+$r$-edge-coloring for which the minimum number of monochromatic trees
+needed to cover its vertices is exactly $r$. The same minimum holds with
+connected monochromatic subgraphs or maximal monochromatic components in
+place of trees.
+\end{corollary}
+
+\begin{proof}
+Take the hypergraph $H$ from Theorem~\ref{thm:main}, with parts
+$V_1,\ldots,V_r$. For $e\in E(H)$, let $x_i(e)$ be the unique vertex in
+$e\cap V_i$. Form the complete graph $K$ with vertex set $E(H)$. For each
+pair of distinct hyperedges $e,f$, choose one index $i$ with
+$x_i(e)=x_i(f)$ and color the edge $ef$ by $i$. Such an index exists
+because $H$ is intersecting. Only one color is chosen even when the two
+hyperedges meet in several parts.
+
+Along a color-$i$ path, the value of $x_i$ is constant. Thus every connected
+color-$i$ subgraph determines a vertex of $V_i$ belonging to every
+hyperedge indexed by that subgraph. For a singleton, choose any color and
+the corresponding vertex of its hyperedge. If $t$ connected monochromatic
+subgraphs cover the vertices of $K$, the at most $t$ hypergraph vertices so
+chosen meet every edge of $H$. Hence $t\ge\tau(H)=r$.
+
+Conversely, fix a vertex of $K$. The $r$ monochromatic stars centered there,
+one for each color and allowing singleton stars, cover all vertices of $K$.
+They give a tree cover of size at most $r$. The lower bound applies to trees
+and to maximal monochromatic components as well, and the stars can be
+enlarged to maximal components. This proves all three stated minima.
+\end{proof}
+
+Thus Gy\'arf\'as's cover conjecture fails at these extension-field ranks.
+The examples admit no cover by $r-1$ connected monochromatic subgraphs even
+without a diameter restriction. They therefore also refute Mili\'cevi\'c's
+bounded-diameter strengthening \cite[Conjecture~6]{Milicevic2017}, whose
+diameter bound may depend on $r$, and the $\alpha=1$ case of the conjecture
+of DeBiasio, Kamel, McCourt, and Sheats
+\cite[Conjecture~2.8]{DeBiasioKamelMcCourtSheats2021}, whose diameter bound
+depends only on the independence number $\alpha$. Indeed, $\alpha(K)=1$,
+so the latter conjecture also demands a cover by at most $r-1$ such
+subgraphs.
+
+\subsection*{Proof strategy}
+
+In the affine-plane model, the parts are indexed by the $q+1$ directions
+in $\F_q^2$, and the vertices of a part are the $q$ lines of that
+direction. A point $x$ defines an edge consisting of its incident line
+in each direction. The edges of two points intersect at their join,
+but a full parallel class always gives a cover of size $q$. We retain
+many of these point edges and add exceptional edges whose labels are
+changed in some directions. The construction has three tasks: preserve
+intersection, restrict small covers of the retained point edges, and
+make the exceptional edges too costly for those covers.
+
+\paragraph{Restricting the ordinary cover.}
+Delete a bounded number of translated shapes in each of $O(q)$
+independent trials. The shapes are lines or curves that meet every
+line in boundedly many points. Thus a shape has bounded intersection
+with a line unless it is that line itself. Section~\ref{sec:linecover} proves that, under explicit
+uniform survival assumptions, every cover of almost all surviving
+points by at most $q$ lines contains almost a pencil. Here a pencil is
+a family of lines through one projective point, including a parallel
+class when that point is at infinity. Surviving points on each
+nondeleted line then force every nondeleted member of this pencil
+into the cover. To obtain the pencil statement over the required
+extension fields, Section~\ref{sec:incidence} first proves an incidence
+saving by the sum--product method of Bourgain, Katz, and
+Tao~\cite{BKT2004}. Its constants are uniform over fields whose proper
+subfields all have order at most $q^{1/3}$.
+
+\paragraph{One configuration in two coordinates.}
+In a fixed prime plane $\F_s^2$, remove one line in each finite nonzero
+direction and call the remaining set $Y$. The change of coordinates
+$f(x,y)=(x^3,y)$ is bijective when $s\equiv2\pmod3$. Original line
+labels determine which ambient point edges must be deleted to
+intersect the exceptional edges. The labels for altered directions
+are instead read from the lines through $f(w)$, for $w\in Y$.
+The map $f$ preserves the horizontal and vertical parallel classes;
+this ensures intersection within the exceptional configuration.
+Its nonlinear action on the other lines makes their images curves
+with at most three intersections with any line. Applying the deletion
+lemma in these new coordinates yields the crucial obstruction:
+covering $f(Y)$ by $a$ lines and $b$ individual points with
+$a+b/2\le s$ requires exactly one full parallel class and no point
+resources. Section~\ref{sec:gadget} proves this statement together
+with the ordinary-coordinate occupancy needed later.
+
+\paragraph{Placing the configurations and spending the cover budget.}
+Embed linearly transformed copies of $Y$ in $\F_q^2$ so that every
+ambient direction is altered in one or two copies. Arbitrary translates
+could create disjoint exceptional edges. We therefore sample a fixed
+number of independent candidate translates for each copy.
+Section~\ref{sec:selection} proves that one can choose compatible
+representatives from these pools, using an independent-transversal
+argument in the tradition of Haxell~\cite{Haxell1995,Haxell2001}.
+Section~\ref{sec:placement} proves the deletion estimates for the
+entire independent pool before making this selection. Keeping only
+the selected deletions restores ordinary points, so the needed
+estimates transfer by deterministic inclusion.
+
+Section~\ref{sec:hypergraph} gives the labels of every edge, proves
+intersection, and excludes a cover of size at most $q$. After the
+ordinary edges force a pencil, its center determines the remaining
+argument. At infinity, the uncovered exceptional configurations
+require more vertices than the missing parallel lines release.
+Inside an embedded configuration, the remaining budget would cover
+$f(Y)$ by at most $s$ lines including two nonparallel axis lines.
+Outside every configuration, each deleted pencil line supplies five
+uncovered exceptional edges, while every additional vertex covers at
+most five. In all three cases the available budget is too small.
+
+\paragraph{Conventions.}
+A direction is a one-dimensional linear subspace of $\F_q^2$; as a subspace
+it includes zero. The projective completion adds one point at infinity per
+direction. A pencil consists of the affine lines through a projective point:
+it has $q+1$ members for a finite center and $q$ members for a center at
+infinity. Thus concurrence always includes parallel lines.
+For a positive integer $j$, write $(a)_j=a(a-1)\cdots(a-j+1)$.
+All asymptotic constants may depend on the parameters explicitly held fixed.
+An event holds with high probability if its probability tends to one along
+the specified sequence of fields. The small field is fixed before the
+ambient field tends to infinity.
+
+\input{incidence}
+\input{linecover}
+\input{selection}
+\input{construction}
+
+\Needspace{32\baselineskip}
+\bibliographystyle{plain}
+\bibliography{references}
+\end{document}

preprints/A-Counterexample-to-Rysers-Covering-Conjecture-September-23-2026/build/references.bib

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+  volume = {161},
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+}
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+}
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+}
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+}
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+}
+
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+  author = {Fox, Jacob and Sudakov, Benny},
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+  url = {https://arxiv.org/abs/0909.3271},
+  note = {\href{https://arxiv.org/abs/0909.3271}{arXiv:0909.3271}}
+}
+
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+  author = {Haxell, P. E.},
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+}
+
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+  author={Freedman, David A.},
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+}
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+  author={Jamison, Robert E.},
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+}
+
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+  author={Brouwer, A. E. and Schrijver, A.},
+  title={The blocking number of an affine space},
+  journal={Journal of Combinatorial Theory, Series A},
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+  doi={10.1016/0097-3165(78)90013-4},
+  note = {\href{https://doi.org/10.1016/0097-3165(78)90013-4}{doi:10.1016/0097-3165(78)90013-4}}
+}
+
+@article{BBS2010,
+  author={Blokhuis, A. and Brouwer, A. E. and Sz{\H o}nyi, T.},
+  title={Covering all points except one},
+  journal={Journal of Algebraic Combinatorics},
+  volume={32}, pages={59--66}, year={2010},
+  doi={10.1007/s10801-009-0204-1},
+  note = {\href{https://doi.org/10.1007/s10801-009-0204-1}{doi:10.1007/s10801-009-0204-1}}
+}
+
+@misc{White2026,
+  author = {White, Patrick},
+  title = {{Tuza}'s {Ryser}-conjecture claim for four-partite hypergraphs with matching number two},
+  year = {2026},
+  eprint = {2609.14281},
+  archivePrefix = {arXiv},
+  primaryClass = {math.CO},
+  note = {Version 1, 13 September 2026},
+  howpublished = {\href{https://arxiv.org/abs/2609.14281v1}{arXiv:2609.14281v1}},
+  url = {https://arxiv.org/abs/2609.14281v1}
+}
+
+
+@misc{OpenAI2026Balanced,
+  author = {{OpenAI}},
+  title = {{Balanced counterexamples to Ryser's conjecture at prime orders}},
+  howpublished = {OpenAI Math Release preprint
+                  \href{https://github.com/openai/math/blob/main/preprints/Balanced-Counterexamples-to-Rysers-Conjecture-at-Prime-Orders-September-27-2026/paper.pdf}{OAI:Balanced-Counterexamples-to-Rysers-Conjecture-at-Prime-Orders-September-27-2026}},
+  year = {2026}
+}
+
+@misc{AbiadGarbePovillSpiegel2025,
+  author = {Abiad, Aida and Garbe, Frederik and Povill, Xavier and Spiegel, Christoph},
+  title = {Infinitely many counterexamples to a conjecture of {Lov{\'a}sz}},
+  year = {2025},
+  eprint = {2506.21286},
+  archivePrefix = {arXiv},
+  primaryClass = {math.CO},
+  note = {Version 2, 14 July 2025},
+  howpublished = {\href{https://arxiv.org/abs/2506.21286v2}{arXiv:2506.21286v2}},
+  url = {https://arxiv.org/abs/2506.21286v2}
+}
+
+@article{KiralyTothmeresz2017,
+  author = {Kir{\'a}ly, Zolt{\'a}n and T{\'o}thm{\'e}r{\'e}sz, Lilla},
+  title = {On {Ryser}'s Conjecture for $t$-Intersecting and Degree-Bounded Hypergraphs},
+  journal = {The Electronic Journal of Combinatorics},
+  volume = {24},
+  number = {4},
+  pages = {P4.40},
+  year = {2017},
+  doi = {10.37236/6448},
+  eprint = {1705.10024},
+  archivePrefix = {arXiv},
+  primaryClass = {math.CO},
+  note = {\href{https://doi.org/10.37236/6448}{doi:10.37236/6448}}
+}
+
+@article{ErdosGyarfasPyber1991,
+  author = {Erd{\H{o}}s, P. and Gy{\'a}rf{\'a}s, A. and Pyber, L.},
+  title = {Vertex coverings by monochromatic cycles and trees},
+  journal = {Journal of Combinatorial Theory, Series B},
+  volume = {51},
+  number = {1},
+  pages = {90--95},
+  year = {1991},
+  doi = {10.1016/0095-8956(91)90007-7},
+  note = {\href{https://doi.org/10.1016/0095-8956(91)90007-7}{doi:10.1016/0095-8956(91)90007-7}},
+  url = {https://doi.org/10.1016/0095-8956(91)90007-7}
+}
+
+@misc{Milicevic2017,
+  author = {Mili{\'c}evi{\'c}, Luka},
+  title = {Covering complete graphs by monochromatically bounded sets},
+  year = {2017},
+  eprint = {1705.09370},
+  archivePrefix = {arXiv},
+  primaryClass = {math.CO},
+  note = {Version 1, 25 May 2017},
+  howpublished = {\href{https://arxiv.org/abs/1705.09370v1}{arXiv:\allowbreak1705.09370v1}},
+  url = {https://arxiv.org/abs/1705.09370v1}
+}
+
+@misc{DeBiasioKamelMcCourtSheats2021,
+  author = {DeBiasio, Louis and Kamel, Yigal and McCourt, Grace and Sheats, Hannah},
+  title = {Generalizations and strengthenings of {Ryser}'s conjecture},
+  year = {2021},
+  eprint = {2009.07239},
+  archivePrefix = {arXiv},
+  primaryClass = {math.CO},
+  note = {Version 3, 3 November 2021},
+  howpublished = {\href{https://arxiv.org/abs/2009.07239v3}{arXiv:2009.07239v3}},
+  url = {https://arxiv.org/abs/2009.07239v3}
+}

preprints/A-Counterexample-to-Rysers-Covering-Conjecture-September-23-2026/build/selection.tex

@@ -0,0 +1,538 @@
+\section{Selecting compatible translates}\label{sec:selection}
+
+The deletion estimates require independent random choices, while
+intersection of the exceptional edges requires geometric compatibility.
+We separate these tasks by sampling several independent translates of
+each bounded configuration and then selecting one from each pool.
+The theorem below supplies the selection step. Its input consists of
+finite point sets $W_i$ and sets $D_i$ of forbidden join directions;
+its output avoids every forbidden join between different selected sets.
+Because a direction is a linear subspace containing zero, the condition
+also rules out a common point between two selected configurations
+whenever at least one of their direction sets is nonempty.
+
+\begin{theorem}[Compatible translates from independent pools]\label{thm:selection}
+For every positive integer $D$ there is a positive integer $K=K(D)$
+with the following property.  Let $q$ tend to infinity through prime
+powers.  For each $q$, let $n\le Dq$, and for $1\le i\le n$ let
+$W_i\subseteq\F_q^2$ and let $D_i$ be a set of directions, with
+\[
+ |W_i|\le D,\qquad |D_i|\le D.
+\]
+Assume that each direction belongs to $D_i$ for at most $D$ indices $i$.
+For distinct indices $i,j$, declare the translates $b+W_i$ and
+$b'+W_j$ to conflict if there are $u\in W_i$, $v\in W_j$, and
+$d\in D_i\cup D_j$ such that
+\[
+ (b+u)-(b'+v)\in d.
+\]
+For each $i$, draw $K$ independent uniform shifts
+$b_{i,1},\ldots,b_{i,K}\in\F_q^2$, independently for all indices.
+With probability tending to one, one can choose $k_i\in\{1,\ldots,K\}$
+for every $i$ so that the translates $b_{i,k_i}+W_i$ have no conflicts.
+The convergence is uniform over all data satisfying the stated bounds.
+\end{theorem}
+
+To prove the theorem, make each candidate a vertex of a graph and
+join candidates from different pools when they conflict. We seek an
+independent set containing one vertex from each pool, called an
+independent transversal. Its nonexistence has a small domination
+certificate. We first prove the form of that assertion we need, then
+show that the independent shifts make every such certificate unlikely.
+The certificate argument is a weaker form of Haxell's
+independent-transversal domination theorem~\cite{Haxell2001},
+whose method originates in~\cite{Haxell1995}. We include its proof.
+Haxell's theorem uses total domination, which also requires neighbors
+for the vertices of the dominating set. Here we use the following
+weaker convention.
+For a graph $G$ and a set $S$ of its vertices, we say that $T\subseteq S$
+\emph{dominates $S$} if every vertex of $S\setminus T$ has a neighbor in
+$T$.  No condition is imposed on the vertices of $T$ themselves.
+
+\begin{lemma}[A domination certificate]\label{sel-domination}
+Let a finite simple graph have a vertex partition into nonempty parts
+$A_1,\ldots,A_n$.  If there is no independent set containing one
+vertex from each part, then there are a nonempty index set $I$ and a set
+\[
+ T\subseteq\bigcup_{i\in I}A_i,
+ \qquad |T|\le 2(|I|-1),
+\]
+such that $T$ dominates $\bigcup_{i\in I}A_i$.
+\end{lemma}
+
+\begin{proof}
+We grow a set of vertices with small size compared with the number
+of parts it reaches. Whenever it does not yet dominate those parts,
+we use an undominated vertex to reach another part. A minimality
+condition on the transversals prevents this growth from stopping
+without the required domination certificate.
+
+Choose an inclusion-minimal nonempty subfamily of parts with no
+independent transversal, and distinguish one part $A_*$ of this
+subfamily.  Let $\mathcal U$ be the family of independent transversals of
+all its other parts.  Minimality makes $\mathcal U$ nonempty.  The
+deficient subfamily has at least two parts, since every part is nonempty.
+
+Choose any $x_1\in A_*$.  Among $U\in\mathcal U$, minimize
+$|N(x_1)\cap U|$, where $N(x)$ denotes the neighborhood of $x$ in $G$.
+Fix a minimizer and write $Y_1=N(x_1)\cap U$.  This set is nonempty:
+otherwise adjoining $x_1$ would give an independent transversal of the
+deficient subfamily.
+
+Inductively, suppose that vertices $x_1,\ldots,x_j$ and sets
+$Y_1,\ldots,Y_j$ have been fixed with the following properties:
+\begin{enumerate}
+ \item The sets $Y_a$ are nonempty and pairwise disjoint, and some
+ $U\in\mathcal U$ jointly realizes all the equalities
+ \[
+   N(x_a)\cap U=Y_a\qquad(1\le a\le j).
+ \]
+ \item At stage $a$, the size $|Y_a|$ was minimized over all
+ $U\in\mathcal U$ satisfying the already fixed equalities for indices
+ smaller than $a$.
+ \item Each $x_a$ lies either in $A_*$ or in a part represented by a
+ vertex of some earlier $Y_b$, with $b<a$.
+\end{enumerate}
+Let $Y=\bigcup_{a=1}^jY_a$, let $I$ consist of the distinguished part
+and all parts represented in $Y$, and put
+\[
+ T=\{x_1,\ldots,x_j\}\cup Y.
+\]
+By joint realization, the vertices of $Y$ occupy distinct parts, none
+of them $A_*$.  Thus $|I|=|Y|+1$, while nonemptiness of the $Y_a$ gives
+$j\le |Y|$.  The third invariant shows that $T$ lies in the union of
+the parts indexed by $I$, and
+\[
+ |T|\le j+|Y|\le 2|Y|=2(|I|-1).
+\]
+If $T$ dominates this union, we have the required certificate.
+
+Otherwise, choose a vertex $x_{j+1}$ in that union, outside $T$, with
+no neighbor in $T$.  Among the transversals realizing all the current
+equalities, choose $U$ minimizing $|N(x_{j+1})\cap U|$, and set
+\[
+ Y_{j+1}=N(x_{j+1})\cap U.
+\]
+This set avoids all the previous $Y_a$, because $x_{j+1}$ has no
+neighbor in $T$.  We claim it is nonempty.
+
+If it were empty and $x_{j+1}\in A_*$, then $U\cup\{x_{j+1}\}$
+would be an independent transversal of the deficient subfamily.
+Otherwise, the part of $x_{j+1}$ contains a unique vertex $y$ of $Y$,
+say $y\in Y_b$.  The transversal $U$ uses $y$ in that part, so
+$x_{j+1}\notin U$.  Replace $y$ by $x_{j+1}$ to form
+\[
+ U'=(U\setminus\{y\})\cup\{x_{j+1}\}.
+\]
+Since $N(x_{j+1})\cap U=\varnothing$, this is again an independent
+transversal of the nondistinguished parts.  For every $a<b$, the vertex
+$y$ does not belong to $Y_a$, and the new vertex is not adjacent to
+$x_a$.  Consequently the exact earlier equalities are preserved:
+\[
+ N(x_a)\cap U'=Y_a\qquad(a<b).
+\]
+For $a=b$, the replacement removes the neighbor $y$ and adds no neighbor,
+so
+\[
+ |N(x_b)\cap U'|=|Y_b|-1.
+\]
+This contradicts the minimum imposed at stage $b$.  Notice that the
+contradiction uses only the equalities preceding stage $b$; preservation
+of later equalities is not required.
+
+Thus $Y_{j+1}$ is nonempty and all the induction invariants continue to
+hold.  Each continuation adds vertices in previously unrepresented
+parts to $Y$, whereas all of $Y$ lies in one transversal of the fixed
+nondistinguished subfamily.  There can be only finitely many
+continuations.  The process must therefore stop with a dominating set
+$T$, proving the lemma.
+\end{proof}
+
+\begin{proof}[Proof of Theorem~\ref{thm:selection}]
+The case $n=0$ is immediate, so suppose $n\ge1$.  Fix $D$ and, for the
+moment, a constant integer $K\ge6$.  We will choose $K$ sufficiently
+large in terms of $D$ at the end.
+
+Form a graph whose vertices are the candidate positions $(i,k)$,
+partitioned into the $n$ pools.  Put an edge between candidates in
+different pools precisely when their translates conflict.  Candidate
+positions are distinct vertices even when their sampled shifts happen
+to coincide.  An independent transversal in this graph is exactly a
+selection as in the theorem.
+
+By Lemma~\ref{sel-domination}, it suffices to show that, with probability
+tending to one, there are no nonempty index sets $I$, with $|I|=m$, and
+sets $T$ of at most $2m$ candidate positions in those pools that dominate
+their union.  We use the slightly weaker bound $2m$ to simplify the
+counting.  Throughout the probability estimates, $I$ and $T$ are fixed
+sets of indices before any shifts are exposed.
+
+The proof splits according to $m$, the number of pools in the
+certificate. When $m$ is small, the forbidden lines associated with
+$T$ occupy only a small part of each candidate's possible shifts.
+When $m$ is proportional to $q$, that direct bound is insufficient.
+Instead, we expose $T$ in a bounded number of chunks and prove that
+a positive average proportion of shifts remains feasible. Crucially,
+this proportion will depend only on $D$, so we may subsequently choose
+$K$ large enough that the unused candidates defeat all certificates.
+
+\paragraph{Forbidden equations and counting possible certificates.}
+For every direction $d$ in use, fix a nonzero linear functional
+$\pi_d:\F_q^2\to\F_q$ whose kernel is $d$.  After exposing the shifts
+of the candidates $e\in T$, call a shift $b\in\F_q^2$ \emph{feasible
+for target $i$} if, for every such $e$, with source-pool index $j$, it
+avoids every equation
+\begin{equation}\label{sel-forbidden}
+ \pi_d(b)=\pi_d(b_e)+\pi_d(v-u),
+ \qquad
+ u\in W_i,\quad v\in W_j,\quad d\in D_i\cup D_j.
+\end{equation}
+We impose these prohibitions even when $i=j$.  This only makes
+feasibility more restrictive.  A feasible candidate outside $T$ has no
+neighbor in $T$, so its existence prevents domination.
+
+Set
+\begin{equation}\label{sel-constants}
+ C=2D^3,\qquad c_1=\frac1{4C},\qquad \alpha=\frac1{2C}.
+\end{equation}
+For one target $i$ and one exposed candidate $e$, there are at most $C$
+equations in \eqref{sel-forbidden}.  This remains true if the two lists
+$D_i$ and $D_j$ are counted separately, retaining duplicate directions.
+Each equation excludes one affine line, containing exactly $q$ shifts.
+
+The number of choices of $I$ is at most
+\[
+ \binom{n}{m}\le\left(\frac{eDq}{m}\right)^m.
+\]
+There is an absolute constant $C_*>0$ such that, for $K\ge6$, the number
+of candidate-index sets $T$ in these pools with $|T|\le2m$ is at most
+\begin{equation}\label{sel-certificate-count}
+ \sum_{t=0}^{2m}\binom{Km}{t}\le (C_*K^2)^m.
+\end{equation}
+For example, the usual bound
+$\sum_{t=0}^{r}\binom Nt\le(eN/r)^r$ for $1\le r\le N/2$
+applies with $N=Km$ and $r=2m$.
+
+\paragraph{The range $1\le m\le c_1q$.}
+Condition on all the exposed shifts in $T$.  A fresh uniform shift for
+any target $i\in I$ is infeasible with probability at most
+\[
+ \frac{C|T|}{q}\le\frac{2Cm}{q}\le\frac12.
+\]
+At least $(K-2)m$ candidate positions in the target pools remain
+unexposed.  Their shifts are conditionally independent and uniform.
+Thus the probability that $T$ dominates, for this fixed $I,T$, is at most
+\begin{equation}\label{sel-small-probability}
+ \left(\frac{2Cm}{q}\right)^{(K-2)m}.
+\end{equation}
+Combining this with the counts of $I$ and $T$, the probability of a
+dominating certificate of size parameter $m$ is at most
+\begin{equation}\label{sel-small-sum}
+ \left[
+  2CeD C_*K^2
+  \left(\frac{2Cm}{q}\right)^{K-3}
+ \right]^m.
+\end{equation}
+Choose $K$ large enough that
+\begin{equation}\label{sel-small-K}
+ 2CeD C_*K^2\,2^{-(K-3)}\le\frac12.
+\end{equation}
+Then the bracket in \eqref{sel-small-sum} is at most $1/2$ throughout
+this range, and tends to zero for every fixed positive $m$.  Summing
+over $m$ gives $o(1)$: for a fixed cutoff, the finite initial sum tends
+to zero, while the remaining terms are bounded by a geometric tail.
+
+\paragraph{A feasible-density estimate for $m>c_1q$.}
+We prove that there is a constant $\delta_1>0$, depending only on $D$
+and not on $K$, such that for every fixed $I,T$ in this range,
+\begin{equation}\label{sel-density-target}
+ \Pr\!\left[
+   \sum_{i\in I}|U_i^{\mathrm{fin}}|<\delta_1mq^2
+ \right]
+ \le \exp\bigl(-\Omega_D(q\log q)\bigr).
+\end{equation}
+Here $U_i^{\mathrm{fin}}$ is the set of shifts feasible for target $i$
+after all candidates in $T$ have been exposed.  The estimate is uniform
+in $I,T$ and the configuration data.  The lower cutoff on $q$ may
+depend on the fixed value of $K$.
+
+Expose the candidate positions of $T$ in a fixed order, divided into
+consecutive chunks of size at most $\lfloor\alpha q\rfloor$, with all
+but the last having that size.  For sufficiently large $q$,
+$\lfloor\alpha q\rfloor\ge\alpha q/2$.  Since $|T|\le2m\le2Dq$,
+the number of chunks is at most
+\begin{equation}\label{sel-chunk-count}
+ J_0=\left\lceil\frac{4D}{\alpha}+1\right\rceil.
+\end{equation}
+In particular, this bound does not involve $K$.  If $T$ is empty, the
+desired feasible-density estimate holds deterministically; otherwise
+there is at least one chunk.
+
+At the start of a chunk, condition on all shifts previously exposed.
+Let $U_i$ be the set of shifts still feasible for target $i$, and write
+\[
+ M=\sum_{i\in I}|U_i|.
+\]
+For the analysis of this chunk we keep all these sets frozen at their
+start values.  They are now deterministic, while the shifts belonging
+to the new chunk are independent uniform vectors.
+
+For an entry $e$ of the chunk, with source-pool index $j$, define $Z_e$
+by summing, over all equations \eqref{sel-forbidden} associated with
+$i\in I$ and $e$, the number of shifts $b\in U_i$ satisfying that
+equation.  Count the $D_i$ and $D_j$ lists separately, including
+duplicates.  A shift deleted more than once is therefore counted more
+than once, which is harmless for an upper bound on loss.  Each line
+contains at most $q$ shifts, and each of its right-hand sides is uniform
+in $\F_q$.  Consequently,
+\begin{equation}\label{sel-Z-basic}
+ 0\le Z_e\le Cmq,
+ \qquad \E Z_e\le \frac{CM}{q}.
+\end{equation}
+All expectations and variances in the following chunk calculation are
+conditional on the fixed past.
+
+\paragraph{The terms whose directions belong to the target.}
+The expectation bound alone does not control the loss in a chunk.
+We estimate its fluctuations by separating the two direction lists
+in each forbidden equation. Bounded reuse of target directions handles
+one list; the affine second moment will handle the source list.
+Write
+\[
+ Z_e=Z_e^{(1)}+Z_e^{(2)},
+\]
+where $Z_e^{(1)}$ uses the list $D_i$ in each target, and $Z_e^{(2)}$
+uses the list $D_j$ of the source.  Group the terms of $Z_e^{(1)}$ by
+direction.  The group for $d$ is a function only of $\pi_d(b_e)$.
+For distinct directions $d,d'$, the functionals $\pi_d,\pi_{d'}$ are
+linearly independent, so
+\[
+ b_e\longmapsto\bigl(\pi_d(b_e),\pi_{d'}(b_e)\bigr)
+\]
+is a bijection from $\F_q^2$ to itself.  The two coordinates are
+therefore independent and uniform.  The direction groups are pairwise
+independent, which suffices for adding their variances.
+
+There are at most $Dm$ such groups.  Each direction belongs to at most
+$D$ target direction sets; for each associated target-source pair
+there are at most $D^2$ choices of $(u,v)$.  Thus every group is bounded
+by $D^3q$, and
+\begin{equation}\label{sel-first-variance}
+ \operatorname{Var}(Z_e^{(1)})\le D^7mq^2.
+\end{equation}
+
+\paragraph{The terms whose directions belong to the source.}
+For each target $i$ and each direction $d$, define the variance of its
+parallel line slices by
+\begin{equation}\label{sel-slice-variance}
+ V_{i,d}=\frac1q\sum_{\ell\parallel d}
+ \left(|U_i\cap\ell|-\frac{|U_i|}{q}\right)^2,
+\end{equation}
+where the sum is over the $q$ affine lines of direction $d$.
+These variances obey the exact identity
+\begin{equation}\label{sel-energy}
+ \sum_d V_{i,d}
+   =|U_i|-\frac{|U_i|^2}{q^2}
+   \le |U_i|\le q^2,
+\end{equation}
+where the sum runs over all $q+1$ directions. This is precisely the
+normalized affine second-moment identity
+in~\cite[Lemma~A.1]{OpenAI2026Balanced}, applied to $U_i\subseteq\F_q^2$.
+That identity holds for every subset of every finite affine plane;
+it has no randomness, characteristic, or subfield hypothesis. Thus it
+applies after conditioning on any history of earlier chunks. This is
+the reason we can control all source directions even when the feasible
+sets $U_i$ have complicated shapes.
+
+Every individual summand of $Z_e^{(2)}$ associated with $i,d,u,v$ is a
+uniformly sampled line-slice count, with its slice index translated by
+a fixed offset.  Its variance is therefore $V_{i,d}$.  There are at
+most $D^3m$ summands, and each pair $(i,d)$ is repeated at most $D^2$
+times.  Cauchy--Schwarz applied to their centered sum, without any
+independence assumption, gives
+\begin{equation}\label{sel-second-variance}
+ \operatorname{Var}(Z_e^{(2)})
+ \le D^5m A_e,
+ \qquad
+ A_e:=\sum_{i\in I}\sum_{d\in D_j}V_{i,d}.
+\end{equation}
+
+Each source pool contributes at most $K$ entries to the chunk, and
+each direction belongs to at most $D$ source direction sets.
+Consequently, by \eqref{sel-energy},
+\begin{equation}\label{sel-exception-energy}
+ \sum_{e\text{ in the chunk}} A_e
+ \le KD\sum_{i\in I}\sum_d V_{i,d}
+ \le KDmq^2.
+\end{equation}
+Call an entry \emph{regular} if $A_e\le mq^{3/2}$.  There are at most
+$KD\sqrt q$ irregular entries.  For a regular entry,
+\eqref{sel-first-variance}, \eqref{sel-second-variance}, and
+$\operatorname{Var}(X+Y)\le2\operatorname{Var}(X)+2\operatorname{Var}(Y)$
+give
+\begin{equation}\label{sel-regular-variance}
+ \operatorname{Var}(Z_e)
+ \le 2D^7mq^2+2D^5m^2q^{3/2}.
+\end{equation}
+For a fixed constant $\gamma>0$, Chebyshev's inequality and
+\eqref{sel-Z-basic} therefore imply
+\begin{align}
+ \Pr\!\left[Z_e>\frac{CM}{q}+\gamma mq\right]
+ &\le \frac{2D^7}{\gamma^2m}
+       +\frac{2D^5}{\gamma^2\sqrt q}
+ \notag\\
+ &\le B_{D,\gamma}q^{-1/2}
+ \qquad\text{for regular }e,
+ \label{sel-regular-tail}
+\end{align}
+where $B_{D,\gamma}$ is independent of $K$.  In the last step we used
+$m>c_1q$.
+
+\paragraph{Conditional independence and loss over one chunk.}
+The preceding variance bounds show that each regular entry has a
+small chance of removing far more than its expected share. We now
+use independence within the chunk to show that few entries do so,
+and convert that statement into a deterministic bound on the total
+remaining feasible set.
+
+The regularity of an entry is determined by the frozen sets $U_i$ and
+its fixed source-pool index.  In particular, it does not depend on its
+fresh shift $b_e$.  After conditioning on the past, the collection of
+regular entries is fixed, and their variables $Z_e$ are independent
+functions of independent shifts.  Their common threshold in
+\eqref{sel-regular-tail} is also fixed under this conditioning.
+
+A union bound over subsets of regular entries, using the chunk size
+bound and \eqref{sel-regular-tail}, shows that the probability of at
+least $\gamma q$ threshold exceedances is at most
+\begin{equation}\label{sel-chunk-tail}
+ 2^{\lceil\alpha q\rceil}
+ \left(B_{D,\gamma}q^{-1/2}\right)^{\lceil\gamma q\rceil}
+ \le \exp(-c_{D,\gamma}q\log q)
+\end{equation}
+for all sufficiently large $q$, with $c_{D,\gamma}>0$ independent of
+$K$.  If there are fewer than $\lceil\gamma q\rceil$ regular entries,
+the event in question is empty and the bound remains valid.
+
+For each fixed $K$, we may also take $q$ sufficiently large that
+\begin{equation}\label{sel-K-cutoff}
+ KD\sqrt q\le\gamma q.
+\end{equation}
+Except on the event in \eqref{sel-chunk-tail}, all but at most
+$2\gamma q$ entries are regular and below the threshold: the exceptions
+include all irregular entries.  The actual loss of feasible shifts
+within the chunk is at most $\sum_eZ_e$, because these counts used the
+larger feasible sets at the start of the chunk.  Bounding each
+exceptional entry by $Cmq$ and all other entries by the threshold gives,
+for the end-of-chunk value $M'$, the deterministic inequality
+\begin{align}
+ M'
+ &\ge M-C\alpha M-\alpha\gamma mq^2-2C\gamma mq^2
+ \notag\\
+ &=(1-C\alpha)M-(\alpha+2C)\gamma mq^2
+ \notag\\
+ &=\frac12M-(\alpha+2C)\gamma mq^2.
+ \label{sel-chunk-recurrence}
+\end{align}
+
+Choose $\gamma>0$, depending only on $D$, so small that
+\begin{equation}\label{sel-gamma-choice}
+ 2(\alpha+2C)\gamma\le 2^{-J_0-1},
+ \qquad
+ \delta_1:=2^{-J_0-1}.
+\end{equation}
+At the beginning, before exposing $T$, every shift is feasible, so
+$M=mq^2$.  If all $J\le J_0$ chunks satisfy
+\eqref{sel-chunk-recurrence}, iteration gives
+\begin{align*}
+ \frac{M_{\mathrm{fin}}}{mq^2}
+ &\ge 2^{-J}-(\alpha+2C)\gamma
+               \sum_{a=0}^{J-1}2^{-a}\\
+ &\ge 2^{-J_0}-2(\alpha+2C)\gamma
+ \ge\delta_1.
+\end{align*}
+The conditional probability bound \eqref{sel-chunk-tail} holds for
+every possible past.  Averaging over the past and taking a union bound
+over the at most $J_0$ chunks proves \eqref{sel-density-target}.
+The constants $J_0$, $\gamma$, and $\delta_1$ depend only on $D$.
+The dependence on $K$ in \eqref{sel-K-cutoff} affects only how large
+$q$ must be, not these constants.
+
+\paragraph{Unused candidates and the large-range union bound.}
+We have established the required positive average feasible density
+with a failure probability of order $\exp(-\Omega_D(q\log q))$.
+There are only exponentially many candidate certificates, so this
+error can be summed over all of them. On the successful event, the
+unused candidates now supply the further exponential saving needed
+to rule out domination.
+
+Condition now on all the shifts belonging to $T$, and suppose the
+feasible-density event in \eqref{sel-density-target} succeeds.  Let
+$t_i$ be the number of exposed candidate positions from pool $i$.
+Every unused candidate from that pool remains independent and uniform,
+and is feasible with probability $|U_i^{\mathrm{fin}}|/q^2$.
+The probability that none of the unused candidates in the target pools
+is feasible is therefore at most
+\begin{align}
+ \prod_{i\in I}
+ \left(1-\frac{|U_i^{\mathrm{fin}}|}{q^2}\right)^{K-t_i}
+ &\le
+ \exp\!\left(-\sum_{i\in I}
+       (K-t_i)\frac{|U_i^{\mathrm{fin}}|}{q^2}\right)
+ \notag\\
+ &\le \exp\bigl(-(K\delta_1-2)m\bigr).
+ \label{sel-unused-candidates}
+\end{align}
+To obtain the last inequality, use
+\[
+ \sum_{i\in I}\frac{|U_i^{\mathrm{fin}}|}{q^2}\ge\delta_1m,
+ \qquad
+ 0\le\frac{|U_i^{\mathrm{fin}}|}{q^2}\le1,
+ \qquad
+ \sum_{i\in I}t_i=|T|\le2m.
+\]
+This argument also covers pools for which every candidate belongs to
+$T$; it does not assume $K-t_i$ is positive in every target pool.
+
+For $m>c_1q$, the number of pairs $I,T$ is at most $A_K^m$, where
+\begin{equation}\label{sel-large-count}
+ A_K=\frac{eD}{c_1}C_*K^2.
+\end{equation}
+For each fixed pair, domination can occur only if the feasible-density
+estimate fails or all unused candidates are infeasible despite its
+success.  The sum of the first failure probabilities over all
+$m,I,T$ in the large range is bounded by
+\[
+ \exp\bigl(O_{D,K}(q)\bigr)
+ \exp\bigl(-\Omega_D(q\log q)\bigr)=o(1).
+\]
+Here the number of choices is at most
+$\sum_{m\le Dq}A_K^m=\exp(O_{D,K}(q))$, with $K$ fixed.
+
+Finally choose $K$ sufficiently large, still depending only on $D$, so
+that both \eqref{sel-small-K} and
+\begin{equation}\label{sel-large-K}
+ A_K\exp(2-K\delta_1)<\frac12
+\end{equation}
+hold.  Such a choice exists because $\delta_1>0$ does not depend on $K$,
+whereas $A_K$ grows only quadratically in $K$.  By
+\eqref{sel-unused-candidates}, the sum of the remaining failure
+probabilities is then at most
+\[
+ \sum_{m>c_1q}
+ \left(A_K\exp(2-K\delta_1)\right)^m
+ \le\sum_{m>c_1q}2^{-m}=o(1).
+\]
+Together with the small-range estimate, this excludes all dominating
+certificates with probability tending to one.  Lemma~\ref{sel-domination}
+therefore supplies an independent transversal, proving the theorem.
+
+The choices have been made in the order
+\[
+ D\ ;\quad C,c_1,\alpha,J_0,\gamma,\delta_1\ ;\quad K\ ;\quad q.
+\]
+In particular, $K$ is fixed before taking $q$ to infinity, and all
+sufficiently-large-$q$ conditions may depend on that fixed $K$.
+Every estimate above is uniform over the configurations and direction
+sets subject to the bounds in the theorem.
+\end{proof}

preprints/A-Counterexample-to-Rysers-Covering-Conjecture-September-23-2026/paper.pdf

@@ -0,0 +1,3 @@
+version https://git-lfs.github.com/spec/v1
+oid sha256:ba76ca1f7af545c68090e0cfa1defba423af058cb4c0ac6087e563b5123b1eb7
+size 544159

preprints/A-Counterexample-to-Walls-D2-Problem-October-6-2026/README.md

@@ -0,0 +1,18 @@
+# [A Counterexample to Wall's D(2) Problem](wall-d2-counterexample.pdf)
+
+**Author:** OpenAI
+
+**Date:** October 6, 2026
+
+## Citation
+
+```bibtex
+@misc{OAI:A-Counterexample-to-Walls-D2-Problem-October-6-2026,
+  author = {{OpenAI}},
+  title = {{A Counterexample to Wall's $D(2)$ Problem}},
+  howpublished = {OpenAI Math Release preprint
+                  \href{https://github.com/openai/math/blob/main/preprints/A-Counterexample-to-Walls-D2-Problem-October-6-2026/wall-d2-counterexample.pdf}{OAI:A-Counterexample-to-Walls-D2-Problem-October-6-2026}},
+  year = {2026}
+}
+```
+

preprints/A-Counterexample-to-Walls-D2-Problem-October-6-2026/build/figures/tower.tex

@@ -0,0 +1,20 @@
+\begin{tikzpicture}[
+  >=Latex,
+  every node/.style={font=\normalsize},
+  every edge/.style={draw,->,semithick},
+  baseline=(current bounding box.center)
+]
+  \node (F) at (0,0) {$F$};
+  \node (next) at (3,1.1) {$A_{i+1}$};
+  \node (cover) at (6.2,1.1) {$\widehat A_i$};
+  \node (base) at (6.2,-1.1) {$A_i$};
+  \path (F) edge node[above,sloped] {$f_{i+1}$} (next);
+  \draw[semithick,{Hooks[right]}-Latex]
+    (next) -- node[above] {inclusion} (cover);
+  \path (cover) edge node[right] {$p_i$} (base);
+  \node[anchor=west,font=\small,align=left] at (6.85,0)
+    {infinite cyclic\\cover};
+  \path (F) edge node[below,sloped] {$f_i$} (base);
+  \node[font=\small] at (2.85,-1.85)
+    {$s(A_i)<s(A_{i+1})\leq s(F)$};
+\end{tikzpicture}

preprints/A-Counterexample-to-Walls-D2-Problem-October-6-2026/build/main.tex

@@ -0,0 +1,20 @@
+\documentclass[11pt]{article}
+\input{preamble}
+\title{A Counterexample to Wall's \texorpdfstring{$D(2)$}{D(2)} Problem}
+\author{OpenAI}
+\date{October 6, 2026}
+\begin{document}
+\maketitle
+\begin{abstract}
+We give a negative answer to Wall's finite $D(2)$ problem. We construct
+a finite connected three-dimensional CW complex satisfying the $D(2)$
+finiteness condition but not homotopy equivalent to any finite CW complex
+of dimension at most two. The example has infinite fundamental group.
+\end{abstract}
+\input{sections/01-introduction}
+\input{sections/02-tower}
+\input{sections/03-plus}
+\input{sections/04-construction}
+\bibliographystyle{alpha}
+\bibliography{references}
+\end{document}

preprints/A-Counterexample-to-Walls-D2-Problem-October-6-2026/build/preamble.tex

@@ -0,0 +1,27 @@
+\usepackage[T1]{fontenc}
+\usepackage{lmodern}
+\usepackage[margin=1in]{geometry}
+\usepackage{amsmath,amssymb,amsthm,mathtools}
+\usepackage{microtype,enumitem,needspace,placeins,xcolor,tikz,booktabs}
+\usetikzlibrary{arrows.meta,positioning,calc}
+\usepackage[colorlinks=true,linkcolor=blue!45!black,citecolor=blue!45!black,urlcolor=blue!45!black]{hyperref}
+\hypersetup{pdftitle={A Counterexample to Wall's D(2) Problem},pdfauthor={OpenAI}}
+\newtheorem{theorem}{Theorem}[section]
+\newtheorem{proposition}[theorem]{Proposition}
+\newtheorem{lemma}[theorem]{Lemma}
+\newtheorem{corollary}[theorem]{Corollary}
+\theoremstyle{definition}
+\newtheorem{definition}[theorem]{Definition}
+\newtheorem{remark}[theorem]{Remark}
+\newcommand{\C}{\mathbb C}
+\newcommand{\Z}{\mathbb Z}
+\newcommand{\Q}{\mathbb Q}
+\newcommand{\RP}{\mathbb{RP}}
+\newcommand{\Hom}{\operatorname{Hom}}
+\newcommand{\id}{\operatorname{id}}
+\newcommand{\im}{\operatorname{im}}
+\newcommand{\ab}{\mathrm{ab}}
+\newcommand{\rk}{\operatorname{rank}}
+\setlist[enumerate]{label=\textup{(\roman*)},leftmargin=*}
+\setlength{\emergencystretch}{1.5em}
+\allowdisplaybreaks[1]

preprints/A-Counterexample-to-Walls-D2-Problem-October-6-2026/build/references.bib

@@ -0,0 +1,148 @@
+@article{Wall65,
+  author = {Wall, C. T. C.},
+  title = {Finiteness conditions for {CW}-complexes},
+  journal = {Annals of Mathematics. Second Series},
+  volume = {81},
+  number = {1},
+  year = {1965},
+  pages = {56--69},
+  doi = {10.2307/1970382}
+}
+
+@incollection{Wall79,
+  author = {Wall, C. T. C.},
+  title = {List of problems},
+  booktitle = {Homological Group Theory},
+  editor = {Wall, C. T. C.},
+  series = {London Mathematical Society Lecture Note Series},
+  volume = {36},
+  publisher = {Cambridge University Press},
+  address = {Cambridge},
+  year = {1979},
+  pages = {369--394}
+}
+
+@book{Johnson03,
+  author = {Johnson, F. E. A.},
+  title = {Stable Modules and the {D(2)}-Problem},
+  series = {London Mathematical Society Lecture Note Series},
+  volume = {301},
+  publisher = {Cambridge University Press},
+  address = {Cambridge},
+  year = {2003},
+  doi = {10.1017/CBO9780511550256}
+}
+
+@incollection{Cohen78,
+  author = {Cohen, J. M.},
+  title = {Complexes of cohomological dimension two},
+  booktitle = {Algebraic and Geometric Topology},
+  series = {Proceedings of Symposia in Pure Mathematics},
+  volume = {32},
+  publisher = {American Mathematical Society},
+  address = {Providence, RI},
+  year = {1978},
+  pages = {221--223},
+  note = {Part 2}
+}
+
+@article{Hambleton18,
+  author = {Hambleton, Ian},
+  title = {Two remarks on {Wall's D2} problem},
+  journal = {Mathematical Proceedings of the Cambridge Philosophical Society},
+  volume = {167},
+  year = {2019},
+  pages = {361--368}
+}
+
+@article{HofmannNicholson27,
+  author = {Hofmann, Tommy and Nicholson, John},
+  title = {Exotic presentations of quaternion groups and {Wall's D2} problem},
+  journal = {Journal of Algebra},
+  volume = {712},
+  year = {2027},
+  pages = {253--310},
+  doi = {10.1016/j.jalgebra.2026.06.040}
+}
+
+@article{Mannan09,
+  author = {Mannan, W. H.},
+  title = {{Quillen's} plus construction and the {D(2)} problem},
+  journal = {Algebraic \& Geometric Topology},
+  volume = {9},
+  number = {3},
+  year = {2009},
+  pages = {1399--1411},
+  doi = {10.2140/agt.2009.9.1399}
+}
+
+@article{Fox53,
+  author = {Fox, Ralph H.},
+  title = {Free differential calculus. {I}. {Derivation} in the free group ring},
+  journal = {Annals of Mathematics. Second Series},
+  volume = {57},
+  number = {3},
+  year = {1953},
+  pages = {547--560},
+  doi = {10.2307/1969736}
+}
+
+@article{Howie81Pairs,
+  author = {Howie, James},
+  title = {On pairs of 2-complexes and systems of equations over groups},
+  journal = {Journal f\"ur die reine und angewandte Mathematik},
+  volume = {324},
+  year = {1981},
+  pages = {165--174},
+  doi = {10.1515/crll.1981.324.165}
+}
+
+@article{LouderWilton17,
+  author = {Louder, Larsen and Wilton, Henry},
+  title = {Stackings and the {W}-cycles conjecture},
+  journal = {Canadian Mathematical Bulletin},
+  volume = {60},
+  number = {3},
+  year = {2017},
+  pages = {604--612}
+}
+
+@incollection{Quillen71,
+  author = {Quillen, Daniel},
+  title = {Cohomology of groups},
+  booktitle = {Actes du Congr\`es International des Math\'ematiciens (Nice, 1970)},
+  volume = {2},
+  publisher = {Gauthier-Villars},
+  address = {Paris},
+  year = {1971},
+  pages = {47--51}
+}
+
+@article{BridsonTweedale07,
+  author = {Bridson, Martin R. and Tweedale, Michael},
+  title = {Deficiency and abelianized deficiency of some virtually free groups},
+  journal = {Mathematical Proceedings of the Cambridge Philosophical Society},
+  volume = {143},
+  year = {2007},
+  pages = {257--264}
+}
+
+@article{MannanPopiel21,
+  author = {Mannan, W. H. and Popiel, Tomasz},
+  title = {An exotic presentation of {$Q_{28}$}},
+  journal = {Algebraic \& Geometric Topology},
+  volume = {21},
+  number = {4},
+  year = {2021},
+  pages = {2065--2084}
+}
+
+@article{Nicholson21,
+  author = {Nicholson, John},
+  title = {On {CW}-complexes over groups with periodic cohomology},
+  journal = {Transactions of the American Mathematical Society},
+  volume = {374},
+  number = {9},
+  year = {2021},
+  pages = {6531--6557}
+}

preprints/A-Counterexample-to-Walls-D2-Problem-October-6-2026/build/sections/01-introduction.tex

@@ -0,0 +1,131 @@
+\section{Introduction}\label{sec:introduction}
+
+Wall's finiteness conditions ask when algebraic restrictions on a CW
+complex force it to have a homotopy model of a prescribed dimension.
+Let $X$ be a finite connected CW complex, let $G=\pi_1(X)$, and let
+$\widetilde X$ be its universal cover. The $D(2)$ condition is
+\begin{equation}\label{eq:d2-condition}
+ H_i(\widetilde X;\Z)=0\quad(i>2),\qquad
+ H^3(X;M)=0\quad\text{for every finitely generated }\Z G\text{-module }M,
+\end{equation}
+where cohomology uses the corresponding local coefficients.
+The finite $D(2)$ problem asks whether every such $X$ is homotopy
+equivalent to a finite CW complex of dimension at most two
+\cite[Section~2]{Wall65}, \cite[Problem~D3]{Wall79}.
+The finiteness of both complexes is part of the question, and no
+asphericity hypothesis is imposed. This is also the formulation in
+Hofmann--Nicholson~\cite[p.~254]{HofmannNicholson27}.
+
+\begin{theorem}\label{thm:main}
+There is a finite connected three-dimensional CW complex $X$ such that
+\[
+ H_i(\widetilde X;\Z)=0\quad(i>2),\qquad
+ H^3(X;M)=0\quad\text{for every }\Z[\pi_1(X)]\text{-module }M,
+\]
+but $X$ is not homotopy equivalent to any finite CW complex of
+dimension at most two.
+\end{theorem}
+
+The example has infinite fundamental group; the finiteness assertion
+concerns its cell structure.
+
+\subsection{Background and approach}
+
+Wall introduced the dimension conditions in his work on finiteness
+for CW complexes~\cite{Wall65}. His reduction produces a finite
+three-dimensional model in the $D(2)$ case; the remaining question is
+whether its three-dimensional cells can also be removed up to homotopy.
+The problem reappeared as Problem~D3 in his 1979 list~\cite{Wall79}.
+It lies at the interface between two-dimensional topology and the
+realization of chain complexes over group rings. Johnson's work
+\cite{Johnson03} develops this algebraic viewpoint and the role of
+cancellation in passing from a stable realization to an actual
+two-dimensional complex.
+
+A stable form of the problem has a positive answer: adjoining
+sufficiently many two-spheres makes a finite $D(2)$ complex homotopy
+equivalent to a finite two-dimensional complex. This goes back to
+Cohen~\cite{Cohen78}; Hambleton~\cite[Lemma~2.1]{Hambleton18} gives
+a proof with the number of added spheres bounded by the number of
+three-dimensional cells. Such stabilization does not answer the
+unstabilized question. There are also substantial positive results
+for particular fundamental groups. Hambleton~\cite[Theorem~B]{Hambleton18}
+proves the $D(2)$ property for finite subgroups of $\operatorname{SO}(3)$.
+More recently, Hofmann and Nicholson~\cite[Theorem~A]{HofmannNicholson27}
+establish the $D(2)$ property for the generalized quaternion groups of
+orders $24$ and $32$ and recover the order-$28$ case previously proved
+using work of Mannan--Popiel and Nicholson~\cite{MannanPopiel21,Nicholson21}.
+In particular,
+they show that the longstanding candidate
+with fundamental group of order $32$, proposed by Cohen and Dyer,
+does have a finite two-dimensional model.
+
+Candidate counterexamples have also come from relation modules.
+A relation gap occurs when the relator subgroup of a free group needs
+more normal generators than its abelianization needs as a module over
+the quotient group ring.
+Bridson and Tweedale~\cite[Sections~3--4]{BridsonTweedale07} use
+coprime annihilation relations to reduce the number of module generators
+and construct finite $D(2)$ complexes with virtually free fundamental
+groups. Their failure to have finite two-dimensional models is conditional
+on a relation-gap assertion.
+
+We apply Quillen's plus construction~\cite{Quillen71}. It kills a
+perfect normal subgroup of the fundamental
+group while preserving homology with coefficients from the quotient.
+Mannan~\cite[Section~2]{Mannan09} shows directly that applying it to a
+finite two-dimensional complex along a finitely normally generated
+perfect subgroup produces a finite $D(2)$ complex. He also proves
+that every finite cohomologically two-dimensional three-dimensional
+complex arises in this way up to homotopy
+\cite[Theorem~3.4]{Mannan09}. Thus this construction supplies the
+dimension condition; the task is to find an obstruction to a
+two-dimensional model of its result.
+
+Our obstruction uses a character. For a connected space $Y$, a
+homomorphism $\rho:\pi_1(Y)\to\C^\times$ determines a
+one-dimensional complex local system $L_\rho$. We prove that, when
+$Y$ is a finite two-dimensional complex,
+\begin{equation}\label{eq:obstruction-overview}
+ H_2(Y;L_\rho)=0
+ \quad\Longrightarrow\quad
+ \rho(g)=1\ \text{for every finite-order }g\in\pi_1(Y).
+\end{equation}
+The proof gives the stronger statement that every map $F\to Y$ from
+a finite connected simplicial complex with $b_1(F;\Q)=0$ pulls $L_\rho$
+back to a system with trivial monodromy. The method is a tower of
+infinite cyclic covers and image subcomplexes, following the cyclic-tower
+method of Howie~\cite{Howie81Pairs}; see also the formulation of
+Louder--Wilton~\cite[Section~3]{LouderWilton17} and its account of
+the earlier work of Papakyriakopoulos and Stallings. The additional input
+is that vanishing of rank-one twisted second homology persists under
+these cyclic covers, by a Laurent-polynomial rank argument.
+
+\subsection{Proof structure}
+
+Section~\ref{sec:tower} establishes this obstruction independently
+of the construction. If a map $F\to Y$ violated it, its finite image
+would have nonpositive Euler characteristic and hence admit an
+infinite cyclic cover. The map lifts because $b_1(F;\Q)=0$.
+Twisted second homology still vanishes upstairs, so the argument
+can be repeated. Each lifted image contains strictly more simplices,
+although all are images of the same fixed triangulated $F$. This
+forces a contradiction.
+
+Section~\ref{sec:plus} presents the finite plus construction with
+its explicit cellular splitting. Section~\ref{sec:construction}
+then gives a five-generator, four-relator presentation and a perfect
+normal subgroup generated normally by two powers. The quotient has
+an element of order two detected by a complex character. An exact
+four-row boundary calculation shows that the presentation complex
+has zero second homology with that character. The plus construction
+preserves this vanishing and gives the required finite $D(2)$
+complex. Equation~\eqref{eq:obstruction-overview} excludes its
+two-dimensional realization.
+
+The geometric obstruction and the finite plus construction play separate
+roles: the former applies to any rank-one local system on a finite
+two-dimensional complex, while the latter is standard once a suitable
+perfect normal subgroup is available. Their combination makes the
+failure of two-dimensional realization visible in a small twisted
+cellular boundary matrix.

preprints/A-Counterexample-to-Walls-D2-Problem-October-6-2026/build/sections/02-tower.tex

@@ -0,0 +1,257 @@
+\section{A tower obstruction in dimension two}
+\label{sec:tower}
+
+We first isolate the obstruction to a finite two-dimensional model.
+Its input is the vanishing of second homology with a rank-one local
+system.  Its conclusion restricts the maps into the complex from finite
+complexes with vanishing first rational Betti number.  The case of a
+Moore complex will then show that the corresponding character cannot
+detect a finite-order element.
+
+A \emph{rank-one complex local system} $L$ on a connected CW complex $T$
+assigns a one-dimensional complex vector space to each point, with
+parallel transport along paths that depends only on homotopy relative to
+the endpoints.  After choosing a basepoint and a basis of its fiber,
+its monodromy is a character
+\[
+  \rho:\pi_1(T)\longrightarrow\C^\times.
+\]
+We use ordinary homology with these coefficients.  In module notation,
+write $\C_\rho$ for the right $\Z[\pi_1(T)]$-module with
+$v\cdot g=\rho(g)v$.  If the universal-cover cellular chains carry the
+left deck action, then
+\[
+  C_*(T;L)
+  =\C_\rho\otimes_{\Z[\pi_1(T)]}C_*(\widetilde T;\Z).
+\]
+Thus cellular chains have finite support, even when $T$ is infinite.
+Equivalently, $C_j(T;L)$ is the direct sum of one coefficient fiber for
+each oriented $j$-cell, with boundary maps defined by parallel transport.
+
+Two elementary consequences will be used throughout the proof.  First,
+if $A\subseteq T$ is a subcomplex and $\dim T\leq2$, then
+\begin{equation}
+  H_2(A;L|_A)\longrightarrow H_2(T;L)
+  \quad\text{is injective}.
+  \label{eq:subcomplex-injection}
+\end{equation}
+Indeed, the cellular chain inclusion is injective, and in dimension at
+most two the second homology is the kernel of the second boundary.
+This argument requires no injectivity of $\pi_1(A)\to\pi_1(T)$.
+Second, on a connected complex the zeroth homology is the space of
+coinvariants,
+\[
+  H_0(T;L)
+  =\C/\langle (\rho(g)-1)v:g\in\pi_1(T),\ v\in\C\rangle.
+\]
+Consequently, nontrivial monodromy implies $H_0(T;L)=0$.
+
+The next lemma ensures that vanishing of second homology also survives
+the covering operation needed in the tower.
+
+\begin{lemma}[Infinite cyclic covers]
+\label{lem:cyclic-cover}
+Let $A$ be a finite connected CW complex of dimension at most two, let
+$L$ be a rank-one complex local system on $A$, and let
+$p:\widehat A\to A$ be a connected infinite cyclic cover.  If
+$H_2(A;L)=0$, then
+\[
+  H_2(\widehat A;p^*L)=0.
+\]
+\end{lemma}
+
+\begin{proof}
+Choose a generator $\tau$ of the deck group and put
+$R=\C[z,z^{-1}]$.  There is a canonical deck action on the pulled-back
+local system: in the fiber description of $p^*L$, it is
+\begin{equation}
+  (\widehat x,v)\longmapsto(\tau\widehat x,v),
+  \qquad v\in L_{p(\widehat x)}.
+  \label{eq:canonical-deck-action}
+\end{equation}
+The equality $p\tau=p$ makes this action compatible with parallel
+transport.  Let $z$ act by this deck transformation.  Choosing one lift
+of each cell of $A$ and a basis in its coefficient fiber identifies
+$C_j(\widehat A;p^*L)$ with a finite free $R$-module, of rank equal to
+the number of $j$-cells of $A$.
+
+Taking coinvariants of this deck action identifies translated cell
+lifts and their coefficient fibers.  It therefore gives an isomorphism
+of chain complexes
+\begin{equation}
+  C_*(\widehat A;p^*L)\otimes_R R/(z-1)
+  \cong C_*(A;L).
+  \label{eq:cyclic-specialization}
+\end{equation}
+In the chosen bases, let $D(z)$ be the matrix of the second boundary
+upstairs, and let $m$ be its number of columns.  The matrix $D(1)$ is
+the second boundary downstairs, so it is injective by the hypothesis.
+If $m=0$, there is nothing to prove.  Otherwise, some $m\times m$ minor
+of $D(1)$ is nonzero.  The corresponding minor of $D(z)$ is consequently
+a nonzero Laurent polynomial.  Thus $D(z)$ has full column rank over
+$\C(z)$ and is injective over $R$, since $R^m$ embeds in $\C(z)^m$.
+There are no cells above dimension two, and hence
+$H_2(\widehat A;p^*L)=\ker D(z)=0$.
+\end{proof}
+
+The following proof adapts the cyclic-tower method of
+Howie~\cite{Howie81Pairs}, in which one alternates subcomplex inclusions
+and infinite cyclic covers; see Louder--Wilton~\cite[Section~3]{LouderWilton17}.
+Lemma~\ref{lem:cyclic-cover} supplies the property that persists along
+the tower here: vanishing of twisted second homology.
+
+\begin{theorem}[Tower obstruction]
+\label{thm:tower}
+Let $Y$ be a finite connected CW complex of dimension at most two, and
+let $L$ be a rank-one complex local system on $Y$ with $H_2(Y;L)=0$.
+There is no map $f:F\to Y$ from a finite connected simplicial complex
+satisfying $b_1(F;\Q)=0$ for which $f^*L$ has nontrivial monodromy.
+\end{theorem}
+
+\begin{proof}
+Suppose that such a map exists.  We will repeatedly lift it to an
+infinite cyclic cover and replace that cover by the image of the lift.
+At every step the image will contain strictly more simplices, although
+the domain retains one fixed finite triangulation.  For a finite
+simplicial complex $A$, write $s(A)$ for its number of nonempty simplices.
+
+We first justify working simplicially without increasing the dimension
+of $Y$.  Its one-skeleton is a finite graph.  Subdivide that graph to
+make it simplicial and homotope each two-cell attaching map to an edge
+path.  After further subdivision if necessary, each nonconstant
+attaching path can be represented by a map from a polygonal circle
+with at least three edges, with every edge mapping homeomorphically
+onto a graph edge.  Replace the attachment by the mapping cylinder of this
+map, capped by a disk along its free boundary circle.  This merely
+inserts a collar into the attached disk.  The resulting cylinder has
+one rectangle for each traversal of a graph edge.  Each rectangle is
+embedded, with its two upper vertices in the separate domain circle
+and its two lower vertices at distinct endpoints of the graph edge.
+After compatible subdivisions of boundary edges, triangulate each
+rectangle from its own interior vertex and each capping disk from a
+new interior vertex.  This yields a finite simplicial complex even
+when different rectangles traverse the same graph edge: their
+interiors and upper edges remain distinct.  A constant attaching map
+contributes a triangulated two-sphere joined to the graph at one
+vertex.  Homotoping the attaching maps preserves the homotopy type,
+so this construction gives a finite simplicial model of dimension at
+most two.
+
+Transfer $L$ and $f$ across this homotopy equivalence.  Homotopy
+invariance preserves both $H_2(Y;L)=0$ and the nontrivial pullback
+monodromy.  After a sufficiently fine subdivision of $F$, simplicial
+approximation makes $f$ simplicial.  Fix this triangulation of $F$
+for the rest of the proof.  Let $A_0=f(F)$, let $f_0:F\to A_0$ be the
+corestriction, and let $L_0=L|_{A_0}$.  The simplicial image $A_0$ is
+a finite connected subcomplex, and
+\eqref{eq:subcomplex-injection} gives $H_2(A_0;L_0)=0$.
+
+Here is the inductive step.  Suppose that $A_i$ is a finite connected
+simplicial complex of dimension at most two, $f_i:F\to A_i$ is a
+simplicial map onto its image $A_i$, and $L_i$ satisfies
+\begin{equation}
+  H_2(A_i;L_i)=0,
+  \qquad f_i^*L_i\text{ has nontrivial monodromy}.
+  \label{eq:tower-invariants}
+\end{equation}
+The second condition implies that $L_i$ itself has nontrivial
+monodromy, so $H_0(A_i;L_i)=0$.  Since rank-one cellular chains have
+the same dimensions as ordinary complex cellular chains, the Euler
+characteristic gives
+\[
+  \chi(A_i)=-\dim_\C H_1(A_i;L_i)\leq0.
+\]
+On the other hand,
+\[
+  \chi(A_i)=1-b_1(A_i;\Q)+b_2(A_i;\Q).
+\]
+It follows that $b_1(A_i;\Q)\geq1$.  The finitely generated
+abelianization of $\pi_1(A_i)$ therefore has a nonzero free part,
+and there is an epimorphism
+$\phi_i:\pi_1(A_i)\twoheadrightarrow\Z$.
+
+Let $p_i:\widehat A_i\to A_i$ be the connected cover corresponding to
+$\ker\phi_i$.  The composite $\phi_i(f_i)_*$ is zero: every
+homomorphism $\pi_1(F)\to\Z$ factors through $H_1(F;\Z)$, which is
+finite because $F$ is finite and $b_1(F;\Q)=0$.
+Thus $f_i$ lifts to $\widetilde f_i:F\to\widehat A_i$.
+Give the cover the lifted triangulation.  The lift is simplicial on
+the already fixed triangulation of $F$, since on each domain simplex
+it is the lift of a simplicial map into a single target simplex.
+No new subdivision of $F$ is needed.
+
+Set
+\[
+  A_{i+1}=\widetilde f_i(F),
+  \qquad L_{i+1}=(p_i^*L_i)|_{A_{i+1}},
+\]
+and corestrict the lift to $f_{i+1}:F\to A_{i+1}$.
+Lemma~\ref{lem:cyclic-cover}, followed by
+\eqref{eq:subcomplex-injection}, gives
+$H_2(A_{i+1};L_{i+1})=0$.  Moreover,
+$f_{i+1}^*L_{i+1}\cong f_i^*L_i$, so the nontrivial monodromy
+persists.  These are precisely the conditions
+\eqref{eq:tower-invariants} needed to continue.  Figure~\ref{fig:tower}
+shows one step.
+
+\begin{figure}[tb]
+  \centering
+  \input{figures/tower}
+  \caption{One step of the tower.  The lifted map factors through its
+  finite image $A_{i+1}\subseteq\widehat A_i$, and the diagram commutes.
+  The domain $F$ keeps the same triangulation at every step; only the
+  indicated subcomplex arrow is an inclusion.}
+  \label{fig:tower}
+\end{figure}
+
+It remains to show that $s(A_i)$ increases strictly.  The restriction
+$p_i|_{A_{i+1}}$ maps onto $A_i$ and maps each
+simplex isomorphically onto a simplex.  Hence
+$s(A_{i+1})\geq s(A_i)$.  If equality held, there would be exactly
+one simplex above each simplex of $A_i$, compatibly with all faces.
+The restriction would then be a simplicial isomorphism.  Its inverse,
+followed by the inclusion into $\widehat A_i$, would give a section
+of $p_i$.  Such a section would make $(p_i)_*$ surjective, contrary
+to
+\[
+  \im(p_i)_*=\ker\phi_i\ne\pi_1(A_i).
+\]
+Thus $s(A_{i+1})>s(A_i)$.  But every simplex of $A_i$ is the image
+of a simplex in the fixed triangulation of $F$, so
+\[
+  s(A_0)<s(A_1)<s(A_2)<\cdots\leq s(F).
+\]
+This is impossible, and the theorem follows.
+\end{proof}
+
+\Needspace{5\baselineskip}
+The following consequence is the form used in the construction.
+
+\begin{corollary}[Characters and torsion]
+\label{cor:torsion}
+Let $Y$ be a finite connected CW complex of dimension at most two.
+If a character $\rho:\pi_1(Y)\to\C^\times$ satisfies
+$H_2(Y;\C_\rho)=0$, then $\rho(g)=1$ for every finite-order
+element $g\in\pi_1(Y)$.
+\end{corollary}
+
+\begin{proof}
+Suppose that $g$ has finite order $n$ and $\rho(g)\ne1$; in
+particular, $n\geq2$.  Let
+\[
+  M_n=S^1\cup_{\deg n}D^2
+\]
+be the Moore complex obtained by attaching a disk along a map of
+degree $n$.  A loop in $Y$ representing $g$ extends to a map
+$M_n\to Y$, because its $n$th power is nullhomotopic.  The cellular
+boundary $C_2(M_n;\Z)\to C_1(M_n;\Z)$ is multiplication by $n$,
+so $H_1(M_n;\Z)=\Z/n$ and $b_1(M_n;\Q)=0$.  The pullback
+character sends the circle generator to $\rho(g)\ne1$.
+
+Use the finite simplicial model for $M_n$ constructed in the proof of
+Theorem~\ref{thm:tower}, and compose its homotopy equivalence to
+$M_n$ with the map to $Y$.  This gives a finite connected simplicial
+domain with vanishing first rational Betti number and nontrivial
+pullback monodromy, contradicting that theorem.  For $n=2$ the
+standard model $M_2$ is $\RP^2$, the case needed below.
+\end{proof}

preprints/A-Counterexample-to-Walls-D2-Problem-October-6-2026/build/sections/03-plus.tex

@@ -0,0 +1,117 @@
+\section{A finite plus construction}\label{sec:plus}
+
+We next recall how a finitely normally generated perfect subgroup
+produces a finite $D(2)$ complex. This is the finite form of Quillen's
+plus construction~\cite{Quillen71} used by Mannan~\cite[Section~2]{Mannan09}.
+We give the construction because its cellular splitting proves both
+the required cohomological vanishing and the preservation of the twisted
+homology used in Section~\ref{sec:tower}.
+
+Recall that a group $K$ is \emph{perfect} if $K=[K,K]$, or equivalently
+if its abelianization $K_{\ab}$ is zero. A normal subgroup of a group
+$P$ is \emph{finitely normally generated} if it is the normal closure
+in $P$ of finitely many elements. This condition does not require
+finite generation as an abstract group.
+
+\begin{proposition}[Finite plus construction]\label{prop:plus}
+Let $B$ be a finite connected CW complex of dimension at most two,
+with $\pi_1(B)=P$. Suppose that the normal closure $K$ of elements
+$t_1,\ldots,t_r\in P$ is perfect, and put $G=P/K$.
+There is a finite complex $X\supseteq B$, obtained by attaching
+$r$ cells of dimension two and $r$ cells of dimension three, such that:
+\begin{enumerate}
+\item the inclusion induces the quotient $P\to G$ on fundamental groups;
+\item $H_i(\widetilde X;\Z)=0$ for every $i>2$, and
+      $H^3(X;M)=0$ for every left $\Z G$-module $M$;
+\item for every right $\Z G$-module $N$, the inclusion induces
+      $H_*(B;N)\cong H_*(X;N)$, with the coefficients on $B$
+      pulled back along $P\to G$.
+\end{enumerate}
+\end{proposition}
+
+\begin{proof}
+Attach two-dimensional cells along loops representing $t_1,\ldots,t_r$,
+and call the resulting complex $B'$. The van Kampen theorem gives
+$\pi_1(B')=G$. Let $\overline B$ be the full preimage of $B$ in the
+universal cover $\widetilde{B'}$. Since $P\to G$ is onto,
+$\overline B$ is connected; it is the covering of $B$ corresponding
+to $K$. Hence
+\[
+ H_1(\overline B;\Z)=K_{\ab}=0.
+\]
+
+Write $R=\Z G$ and regard cellular chains of these covers as left
+$R$-modules. The relative cellular complex
+$C_*(\widetilde{B'},\overline B)$ consists of $R^r$ in degree two
+and zero in every other degree. The homology exact sequence of the
+pair therefore gives a surjection
+\begin{equation}\label{eq:relative-surjection}
+ H_2(\widetilde{B'};\Z)
+ \longrightarrow H_2(\widetilde{B'},\overline B;\Z)=R^r.
+\end{equation}
+Choose classes $\alpha_1,\ldots,\alpha_r$ mapping to the standard
+basis vectors, where the basis records chosen lifts and orientations
+of the new cells. The space $\widetilde{B'}$ is simply connected, so
+the degree-two Hurewicz theorem represents each $\alpha_j$ by a map
+$S^2\to\widetilde{B'}$. Project these maps to $B'$ and attach one
+three-dimensional cell along each projected map. This defines $X$.
+Only finitely many cells have been added, and attaching the last cells
+does not change the fundamental group.
+
+The preimage of $B'$ in $\widetilde X$ is its universal cover, so we
+may retain the notation $\widetilde{B'}$ and $\overline B$. In the
+chosen cellular bases,
+\[
+ C_3(\widetilde X)=R^r,
+ \qquad C_2(\widetilde X)=C_2(\overline B)\oplus R^r.
+\]
+Let $q:C_2(\widetilde X)\to R^r$ be projection onto the second
+summand. The choice in \eqref{eq:relative-surjection} says exactly that
+\begin{equation}\label{eq:split-boundary}
+ q\,\partial_3=\id_{R^r}.
+\end{equation}
+Indeed, the relative cellular boundary of the $j$th three-dimensional
+cell is the $j$th basis vector; in the two-dimensional complex
+$\widetilde{B'}$, a homology class is represented by its cellular
+two-cycle without any ambiguity from three-boundaries.
+
+Equation~\eqref{eq:split-boundary} proves that $\partial_3$ is
+injective. There are no cells above dimension three, so
+$H_i(\widetilde X;\Z)=0$ for all $i>2$. It also gives the
+cohomological assertion directly. For any $R$-linear map
+$\varphi:C_3(\widetilde X)\to M$, its extension
+$\varphi q:C_2(\widetilde X)\to M$ satisfies
+\[
+ (\varphi q)\partial_3=\varphi.
+\]
+Thus the cochain boundary
+$\Hom_R(C_2(\widetilde X),M)\to\Hom_R(C_3(\widetilde X),M)$
+is onto, and $H^3(X;M)=0$.
+
+Finally, the relative cellular complex of $(\widetilde X,\overline B)$
+is
+\begin{equation}\label{eq:relative-complex}
+ 0\longrightarrow R^r\xrightarrow{\,\id\,}R^r
+   \longrightarrow0,
+\end{equation}
+in degrees three and two. It is contractible. The short exact sequence
+of absolute and relative cellular complexes is split in each degree,
+since the cells of $\overline B$ form subsets of the chosen cellular
+bases. Tensoring with any right $R$-module $N$ therefore leaves a short
+exact sequence of complexes whose relative term is still contractible.
+Its homology exact sequence gives
+\[
+ H_*(N\otimes_R C_*(\overline B))
+   \cong H_*(N\otimes_R C_*(\widetilde X)).
+\]
+The left side computes $H_*(B;N)$: the coefficient action of $P$
+factors through the deck group $G$ of $\overline B\to B$.
+The right side computes $H_*(X;N)$, proving the last assertion.
+\end{proof}
+
+The splitting in \eqref{eq:split-boundary} is stronger than ordinary
+third-homology vanishing: it makes the cohomology calculation work for
+every coefficient module. In the application, we will choose $B$ and
+$K$ so that a character of $G$ has vanishing second homology on $B$,
+yet detects a torsion element of $G$. Proposition~\ref{prop:plus}
+preserves that homology vanishing when passing to $X$.

preprints/A-Counterexample-to-Walls-D2-Problem-October-6-2026/build/sections/04-construction.tex

@@ -0,0 +1,189 @@
+\section{The counterexample}\label{sec:construction}
+
+We now construct the presentation complex and perfect normal subgroup
+needed in Proposition~\ref{prop:plus}. The quotient must contain torsion
+detected by a character, and the corresponding local system on the
+presentation complex must have vanishing second homology.
+Perfectness will follow from
+annihilation relations with coprime coefficients in the subgroup's
+abelianization.
+Related uses of coprime annihilators for relation modules appear in
+Bridson--Tweedale~\cite[Proposition~3.3]{BridsonTweedale07}.
+
+Let $B$ be the presentation complex of
+\begin{equation}\label{eq:presentation}
+ P=\left\langle x_1,a_1,x_2,a_2,s\ \middle|\
+ \begin{aligned}
+ x_1a_1x_1^{-1}&=a_1^4,&
+ x_2a_2x_2^{-1}&=a_2^3,\\
+ x_1&=a_2^{13},&
+ s x_2 s^{-1}&=a_1^5
+ \end{aligned}\right\rangle.
+\end{equation}
+Thus $B$ has one vertex, five edges, and four two-dimensional cells.
+Define
+\[
+ t_1=x_1^2,\qquad t_2=x_2^3,\qquad
+ K=\langle\!\langle t_1,t_2\rangle\!\rangle_P,
+ \qquad G=P/K,
+\]
+where the double brackets denote normal closure in $P$.
+Sending $s$ to $1\in\Z$ and the other four generators to zero
+respects all these relations and descends to an epimorphism
+$G\to\Z$. In particular, $G$ is infinite.
+We use the commutator convention $[g,h]=ghg^{-1}h^{-1}$.
+
+\begin{lemma}\label{lem:perfect}
+The normal subgroup $K$ is perfect.
+\end{lemma}
+
+\begin{proof}
+The exponents are matched by $4^2-1=15=3\cdot5$ and
+$3^3-1=26=2\cdot13$, with $\gcd(15,26)=1$.
+Conjugation by $t_1=x_1^2$ sends $a_1$ to $a_1^{16}$, while
+conjugation by $t_2=x_2^3$ sends $a_2$ to $a_2^{27}$.
+Squaring $x_1=a_2^{13}$ and cubing $s x_2s^{-1}=a_1^5$ now give
+\begin{equation}\label{eq:commutators}
+ t_1=a_2^{26}=[t_2,a_2],
+ \qquad
+ s t_2 s^{-1}=a_1^{15}=[t_1,a_1].
+\end{equation}
+Write $K_{\ab}$ additively and let $u_j$ be the image of $t_j$.
+Conjugation makes $K_{\ab}$ a left $\Z G$-module: every element
+of $K$ acts trivially on its own abelianization. For $t\in K$,
+the image of $[t,a]$ is $(1-a)[t]$. Consequently
+\begin{equation}\label{eq:module-relations}
+ u_1=(1-a_2)u_2,
+ \qquad s u_2=(1-a_1)u_1.
+\end{equation}
+
+Since $t_1=a_2^{26}$, the element $a_2$ fixes $u_1$ and
+$a_2^{26}$ acts trivially on all of $K_{\ab}$. Applying
+$1+a_2+\cdots+a_2^{25}$ to the first equality gives
+\[
+ 26u_1=(1-a_2^{26})u_2=0.
+\]
+Likewise, $a_1$ fixes $s u_2$, because $s t_2s^{-1}=a_1^{15}$;
+and $a_1^{15}$ acts trivially on $K_{\ab}$. Applying
+$1+a_1+\cdots+a_1^{14}$ to the second equality gives
+\[
+ 15s u_2=(1-a_1^{15})u_1=0.
+\]
+The action of $s$ is invertible, so $15u_2=0$, and the first
+equality of \eqref{eq:module-relations} then gives $15u_1=0$.
+Since $15$ and $26$ are relatively prime, $u_1=0$. The second
+equality gives $s u_2=0$, hence $u_2=0$.
+
+Every element of $K$ is a product of $P$-conjugates of $t_1^{\pm1}$
+and $t_2^{\pm1}$. Thus $u_1,u_2$ generate $K_{\ab}$ under
+addition and the $P$-action. Their vanishing proves $K_{\ab}=0$.
+\end{proof}
+
+By Proposition~\ref{prop:plus}, there is a finite complex $X\supseteq B$
+with fundamental group $G$, obtained by adding two two-dimensional
+and two three-dimensional cells, such that
+\begin{equation}\label{eq:d2-for-x}
+ H_i(\widetilde X;\Z)=0\quad(i>2),\qquad
+ H^3(X;M)=0\quad\text{for every left }\Z G\text{-module }M.
+\end{equation}
+It remains to find the character that obstructs a two-dimensional model.
+
+Choose a primitive third root of unity $\zeta\in\C$. The assignments
+\begin{equation}\label{eq:character}
+ \rho(x_1)=-1,\quad \rho(a_1)=\zeta,\quad
+ \rho(x_2)=\zeta^2,\quad \rho(a_2)=-1,\quad \rho(s)=1
+\end{equation}
+satisfy the four relations of \eqref{eq:presentation}, since
+\[
+ \zeta=\zeta^4,\qquad -1=(-1)^3,\qquad
+ -1=(-1)^{13},\qquad \zeta^2=\zeta^5.
+\]
+They also send $t_1$ and $t_2$ to $1$. Hence they define a character
+$\rho:G\to\C^\times$. Denote its local system on $X$ and its
+pullback to $B$ by $L_\rho$.
+
+\Needspace{6\baselineskip}
+\begin{lemma}\label{lem:twisted-vanishing}
+The local system $L_\rho$ satisfies $H_2(X;L_\rho)=0$.
+\end{lemma}
+
+\begin{proof}
+Proposition~\ref{prop:plus} reduces the computation to $B$.
+Its twisted cellular groups in degrees two and one are $\C^4$ and
+$\C^5$, respectively. We compute the second boundary by the lifted
+edge-path sum, equivalently by Fox differentiation~\cite{Fox53}.
+In this convention the derivative of a word obeys
+\[
+ \frac{\partial(uv)}{\partial g}
+   =\frac{\partial u}{\partial g}
+      +u\frac{\partial v}{\partial g},\qquad
+ \frac{\partial g}{\partial g}=1,\qquad
+ \frac{\partial g^{-1}}{\partial g}=-g^{-1},
+\]
+and the derivatives with respect to other generators are zero.
+Evaluate their group-ring coefficients at $\rho$.
+
+Use the relators
+\[
+ \begin{split}
+ r_1&=x_1a_1x_1^{-1}a_1^{-4},\qquad
+ r_2=x_2a_2x_2^{-1}a_2^{-3},\\
+ r_3&=x_1a_2^{-13},\qquad
+ r_4=sx_2s^{-1}a_1^{-5}.
+ \end{split}
+\]
+With the edges ordered as $x_1,a_1,x_2,a_2,s$, the four boundary
+vectors, written as rows, are
+\begin{equation}\label{eq:fox-matrix}
+ D=\begin{pmatrix}
+ 1-\zeta & -2 & 0 & 0 & 0\\
+ 0&0&2&\zeta^2-1&0\\
+ 1&0&0&-1&0\\
+ 0&-(1+\zeta)&1&0&1-\zeta^2
+ \end{pmatrix}.
+\end{equation}
+Here is a direct check of the entries. For a relator
+$xax^{-1}a^{-n}$, the evaluated entries in the $x$ and $a$
+coordinates are
+\[
+ 1-\rho(a)^n,
+ \qquad \rho(x)-\sum_{j=0}^{n-1}\rho(a)^j.
+\]
+For $r_1$, the latter sum is
+$1+\zeta+\zeta^2+\zeta^3=1$; for $r_2$, it is $1-1+1=1$.
+For $r_3$, the alternating sum of thirteen terms is $1$.
+For $r_4$, the $a_1$ entry is
+$-\sum_{j=0}^4\zeta^j=-(1+\zeta)$, the $x_2$ entry is
+$\rho(s)=1$, and the $s$ entry is $1-\rho(x_2)=1-\zeta^2$.
+These give \eqref{eq:fox-matrix}.
+
+The minor using columns $x_1,a_1,x_2,s$ has determinant
+\begin{equation}\label{eq:nonzero-minor}
+ -4(1-\zeta^2)\ne0.
+\end{equation}
+Thus the four rows are independent, so the boundary
+$C_2(B;L_\rho)\to C_1(B;L_\rho)$ is injective.
+Since $B$ has no three-dimensional cells, $H_2(B;L_\rho)=0$,
+as required.
+\end{proof}
+
+\begin{proof}[Proof of Theorem~\ref{thm:main}]
+The complex $X$ constructed above is finite and satisfies
+\eqref{eq:d2-for-x}. In $G$, the element $x_1$ has square one,
+and it is nonidentity because $\rho(x_1)=-1$. Thus its order is
+exactly two.
+
+Suppose that $X$ were homotopy equivalent to a finite connected
+complex $Y$ of dimension at most two. Transfer $\rho$ along the
+induced isomorphism of fundamental groups, giving a local system
+$L$ on $Y$. Homotopy invariance and Lemma~\ref{lem:twisted-vanishing}
+give $H_2(Y;L)=0$. But the transferred character takes the value
+$-1$ on an element of order two, contrary to
+Corollary~\ref{cor:torsion}.
+
+Concretely, a loop representing that element extends to a map
+$\RP^2\to Y$: the attaching loop of the two-dimensional cell of
+$\RP^2$ is its generator traversed twice. The pulled-back local
+system has monodromy $-1$, whereas $b_1(\RP^2;\Q)=0$.
+This is precisely the map excluded by Theorem~\ref{thm:tower}.
+\end{proof}

preprints/A-Counterexample-to-Walls-D2-Problem-October-6-2026/wall-d2-counterexample.pdf

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+version https://git-lfs.github.com/spec/v1
+oid sha256:e34de3716e64bba49880fff61808d5065477785596703bc0debf2379ad246c0c
+size 338287

preprints/A-Counterexample-to-the-Group-Ring-Determinant-Conjecture-September-23-2026/README.md

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+# [A Counterexample to the Group-Ring Determinant Conjecture](paper.pdf)
+
+**Author:** OpenAI
+
+**Date:** September 23, 2026
+
+## Citation
+
+```bibtex
+@misc{OAI:A-Counterexample-to-the-Group-Ring-Determinant-Conjecture-September-23-2026,
+  author = {{OpenAI}},
+  title = {{A Counterexample to the Group-Ring Determinant Conjecture}},
+  howpublished = {OpenAI Math Release preprint
+                  \href{https://github.com/openai/math/blob/main/preprints/A-Counterexample-to-the-Group-Ring-Determinant-Conjecture-September-23-2026/paper.pdf}{OAI:A-Counterexample-to-the-Group-Ring-Determinant-Conjecture-September-23-2026}},
+  year = {2026}
+}
+```
+

preprints/A-Counterexample-to-the-Group-Ring-Determinant-Conjecture-September-23-2026/build/main.tex

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+\documentclass[11pt]{article}
+\usepackage[T1]{fontenc}
+\usepackage{lmodern}
+\usepackage[margin=1in]{geometry}
+\usepackage{amsmath,amssymb,amsthm,mathtools}
+\usepackage{microtype}
+\usepackage{xcolor}
+\PassOptionsToPackage{hyphens}{url}
+\usepackage{hyperref}
+\usepackage{bookmark}
+\hypersetup{colorlinks=true,linkcolor=blue!45!black,citecolor=blue!45!black,
+  urlcolor=blue!45!black,pdfauthor={OpenAI},
+  pdftitle={A Counterexample to the Group-Ring Determinant Conjecture}}
+\input{glyphtounicode-cmex.tex}
+\pdfgentounicode=1
+\newtheorem{theorem}{Theorem}[section]
+\newtheorem{proposition}[theorem]{Proposition}
+\newtheorem{lemma}[theorem]{Lemma}
+\theoremstyle{remark}
+\newtheorem{remark}[theorem]{Remark}
+\numberwithin{equation}{section}
+\newcommand{\F}{\mathbb F}
+\newcommand{\Z}{\mathbb Z}
+\newcommand{\Q}{\mathbb Q}
+\newcommand{\R}{\mathbb R}
+\newcommand{\C}{\mathbb C}
+\newcommand{\Mat}{\operatorname{M}}
+\newcommand{\diag}{\operatorname{diag}}
+\newcommand{\N}{\mathcal N}
+\setlength{\emergencystretch}{2em}
+\title{A Counterexample to the Group-Ring Determinant Conjecture}
+\author{OpenAI}
+\date{September 23, 2026}
+
+\begin{document}
+\maketitle
+\begin{abstract}
+We disprove the unrestricted group-ring Determinant Conjecture. We construct a finitely generated group $G$ and a square matrix over $\mathbb Z[G]$ that is invertible over $\mathbb Q[G]$ and has Fuglede--Kadison determinant strictly between zero and one. The logarithmic integral defining the determinant is finite.
+\end{abstract}
+
+\section{Introduction and proof overview}
+\label{sec:introduction}
+
+A nonsingular integer matrix has determinant of absolute
+value at least one. The group-ring Determinant Conjecture asks for an
+analogue when the entries are finite integer combinations of elements
+of a discrete group. The relevant determinant, introduced by Fuglede
+and Kadison~\cite{FK1952}, uses the trace of a logarithm rather than
+a finite-dimensional determinant polynomial.
+
+Here are the conventions that fix this question. Let $G$ be a discrete
+group and let $\Z[G]$ denote its ordinary group ring of finite sums.
+Use the left regular representation
+\[
+ (L_g\xi)(h)=\xi(g^{-1}h),\qquad \xi\in\ell^2(G).
+\]
+For $A\in\Mat_n(\Z[G])$, write $T_A$ for its operator on
+$\ell^2(G)^n$ and $e_i$ for the vector with $\delta_1$ in component
+$i$. The probability spectral measure of the positive square is
+\[
+ \mu_A(E)=\frac1n\sum_{i=1}^n
+ \langle \mathbf1_E(T_A^*T_A)e_i,e_i\rangle
+ \quad(E\subseteq[0,\infty)\text{ Borel}).
+\]
+The square-matrix instance of the conjecture is
+\begin{equation}\label{eq:conjecture}
+ \int_{(0,\infty)}\log t\,d\mu_A(t)\ge0.
+\end{equation}
+The zero atom is omitted and the general integral may be $-\infty$.
+The usual conjecture also permits rectangular matrices
+\cite[Conjecture 13.2]{Luck2002}; a square counterexample already
+contradicts it. Our example will have a bounded inverse, so its
+spectrum stays away from zero.
+
+Write $\N(G)$ for the group von Neumann algebra and use the
+\emph{unnormalized} matrix trace
+$\tau_n(T)=\sum_i\langle Te_i,e_i\rangle$, so $\tau_n(I_n)=n$.
+For invertible $T_A$ the determinant convention is
+\begin{equation}\label{eq:normalization}
+ \log\det_{\N(G)}(T_A)
+ =\frac12\tau_n\bigl(\log(T_A^*T_A)\bigr)
+ =\frac n2\int_{(0,\infty)}\log t\,d\mu_A(t).
+\end{equation}
+This extends the absolute value of the ordinary determinant.
+The inversion unitary $\delta_g\mapsto\delta_{g^{-1}}$ identifies
+our integral-matrix convention with right multiplication by $A^*$
+on row vectors. It fixes the identity vectors, and therefore
+preserves the corresponding positive-square spectral measures.
+
+\begin{theorem}\label{thm:main}
+There are a finitely generated discrete group $G_D$, an integer
+$n\ge1$, and $A\in\Mat_n(\Z[G_D])$ invertible over $\Q[G_D]$ such
+that
+\[
+ 0<\det_{\N(G_D)}(T_A)<1.
+\]
+The integral in Equation~\eqref{eq:conjecture} is finite and strictly
+negative. Thus the unrestricted group-ring Determinant Conjecture
+has a negative resolution.
+\end{theorem}
+
+\paragraph{The exact existence input.}
+A unital ring is directly finite if $ab=1$ implies $ba=1$.
+We use the following scalar failure of direct finiteness from the
+complete characteristic-two companion.
+
+\begin{theorem}[{\cite[Theorem 1.1]{CharacteristicTwo}}]
+\label{thm:input}
+There exist a finite field $K_2$ of characteristic two, a finitely
+generated group $H_2$, and finite sums $a_2,b_2\in K_2[H_2]$ with
+$a_2b_2=1$ and $b_2a_2\ne1$.
+\end{theorem}
+
+The companion's Section~6 specifies its field, group presentation
+and scalar sums by a terminating finite prescription. We use the
+existence theorem, not an executed numerical instance or a
+group-word decision procedure. The coefficient field is finite but
+is not asserted to be $\F_2$. There is no torsion-free hypothesis;
+the companion construction uses torsion. All steps after this exact
+input are proved below. In particular, a result only in odd
+characteristic would not supply the binary premise used here.
+
+\paragraph{Context and mathematical ancestry.}
+Determinant bounds are part of the theory of $L^2$-invariants
+\cite{Luck2002}. Approximation and closure arguments established
+such bounds for large classes of groups, including amenable groups
+\cite[Theorem 1.21, with its erratum]{Schick2001}; Elek and Szab\'o proved the
+determinant bound for sofic groups
+\cite[Theorem 5]{ElekSzabo2005}. Their proof uses finite graph
+approximations and the integrality of products of nonzero eigenvalues
+of positive integral matrices. Our construction instead starts from
+failure of direct finiteness in characteristic two. The cited sofic theorem implies, as a
+separate consequence, that its resulting group is not sofic.
+
+The lower-bound question is distinct from determinant
+\emph{approximation}. Kammeyer's counterexample to a finite-quotient
+limiting formula has determinant $2$ over the discrete Heisenberg
+group, while its principal congruence quotients of levels $3^i$, $i\ge1$,
+have determinants at most $5^{1/3}$
+\cite[Theorem 2 and Section 2]{Kammeyer2026}. That discrepancy does not violate
+the lower bound one.
+
+The finite-coordinate projection construction below is related to
+the commuting wreath-product projections of Dicks and Schick
+\cite[Section 3]{DicksSchick2002} and of Lehner, Neuhauser and
+Woess~\cite[Section 2]{LehnerNeuhauserWoess2008}. Here the
+projections separate one coefficient of a binary defect; we do not
+import a spectral-distribution theorem from those works. The
+analytic step is the classical tracial logarithmic derivative
+\cite[Lemma 2]{FK1952}, also present in the path-integral determinant
+viewpoint of de la Harpe and Skandalis
+\cite[Lemma 1 and Proposition 2(d)]{deLaHarpeSkandalis1984}. We give its proof and its
+application to block shears locally.
+
+\paragraph{How a binary defect produces a real trace gap.}
+Set $d=[K_2:\F_2]$. Representing multiplication by field elements
+on the $d$-dimensional binary vector space $K_2$ gives matrices
+$X,Y\in\Mat_d(\F_2[H_2])$ with $XY=I_d$ and a nonzero idempotent
+$C=I_d-YX$. One entry of $C$ has a nonzero coefficient. We enlarge
+the group to
+\[
+ G_D=\left(\bigoplus_{h\in H_2}C_3\right)\rtimes H_2,
+\]
+where $H_2$ acts by left translation on the coordinate labels. Averaging over the
+cyclic group at one coordinate and taking complementary averages
+at finitely many other coordinates isolates that coefficient.
+The result is a rational orthogonal projection $P$ of size $d+1$
+with trace $d+\theta$, for a positive rational number $\theta$.
+Its reduction modulo two nevertheless factors through $d$ columns.
+Section~\ref{sec:projection} proves the coefficient separation and
+this factorization in full.
+
+\paragraph{How the gap becomes an integral determinant.}
+For a general rational projection $P$ of size $m$, suppose an odd
+integer $D$ clears its denominators and $\bar P=uv$ factors through
+$r$ columns over $\F_2[G]$. Choose a power $q$ of two congruent to
+one modulo $D$. Section~\ref{sec:lifting} lifts the factorization modulo $q$
+by a finite geometric series in a corner ring. Arbitrary integral
+lifts then give an integral block matrix with an explicit rational
+inverse. Section~\ref{sec:analysis} removes its two block shears
+without changing the trace of the logarithm of its positive square.
+The remaining diagonal matrix expands the first $r$ coordinate
+summands by $q$ and contracts the range of
+$P$ by $q^{-1}$, giving
+\[
+ \log\det(T_A)=(r-\tau_m(P))\log q.
+\]
+Finally Section~\ref{sec:assembly} uses $r=d$ and
+$\tau_{d+1}(P)=d+\theta$ to obtain the exact value $q^{-\theta}$.
+The projection construction, modular lifting and trace calculation
+have distinct roles; none replaces the scalar existence input.
+
+\section{From the scalar defect to a projection}
+\label{sec:projection}
+
+We first produce the trace gap. All group-ring sums in this section
+are finite, and matrix multiplication keeps the indicated order.
+For a group-ring sum $c$, write $[1]c$ for its identity coefficient.
+The involution $*$ inverts group elements and conjugates coefficients;
+on matrices it also transposes the indices. Thus an orthogonal
+projection satisfies $P=P^*=P^2$ in the regular representation.
+
+\begin{lemma}\label{lem:binary}
+Let $K_2$ be a finite field of characteristic two and $H_2$ a group.
+For $d=[K_2:\F_2]$, there is an injective unital ring homomorphism
+\[
+ \rho:K_2[H_2]\longrightarrow\Mat_d(\F_2[H_2]).
+\]
+For $a_2,b_2\in K_2[H_2]$ with $a_2b_2=1\ne b_2a_2$, their images $X,Y$ satisfy
+$XY=I_d$ and $C:=I_d-YX\ne0$, with
+\begin{equation}\label{eq:defect}
+ C^2=C,\qquad XC=0,\qquad CY=0.
+\end{equation}
+\end{lemma}
+
+\begin{proof}
+Choose a binary basis of $K_2$. The matrix of multiplication by
+$\lambda\in K_2$ is a unital ring representation; it is injective
+because its value on $1\in K_2$ recovers $\lambda$. Apply this map
+coefficientwise to a group-ring sum. The coefficient matrices
+commute with the formal group symbols, so products are preserved.
+Comparing the coefficient matrix at each group element proves
+injectivity. Thus $XY=I_d$ and $I_d-YX\ne0$. Now
+$(YX)^2=Y(XY)X=YX$, $X(I_d-YX)=X-(XY)X=0$ and
+$(I_d-YX)Y=Y-Y(XY)=0$, proving all the identities.
+\end{proof}
+
+\begin{proposition}\label{prop:separation}
+Let $X,Y\in\Mat_d(\F_2[H_2])$, $d\ge1$, satisfy
+$XY=I_d\ne YX$. In the group
+$G_D=(\bigoplus_{h\in H_2}C_3)\rtimes H_2$, where $H_2$ acts
+by left translation on coordinate labels, there is an orthogonal
+projection $P\in\Mat_{d+1}(\Z[1/3][G_D])$ and a finite set
+$S\subseteq H_2\setminus\{1\}$ such that
+\[
+ \tau_{d+1}(P)=d+\theta,\qquad
+ \theta=\frac13\left(\frac23\right)^{|S|}>0,
+ \qquad \bar P=uv,
+\]
+where $u$ has size $(d+1)\times d$ and $v$ has size $d\times(d+1)$
+over $\F_2[G_D]$. The integer $D=3^{1+|S|}$ clears all coefficients
+of $P$.
+\end{proposition}
+
+\begin{proof}
+Put $C=I_d-YX$. Choose an entry $C_{ij}$ and a group element
+$h_0$ whose coefficient in that entry is nonzero. The coefficient
+is $1$ because the field is $\F_2$. Left multiplication by
+$h_0^{-1}$ therefore gives
+\begin{equation}\label{eq:support}
+ h_0^{-1}C_{ij}=1+\sum_{s\in S}s
+\end{equation}
+for a finite $S\subseteq H_2\setminus\{1\}$, possibly empty.
+
+To isolate the identity term in Equation~\eqref{eq:support}, we
+seek a nonzero rational projection $p$ with $psp=0$ for every
+$s\in S$. The cyclic coordinates provide complementary projections
+that enforce these vanishing products. Let $z_h$ generate the cyclic
+coordinate labelled by $h$ and put
+\[
+ p_h=\frac{1+z_h+z_h^2}{3},\qquad
+ p=p_1\prod_{s\in S}(1-p_s).
+\]
+The $p_h$ are self-adjoint idempotents: averaging the nine products
+in $(1+z_h+z_h^2)^2$ gives three copies of each element of $C_3$.
+They commute because distinct coordinates commute. Consequently
+$p=p^*=p^2$, and the displayed product shows that $3^{1+|S|}p$
+is integral. This is the elementary commuting-coordinate method
+in the works cited above; compare
+\cite[Lemmas 3.1 and 3.4]{DicksSchick2002}.
+
+Conjugation obeys $s p_hs^{-1}=p_{sh}$. For $s\in S$, the
+product $sps^{-1}$ contains $p_s$, while $p$ contains $1-p_s$.
+All these coordinate factors commute, so
+\begin{equation}\label{eq:separation}
+ psp=p(sps^{-1})s=0.
+\end{equation}
+This argument does not need disjointness of the two collections of
+coordinate labels. Inverse pairs or finite-order elements in $S$
+cause no change.
+
+The original coordinates $\{1\}\cup S$ are distinct. The coefficient
+of the identity in their product is therefore the product of the
+individual identity coefficients. Hence
+\begin{equation}\label{eq:theta}
+ \tau_1(p)=[1]p=\frac13\prod_{s\in S}\frac23=\theta>0.
+\end{equation}
+For empty $S$ this means $p=p_1$ and $\theta=1/3$.
+
+The map $\Z[1/3]\to\F_2$ is defined because $3$ is odd. Reduce
+the projection modulo two and combine
+Equations~\eqref{eq:support} and~\eqref{eq:separation} to obtain
+\[
+ \bar p h_0^{-1} C_{ij}\bar p=\bar p.
+\]
+Let $f$ be the row of length $d$ with entry $\bar p h_0^{-1}$
+at position $i$ and zero elsewhere. Let $w$ be the column with
+entry $\bar p$ at position $j$ and zero elsewhere. Thus
+$fCw=\bar p$. Define
+\[
+ P=\diag(I_d,p),\qquad
+ u=\begin{pmatrix}X\\fC\end{pmatrix},\qquad
+ v=\begin{pmatrix}Y&Cw\end{pmatrix}.
+\]
+Using Equation~\eqref{eq:defect} in this order gives
+\[
+ uv=\begin{pmatrix}XY&XCw\\fCY&fC^2w\end{pmatrix}
+   =\diag(I_d,\bar p)=\bar P.
+\]
+The rational trace is $d+\theta$ by Equation~\eqref{eq:theta}.
+It is the rational orthogonal projection, not its binary reduction,
+that has this strictly positive extra trace.
+\end{proof}
+
+We have obtained a projection of real trace greater than $d$ but a
+binary factorization through $d$ columns. The next step produces
+a factorization through the same number of columns modulo a power of two, so
+that it can be inserted into an integral matrix.
+
+\section{Lifting the factorization and preserving integrality}
+\label{sec:lifting}
+
+We now work with any group $G$, an orthogonal rational projection
+$P\in\Mat_m(\Q[G])$, and a factorization of $\bar P$ through $r$
+columns over $\F_2[G]$. Assume that an odd integer $D$ clears the
+denominators of $P$. Here orthogonality means $P^2=P=P^*$, where
+$*$ transposes the matrix and inverts group elements.
+
+For a power $q>1$ of two, the trace gap suggests the absolute value
+of the operator we want: expand $r$ coordinate summands by $q$ and
+contract the range of $P$ by $q^{-1}$. Its
+logarithmic determinant should be $(r-\tau_m(P))\log q$, which is
+negative when $\tau_m(P)>r$. The difficulty is to realize this
+determinant by an integral matrix. We will add block shears whose
+off-diagonal entries come from a factorization of $P$ modulo $q$.
+The analytic justification that these shears preserve the determinant
+is given in Section~\ref{sec:analysis}.
+
+The first lemma lifts the binary factorization to every
+power $q$ of two; only idempotence is needed for this step.
+
+\begin{lemma}\label{lem:lift}
+Let $G$ be a group, $m,r\ge1$, $D$ an odd positive integer, and
+$P\in\Mat_m(\Q[G])$ satisfy $P^2=P$ and
+$DP\in\Mat_m(\Z[G])$. Suppose $\bar P=uv$ over $\F_2[G]$, where $u$ and $v$
+have respective sizes $m\times r$ and $r\times m$. For $q=2^k$,
+$k\ge1$, the reduction $e=P_q$ has a factorization
+$e=u_qv_q$ of the same sizes over $(\Z/q\Z)[G]$.
+\end{lemma}
+
+\begin{proof}
+Since $D$ is odd, every coefficient of $P$ has a well-defined
+reduction modulo $q$. Set $\mathcal R=(\Z/q\Z)[G]$ and choose
+arbitrary finite lifts $u_0,v_0$ of $u,v$ over $\mathcal R$.
+In the corner ring $e\Mat_m(\mathcal R)e$, whose identity is $e$,
+put
+\[
+ Z=eu_0v_0e,\qquad W=e-Z.
+\]
+Modulo two, $Z$ is $\bar Puv\bar P=\bar P^3=\bar P$.
+Every coefficient of $W$ is therefore even. A product of $k$
+such coefficients is divisible by $2^k$, giving $W^k=0$.
+This is nilpotence in the even ideal of the corner ring; the
+corner ring itself need not be nilpotent.
+
+Because $Z=e-W$, the finite geometric sum
+\[
+ Q=e+W+\cdots+W^{k-1}
+\]
+satisfies $QZ=ZQ=e-W^k=e$. When $k=1$, this sum is just $e$.
+Thus $Q$ is a corner inverse of $Z$, not a claimed inverse in
+the entire matrix ring. Set
+\[
+ u_q=Qeu_0,\qquad v_q=v_0e.
+\]
+Then $u_qv_q=Qeu_0v_0e=QZ=e$. Only finite sums and products
+were used, so the resulting supports remain finite.
+\end{proof}
+
+For the integral construction, choose $q=2^k>1$ with
+$q\equiv1\pmod D$. Such a power exists because two is a unit
+modulo the odd integer $D$; for $D=1$, any $2^k>1$ works.
+Choose integral lifts $U,V$ of the factors from Lemma~\ref{lem:lift}.
+The diagonal operator with the desired expansion and contraction is
+\[
+ J=-q^{-1}P-(I_m-P),\qquad B=\diag(qI_r,J).
+\]
+The minus signs leave its positive square unchanged and will produce
+the subtraction $UV-P$ needed for integrality. Prescribe the
+upper-left block $qI_r$ and off-diagonal blocks $V,U$. Eliminating
+these off-diagonal blocks subtracts $U(qI_r)^{-1}V=UV/q$ from the
+lower-right block. To leave $J$ as the Schur complement, that block
+must therefore be $UV/q+J$. Thus the factorization modulo $q$ is exactly what is
+needed to cancel the denominators in this block.
+
+\begin{lemma}\label{lem:integral}
+Suppose $P=P^*=P^2\in\Mat_m(\Q[G])$, an odd $D>0$ clears its
+denominators, and $\bar P=uv$ factors through $r\ge1$ columns.
+Let $q=2^k>1$ satisfy $q\equiv1\pmod D$. Choose arbitrary
+integral lifts $U,V$ of the factors supplied by Lemma~\ref{lem:lift}.
+Then
+\begin{equation}\label{eq:integral-matrix}
+ A=\begin{pmatrix}
+ qI_r&V\\
+ U&q^{-1}\bigl(UV-P-q(I_m-P)\bigr)
+ \end{pmatrix}
+ \in\Mat_{r+m}(\Z[G])
+\end{equation}
+has an inverse in $\Mat_{r+m}(\Q[G])$.
+\end{lemma}
+
+\begin{proof}
+The matrix $P+q(I_m-P)=qI_m+(1-q)P$ is integral because
+$D\mid(q-1)$. Its reduction modulo $q$ is $e=P_q$, which is
+also the reduction of $UV$. Thus every coefficient of
+$UV-P-q(I_m-P)$ is an integer divisible by $q$; the bottom-right
+block of Equation~\eqref{eq:integral-matrix} is integral.
+
+For $J$ and $B$ defined above, the complementary projection identities give
+$J^{-1}=-qP-(I_m-P)$. For $x\in[0,1]$ put
+\[
+ L_x=\begin{pmatrix}I_r&0\\xU/q&I_m\end{pmatrix},\qquad
+ R_x=\begin{pmatrix}I_r&xV/q\\0&I_m\end{pmatrix}.
+\]
+Their inverses negate the off-diagonal blocks, and multiplication
+without interchanging any factors gives
+\begin{equation}\label{eq:path}
+ M(x):=L_xBR_x=
+ \begin{pmatrix}qI_r&xV\\xU&x^2UV/q+J\end{pmatrix}.
+\end{equation}
+In particular $A=M(1)$, with the rational group-ring inverse
+$R_1^{-1}B^{-1}L_1^{-1}$. For general real $x$, both $M(x)$ and
+its inverse lie in the real group ring and depend continuously
+differentiably on $x$. No identities $PU=U$, $VP=V$, or
+$V=U^*$ are required of the chosen integer lifts.
+\end{proof}
+
+The construction is integral at its endpoint, though its simplifying
+path need not be integral. We
+now show that this path preserves the logarithmic determinant.
+
+\section{The trace calculation}
+\label{sec:analysis}
+
+Let $\mathcal A_n$ be the operator-norm closure of
+$\Mat_n(\C[G])$ in its left regular representation. For finite
+group-ring matrices define
+\[
+ \tau_n(C)=\sum_{i=1}^n[1]C_{ii}.
+\]
+The identity-coefficient formula gives
+$\tau_n(CD)=\tau_n(DC)$ by summing
+$(C_{ij})_g(D_{ji})_{g^{-1}}$ over $i,j,g$. Moreover
+$|\tau_n(C)|\le n\|C\|$ and
+$\tau_n(C^*)=\overline{\tau_n(C)}$. Thus it extends to a
+continuous trace on the unital $C^*$-algebra $\mathcal A_n$.
+
+\begin{lemma}\label{lem:logderivative}
+For a continuously differentiable path
+$M:[0,1]\to\mathcal A_n$ of invertible elements,
+\[
+ \frac{d}{dx}\tau_n\bigl(\log(M(x)^*M(x))\bigr)
+ =2\operatorname{Re}\tau_n\bigl(M'(x)M(x)^{-1}\bigr).
+\]
+\end{lemma}
+
+\begin{proof}
+Put $N=M^*M$. Continuity of $M$ and its inverse on the compact
+interval gives constants $0<\delta<c$ with
+$\delta I\le N(x)\le cI$ for all $x$. Write
+$W=I-N/c$, so $\|W\|\le\rho=1-\delta/c<1$. Functional calculus
+gives the uniformly norm-convergent series
+\[
+ \log N=(\log c)I-\sum_{j\ge1}\frac{W^j}{j}.
+\]
+Its derivative series is also uniformly norm-convergent, since
+\[
+ \left\|\frac{d}{dx}\frac{W^j}{j}\right\|
+ \le\sup_{y\in[0,1]}\|W'(y)\|\rho^{j-1}.
+\]
+In the derivative of $W^j$, cyclicity makes the traces of its
+$j$ terms equal; commutation of $N$ and $N'$ is not assumed.
+Using $W'=-N'/c$ therefore gives
+\[
+ \frac{d}{dx}\tau_n(\log N)
+ =\frac1c\sum_{j\ge1}\tau_n(W^{j-1}N')
+ =\tau_n(N^{-1}N').
+\]
+Since $N'=M'^*M+M^*M'$, the second summand contributes
+$\tau_n(M'M^{-1})$. Cyclicity and
+$MN^{-1}=(M^*)^{-1}$ turn the first into
+\[
+ \tau_n((M^*)^{-1}M'^*)
+ =\overline{\tau_n(M'M^{-1})}.
+\]
+Their sum is the asserted real part. This is the tracial
+logarithmic derivative of~\cite[Lemma 2]{FK1952}, written with
+the unnormalized trace used here.
+\end{proof}
+
+\begin{lemma}\label{lem:shears}
+Let $r,m\ge1$ and let $B\in\mathcal A_{r+m}$ be invertible.
+Let $S,T$ have respective sizes $m\times r$ and $r\times m$ over
+the operator-norm closure of $\C[G]$. For $x\in[0,1]$, let
+$L_x,R_x$ be lower and upper block shears with diagonal blocks
+$I_r,I_m$ and off-diagonal blocks $xS,xT$, respectively.
+Then $\tau_{r+m}(\log((L_xBR_x)^*(L_xBR_x)))$ is constant.
+\end{lemma}
+
+\begin{proof}
+Write $M=L_xBR_x$. Both shears have inverses obtained by negating
+their off-diagonal blocks. The ordered product rule gives
+\[
+ M'M^{-1}=L_x'L_x^{-1}
+ +(L_xB)(R_x'R_x^{-1})(L_xB)^{-1}.
+\]
+The matrices $L_x'L_x^{-1}$ and $R_x'R_x^{-1}$ have zero diagonal
+blocks, so each has trace zero. Cyclicity gives trace zero also for
+the conjugate $(L_xB)(R_x'R_x^{-1})(L_xB)^{-1}$.
+Lemma~\ref{lem:logderivative} now gives zero derivative.
+\end{proof}
+
+\begin{proposition}\label{prop:criterion}
+Let $G$ be a discrete group, $m,r\ge1$, and let
+$P=P^*=P^2\in\Mat_m(\Q[G])$. Suppose an odd $D>0$ clears its
+denominators and $\bar P=uv$ with
+$u\in\Mat_{m,r}(\F_2[G])$ and
+$v\in\Mat_{r,m}(\F_2[G])$. For every $q=2^k>1$ congruent to one
+modulo $D$, the matrix $A$ of Lemma~\ref{lem:integral} satisfies
+\begin{align}
+ \tau_{r+m}(\log(T_A^*T_A))
+   &=2(r-\tau_m(P))\log q,\label{eq:criterion-trace}\\
+ \log\det_{\N(G)}(T_A)&=(r-\tau_m(P))\log q.
+ \label{eq:criterion-det}
+\end{align}
+It is integral and rationally invertible, and its positive-square
+spectral measure has support in a compact subinterval of $(0,\infty)$.
+\end{proposition}
+
+\begin{proof}
+Lemma~\ref{lem:integral} has already established integrality and
+the finite-support rational inverse. Apply Lemma~\ref{lem:shears}
+to Equation~\eqref{eq:path}. At $x=0$ the matrix is
+$B=\diag(qI_r,-q^{-1}P-(I_m-P))$. Orthogonality gives
+\[
+ B^*B=\diag(q^2I_r,q^{-2}P+I_m-P),\qquad
+ \log(B^*B)=2\log q\,\diag(I_r,-P).
+\]
+Taking the trace and using shear invariance proves
+Equation~\eqref{eq:criterion-trace}; half this trace is
+Equation~\eqref{eq:criterion-det}. The regular representation
+of the rational group-ring inverse is bounded, so
+\[
+ \|T_A^{-1}\|^{-2}I\le T_A^*T_A\le\|T_A\|^2I.
+\]
+Its logarithm is therefore bounded. In the probability measure
+convention from the introduction this also gives
+\begin{equation}\label{eq:criterion-integral}
+ \int_{(0,\infty)}\log t\,d\mu_A(t)
+ =\frac{2(r-\tau_m(P))\log q}{r+m}.
+\end{equation}
+\end{proof}
+
+\section{Assembly and exact determinant}
+\label{sec:assembly}
+
+\begin{proof}[Proof of Theorem~\ref{thm:main}]
+Take $K_2,H_2,a_2,b_2$ from Theorem~\ref{thm:input}. Apply
+Lemma~\ref{lem:binary} with $d=[K_2:\F_2]\ge1$ and then
+Proposition~\ref{prop:separation}. It gives the group
+$G_D=(\bigoplus_{H_2}C_3)\rtimes H_2$ and a projection $P$
+of size $d+1$ whose binary reduction factors through $d$ columns.
+Its trace is $d+\theta$, where
+\[
+ \theta=\frac13\left(\frac23\right)^{|S|}>0,
+ \qquad D=3^{1+|S|}.
+\]
+The odd integer $D$ clears the coefficients of $P$. Choose
+$k\ge1$ with $q=2^k\equiv1\pmod D$.
+Proposition~\ref{prop:criterion}, with $r=d$ and $m=d+1$,
+produces $A\in\Mat_{2d+1}(\Z[G_D])$ with a rational group-ring
+inverse and
+\[
+ \log\det_{\N(G_D)}(T_A)=-\theta\log q.
+\]
+Since $q>1$ and $\theta>0$, this proves
+\begin{equation}\label{eq:final-det}
+ 0<\det_{\N(G_D)}(T_A)=q^{-\theta}<1.
+\end{equation}
+With $n=2d+1$, Equation~\eqref{eq:criterion-integral} reads
+\begin{equation}\label{eq:final-integral}
+ \int_{(0,\infty)}\log t\,d\mu_A(t)
+ =-\frac{2\theta\log q}{2d+1}<0.
+\end{equation}
+It is finite because of the spectral bounds in
+Proposition~\ref{prop:criterion}. Finally, a finite generating
+set of $H_2$ together with $z_1$ generates $G_D$, since
+$z_h=hz_1h^{-1}$. Thus $G_D$ is finitely generated.
+\end{proof}
+
+The target group differs from the companion's group and contains
+the cyclic coordinates of order three. The construction has a
+finite negative logarithmic determinant and no zero spectral mass;
+it is a failure of the asserted lower bound, not of determinant
+class. Applying the sofic determinant bound of
+\cite[Theorem 5]{ElekSzabo2005} to the positive integral matrix
+$A^*A$ proves that $G_D$ is not sofic. This last implication is an
+external consequence, not an ingredient in constructing $A$.
+
+\begingroup
+\raggedright
+\bibliographystyle{plain}
+\bibliography{references}
+\endgroup
+\end{document}

preprints/A-Counterexample-to-the-Group-Ring-Determinant-Conjecture-September-23-2026/build/references.bib

@@ -0,0 +1,108 @@
+@book{Luck2002,
+  author = {Wolfgang L{\"u}ck},
+  title = {{$L^2$-Invariants: Theory and Applications to Geometry and $K$-Theory}},
+  series = {Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge},
+  volume = {44},
+  publisher = {Springer},
+  year = {2002},
+  doi = {10.1007/978-3-662-04687-6}
+}
+
+@article{ElekSzabo2005,
+  author = {G{\'a}bor Elek and Endre Szab{\'o}},
+  title = {Hyperlinearity, essentially free actions and {$L^2$}-invariants.
+           {The} sofic property},
+  journal = {Mathematische Annalen},
+  volume = {332},
+  pages = {421--441},
+  year = {2005},
+  doi = {10.1007/s00208-005-0640-8}
+}
+
+@article{FK1952,
+  author = {Bent Fuglede and Richard V. Kadison},
+  title = {Determinant theory in finite factors},
+  journal = {Annals of Mathematics},
+  series = {2},
+  volume = {55},
+  number = {3},
+  pages = {520--530},
+  year = {1952},
+  doi = {10.2307/1969645}
+}
+
+@misc{CharacteristicTwo,
+  author = {{OpenAI}},
+  title = {{A Counterexample to Kaplansky's Direct-Finiteness Conjecture in Characteristic Two}},
+  howpublished = {OpenAI Math Release preprint
+                  \href{https://github.com/openai/math/blob/main/preprints/A-Counterexample-to-Kaplanskys-Direct-Finiteness-Conjecture-in-Characteristic-Two-September-23-2026/paper.pdf}{OAI:A-Counterexample-to-Kaplanskys-Direct-Finiteness-Conjecture-in-Characteristic-Two-September-23-2026}},
+  year = {2026},
+  note = {Theorem 1.1 and Section 6}
+}
+
+@misc{Kammeyer2026,
+  author = {Holger Kammeyer},
+  title = {A counterexample to the determinant approximation conjecture},
+  year = {2026},
+  eprint = {2609.15567},
+  howpublished = {\href{https://arxiv.org/abs/2609.15567v1}{arXiv:2609.15567v1}},
+  archivePrefix = {arXiv},
+  primaryClass = {math.GR},
+  note = {Version 1, September 14, 2026},
+  url = {https://arxiv.org/abs/2609.15567v1}
+}
+
+@article{DicksSchick2002,
+  author = {Warren Dicks and Thomas Schick},
+  title = {The spectral measure of certain elements of the complex group ring of a wreath product},
+  journal = {Geometriae Dedicata},
+  volume = {93},
+  pages = {121--137},
+  year = {2002},
+  doi = {10.1023/A:1020381532489},
+  eprint = {math/0107145},
+  archivePrefix = {arXiv},
+  url = {https://arxiv.org/abs/math/0107145}
+}
+
+@article{LehnerNeuhauserWoess2008,
+  author = {Franz Lehner and Markus Neuhauser and Wolfgang Woess},
+  title = {On the spectrum of lamplighter groups and percolation clusters},
+  journal = {Mathematische Annalen},
+  volume = {342},
+  number = {1},
+  pages = {69--89},
+  year = {2008},
+  doi = {10.1007/s00208-008-0222-7},
+  eprint = {0712.3135},
+  archivePrefix = {arXiv},
+  url = {https://arxiv.org/abs/0712.3135}
+}
+
+@article{deLaHarpeSkandalis1984,
+  author = {de la Harpe, Pierre and Skandalis, Georges},
+  title = {D\'eterminant associ\'e \`a une trace sur une alg\`ebre de {Banach}},
+  journal = {Annales de l'Institut Fourier},
+  volume = {34},
+  number = {1},
+  pages = {241--260},
+  year = {1984},
+  doi = {10.5802/aif.958},
+  url = {https://www.numdam.org/articles/10.5802/aif.958/}
+}
+
+@article{Schick2001,
+  author = {Thomas Schick},
+  title = {{$L^2$}-determinant class and approximation of {$L^2$}-{Betti} numbers},
+  journal = {Transactions of the American Mathematical Society},
+  volume = {353},
+  number = {8},
+  pages = {3247--3265},
+  year = {2001},
+  doi = {10.1090/S0002-9947-01-02699-X},
+  eprint = {math/9807032},
+  archivePrefix = {arXiv},
+  url = {https://arxiv.org/abs/math/9807032v2},
+  note = {With the appended erratum in
+          \href{https://arxiv.org/abs/math/9807032v2}{arXiv:math/9807032v2}}
+}

preprints/A-Counterexample-to-the-Group-Ring-Determinant-Conjecture-September-23-2026/paper.pdf

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+size 311036

preprints/A-Counterexample-to-the-Infinite-Matroid-Packing-Covering-Conjecture-September-24-2026/README.md

@@ -0,0 +1,20 @@
+# [A Counterexample to the Infinite Matroid Packing/Covering Conjecture](paper.pdf)
+
+**Author:** OpenAI
+
+**Date:** September 24, 2026
+
+## Citation
+
+To cite this manuscript:
+
+```bibtex
+@misc{OAI:A-Counterexample-to-the-Infinite-Matroid-Packing-Covering-Conjecture-September-24-2026,
+  author = {{OpenAI}},
+  title = {{A Counterexample to the Infinite Matroid Packing/Covering Conjecture}},
+  howpublished = {OpenAI Math Release preprint
+                  \href{https://github.com/openai/math/blob/main/preprints/A-Counterexample-to-the-Infinite-Matroid-Packing-Covering-Conjecture-September-24-2026/paper.pdf}{OAI:A-Counterexample-to-the-Infinite-Matroid-Packing-Covering-Conjecture-September-24-2026}},
+  year = {2026}
+}
+```
+

preprints/A-Counterexample-to-the-Infinite-Matroid-Packing-Covering-Conjecture-September-24-2026/build/figures/double-ray.tex

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+\usepackage[colorlinks=true,linkcolor=blue!45!black,citecolor=blue!45!black,urlcolor=blue!45!black]{hyperref}
+\hypersetup{pdftitle={A Counterexample to the Infinite Matroid Packing/Covering Conjecture},pdfauthor={OpenAI}}
+\newtheorem{theorem}{Theorem}[section]
+\newtheorem{proposition}[theorem]{Proposition}
+\newtheorem{lemma}[theorem]{Lemma}
+\newtheorem{corollary}[theorem]{Corollary}
+\theoremstyle{definition}
+\newtheorem{definition}[theorem]{Definition}
+\theoremstyle{remark}
+\newtheorem{remark}[theorem]{Remark}
+\numberwithin{equation}{section}
+\DeclareMathOperator{\cl}{cl}
+\setlist{itemsep=3pt,topsep=5pt}
+\title{A Counterexample to the Infinite Matroid\newline Packing/Covering Conjecture}
+\author{OpenAI}
+\date{September 24, 2026}
+\begin{document}
+\maketitle
+\begin{abstract}
+We construct in ZFC two self-dual partitional matroids on a countably infinite
+common ground set that admit neither a packing/covering partition nor an
+intersection witness. This disproves the unrestricted infinite matroid
+packing/covering and intersection conjectures and answers Jo\'o's question
+for two partitional matroids negatively. The examples are neither finitary
+nor cofinitary.
+\end{abstract}
+\begingroup
+\small
+\tableofcontents
+\endgroup
+\clearpage
+\input{sections/01-introduction}
+\input{sections/02-matroids}
+\input{sections/03-filters}
+\input{sections/04-local-bases}
+\input{sections/05-assembly}
+\input{sections/06-counterexample}
+\input{sections/07-consequences}
+\begingroup\small
+\bibliographystyle{plain}
+\bibliography{references}
+\endgroup
+\end{document}

preprints/A-Counterexample-to-the-Infinite-Matroid-Packing-Covering-Conjecture-September-24-2026/build/references.bib

@@ -0,0 +1,83 @@
+@article{BDKPW2013,
+  author = {Bruhn, Henning and Diestel, Reinhard and Kriesell, Matthias and Pendavingh, Rudi and Wollan, Paul},
+  title = {Axioms for infinite matroids},
+  journal = {Advances in Mathematics},
+  volume = {239},
+  year = {2013},
+  pages = {18--46},
+  doi = {10.1016/j.aim.2013.01.011}
+}
+
+@article{BowlerCarmesin2015,
+  author = {Bowler, Nathan and Carmesin, Johannes},
+  title = {Matroid intersection, base packing and base covering for infinite matroids},
+  journal = {Combinatorica},
+  volume = {35},
+  number = {2},
+  year = {2015},
+  pages = {153--180},
+  doi = {10.1007/s00493-014-2953-2},
+  note = {Numbered locators refer to the \href{https://www.math.uni-hamburg.de/spag/dm/papers/PackingCovering.pdf}{author version dated 25 June 2014}}
+}
+
+@article{BowlerGeschke2016,
+  author = {Bowler, Nathan and Geschke, Stefan},
+  title = {Self-dual uniform matroids on infinite sets},
+  journal = {Proceedings of the American Mathematical Society},
+  volume = {144},
+  number = {2},
+  year = {2016},
+  pages = {459--471},
+  doi = {10.1090/proc/12667},
+  note = {Numbered locators refer to the \href{https://www.math.uni-hamburg.de/home/geschke/papers/UniformMatroid7.pdf}{author version dated 6 October 2014}}
+}
+
+@misc{GollinJoo2025,
+  author = {Gollin, J. Pascal and Jo{\'o}, Attila},
+  title = {Wild generalised truncation of infinite matroids},
+  year = {2025},
+  note = {\href{https://arxiv.org/abs/2504.05064v1}{arXiv:2504.05064v1}, 7 April 2025}
+}
+
+@article{Joo2024Packing,
+  author = {Jo{\'o}, Attila},
+  title = {On the packing/covering conjecture of infinite matroids},
+  journal = {Israel Journal of Mathematics},
+  volume = {261},
+  number = {2},
+  year = {2024},
+  pages = {765--790},
+  doi = {10.1007/s11856-023-2595-4}
+}
+
+@article{Joo2021Partitional,
+  author = {Jo{\'o}, Attila},
+  title = {Intersection of a partitional and a general infinite matroid},
+  journal = {Discrete Mathematics},
+  volume = {344},
+  number = {9},
+  year = {2021},
+  pages = {112514},
+  doi = {10.1016/j.disc.2021.112514}
+}
+
+@article{Joo2021Countable,
+  author = {Jo{\'o}, Attila},
+  title = {Proof of {Nash-Williams}' intersection conjecture for countable matroids},
+  journal = {Advances in Mathematics},
+  volume = {380},
+  year = {2021},
+  pages = {107608},
+  doi = {10.1016/j.aim.2021.107608}
+}
+
+@article{AharoniZiv1998,
+  author = {Aharoni, Ron and Ziv, Ran},
+  title = {The intersection of two infinite matroids},
+  journal = {Journal of the London Mathematical Society},
+  volume = {58},
+  number = {3},
+  year = {1998},
+  pages = {513--525},
+  doi = {10.1112/S0024610798006723}
+}

preprints/A-Counterexample-to-the-Infinite-Matroid-Packing-Covering-Conjecture-September-24-2026/build/sections/01-introduction.tex

@@ -0,0 +1,202 @@
+\section{Introduction}\label{sec:introduction}
+
+A \emph{packing} of a family of matroids on a common ground set is a choice
+of pairwise disjoint spanning sets, one for each matroid. A \emph{covering}
+is a choice of independent sets, one for each matroid, whose union is the
+ground set. The infinite matroid packing/covering conjecture asks whether
+every family admits a partition of the ground set on which these two
+requirements can be satisfied in the following complementary senses.
+
+Throughout, matroids are standard infinite matroids, with the maximal
+extension axiom recalled in Section~\ref{sec:matroids}. For a matroid $M$
+on $E$ and $X\subseteq E$, write $M\!\upharpoonright X$ for restriction
+and
+\[
+ M.X:=M/(E\setminus X)=(M^*\!\upharpoonright X)^*
+\]
+for contraction onto $X$. A set is spanning if its closure is the whole
+ground set. A \emph{packing/covering partition} for $(M_i:i\in\Theta)$ is
+a partition $E=P\dot\cup C$ together with sets $S_i\subseteq P$ and
+$I_i\subseteq C$ such that
+\begin{align*}
+ S_i\cap S_j&=\varnothing &&(i\ne j),\\
+ \cl_{M_i\upharpoonright P}(S_i)&=P &&(i\in\Theta),\\
+ I_i&\text{ is independent in }M_i.C &&(i\in\Theta),\\
+ \bigcup_{i\in\Theta}I_i&=C.
+\end{align*}
+Bowler and Carmesin formulated the assertion that every set-indexed family
+has such a partition and established its relationship with infinite
+matroid intersection~\cite[Conjecture~1.3 and Corollary~3.9]{BowlerCarmesin2015}. We give a counterexample
+with two matroids.
+
+\begin{theorem}\label{thm:main}
+In ZFC, there exist a countably infinite set $E$ and two infinite matroids
+$M_0,M_1$ on $E$ such that $M_i^*=M_i$ on the same labelled ground set for
+$i=0,1$, and
+\[
+ I_0\cup I_1\ne E
+ \quad\text{for every independent }I_0\text{ in }M_0
+ \text{ and independent }I_1\text{ in }M_1.
+\]
+The pair $(M_0,M_1)$ has no packing/covering partition. Consequently, the
+infinite matroid packing/covering conjecture is false.
+\end{theorem}
+
+The last conclusion needs a separate argument: failure of a covering of
+$E$ alone does not rule out a mixed partition. For a self-dual pair,
+however, the spanning sets on $P$ and the independent sets in the
+contractions onto $C$ give disjoint spanning sets on $E$. Their complements
+are an independent covering. We prove this implication in
+Section~\ref{sec:matroids} before constructing the counterexample.
+
+\subsection*{Context and the local construction}
+
+The axioms of Bruhn, Diestel, Kriesell, Pendavingh and
+Wollan~\cite{BDKPW2013} provide an infinite matroid theory with restriction,
+contraction and duality. In particular, independent sets must extend to
+maximal independent sets inside \emph{every} prescribed subset of the
+ground set. This condition is a central part of our construction.
+
+A matroid is \emph{uniform} if replacing any element of any basis by any
+element outside that basis again gives a basis. Bowler and
+Geschke~\cite[Theorem~6 and the proof of Theorem~15]{BowlerGeschke2016} developed a basis criterion for infinite
+uniform matroids and constructed countable self-dual examples under
+Martin's axiom for countable partially ordered sets, and hence under the
+continuum hypothesis. Their criterion combines an
+antichain of bases, invariance under balanced finite changes, and a basis
+existence condition for intervals of subsets. Our assembly also uses the
+complementary interval-recursion pattern of their Lemma~12, while
+verifying the full independent-set axioms directly.
+
+Bowler and Geschke left open whether an infinite self-dual uniform matroid
+can be constructed in ZFC. Even without self-duality, ZFC existence of a
+uniform matroid of infinite rank and corank is recorded as open by Gollin
+and Jo\'o~\cite[Section~1]{GollinJoo2025}.
+
+Our local construction gives a self-dual uniform matroid $Q$ on a
+countable set in ZFC, with every basis and its complement infinite.
+It therefore answers both of these existence questions affirmatively.
+Its bases satisfy an additional ordinal inequality that obstructs
+independent coverings when copies of $Q$ are placed along a doubly
+infinite path. The two matroids in Theorem~\ref{thm:main} are countably
+infinite direct sums of these copies. In Jo\'o's terminology, a direct sum
+of uniform matroids is \emph{partitional}. His definition of uniformity is
+equivalent to requiring every subset to be independent or spanning
+\cite[Definition~1.2 and Proposition~2.2]{Joo2021Partitional}, the property
+proved for $Q$ in Remark~\ref{rem:uniform-local}. Thus both matroids in our
+pair are partitional. This distinction matters: Jo\'o proved an intersection
+theorem for an arbitrary infinite matroid paired with a direct sum of
+\emph{finitely many} uniform matroids, and asked whether intersection holds
+for two partitional matroids on a common countable ground set
+\cite[Theorem~1.4 and Question~1.5]{Joo2021Partitional}.
+
+For matroids $M,N$ on a common ground set $E$, the pair \emph{satisfies
+intersection} if there is a set $J$ independent in both matroids and a
+partition $J=J_M\dot\cup J_N$ such that
+\[
+ \cl_M(J_M)\cup\cl_N(J_N)=E.
+\]
+The unrestricted infinite Matroid Intersection Conjecture asserts that
+every pair of matroids on a common, possibly infinite, ground set satisfies this
+property~\cite[Definition~3.5 and Section~7]{BowlerCarmesin2015}.
+
+\begin{corollary}\label{cor:intersection}
+The same pair $(M_0,M_1)$ constructed in Theorem~\ref{thm:main} consists of
+partitional matroids and does not satisfy intersection. More explicitly,
+for every set $J$ independent in both $M_0$ and $M_1$ and every partition
+$J=J_0\dot\cup J_1$,
+\[
+ \cl_{M_0}(J_0)\cup\cl_{M_1}(J_1)\ne E.
+\]
+Consequently, the unrestricted infinite Matroid Intersection Conjecture is
+false in ZFC. The pair also gives a negative ZFC answer to Jo\'o's
+Question~1.5 for two partitional matroids on a common countable ground set.
+\end{corollary}
+
+\begin{proof}
+Each $M_i$ is a direct sum of countably infinitely many copies of the uniform
+matroid $Q$, so it is partitional. Bowler and Carmesin
+\cite[Proposition~3.6]{BowlerCarmesin2015} prove that, for any matroids $M,N$
+on the same ground set,
+\[
+ (M,N)\text{ satisfies intersection}
+ \quad\Longleftrightarrow\quad
+ (M,N^*)\text{ has a packing/covering partition}.
+\]
+Apply this equivalence with $M=M_0$ and $N=M_1$. Since $M_1^*=M_1$ on the same labelled ground set,
+the right-hand side concerns the identical pair $(M_0,M_1)$ and is false by
+Theorem~\ref{thm:main}. Its common ground set is countable, so it also
+satisfies the hypotheses of Jo\'o's question.
+\end{proof}
+
+Bowler and Carmesin also formulated separate Covering and Packing
+conjectures. Their universal equivalences imply that both conjectures are
+false in ZFC; Section~\ref{sec:separate-consequences} gives the precise
+formulations and deduction. This consequence concerns universal statements
+about matroid families and does not identify their witnessing families
+with the pair $(M_0,M_1)$.
+
+The examples are neither finitary nor cofinitary. Here a matroid is
+\emph{finitary} if a set is independent whenever all its finite subsets
+are independent, and \emph{cofinitary} if its dual is finitary. Nash-Williams' original intersection conjecture concerned finitary
+matroids; Aharoni and Ziv~\cite{AharoniZiv1998} gave an early published
+formulation. Jo\'o proved intersection for two finitary matroids on a countable
+ground set~\cite[Theorem~1.3]{Joo2021Countable}.
+His subsequent packing/covering theorem allows every set-indexed family
+on a countable ground set whose members are direct sums of finitary and
+cofinitary matroids~\cite[Theorem~1.2]{Joo2024Packing}.
+The examples here have no nonempty finitary or cofinitary direct summand,
+as we show in Section~\ref{sec:counterexample}, so they lie outside this
+mixed direct-sum class. Countability of the ground set is distinct from
+these structural hypotheses; the present result concerns unrestricted
+infinite matroids and does not refute Nash-Williams' original finitary
+conjecture.
+
+\subsection*{Proof overview}
+
+We first construct an ultrafilter $\mathcal U$ on a countable set $D$ and
+an ordinal-valued function $\rho$ on its members. Call a subset of $D$
+\emph{large} if it belongs to $\mathcal U$. This auxiliary ordinal rank, distinct from matroid rank, satisfies
+$\rho(X)\ge\rho(Y)$ whenever $X\subseteq Y$ are large. On two labelled copies $L,R$ of $D$, we construct
+a self-dual matroid $Q$ with the following property: if $B$ is a basis
+and its slice $B_L\subseteq D$ is large, then $D\setminus B_R$ is large
+and
+\[
+ \rho(B_L)>\rho(D\setminus B_R).
+\]
+This is the property used at the end of the proof. To build the pair,
+join consecutive integer vertices by \emph{bundles}, each a labelled
+copy of $D$, and place $Q$ on the two incident bundles at each vertex.
+Its $L$-coordinate points toward the central bundle between $-1$ and $0$,
+and its $R$-coordinate points outward. The copies at even and odd vertices
+form $M_0$ and $M_1$, respectively. In a putative independent covering, one starts with the large side of a
+central bundle and follows the path outward. Each successive copy of
+$Q$ strictly decreases the ordinal rank, an impossibility.
+
+Most of the proof constructs $Q$. Partition $D$ into rapidly growing
+finite blocks. The first ideal consists of subsets whose proportions
+in these blocks tend to zero. Within each equivalence class modulo this
+ideal, we define real-valued probes of a subset; a half-open interval
+condition on their lower or upper limits selects the bases. Finite
+changes shift these limits by their net cardinality, making basis
+exchange immediate. A separate interpolation argument supplies bases
+between suitable endpoints, including endpoints with infinite limits.
+This argument is the key to maximal extension.
+
+A second, larger ideal permits the insertion of sets that preserve the
+ordinal inequalities while separating previously chosen classes.
+Independent binary columns supply these separating sets simultaneously
+for fewer than continuum many requirements. We then address every
+interval of subsets in a transfinite recursion, inserting complementary
+classes of bases when necessary. At each stage there are fewer than
+continuum many earlier requirements; no regularity assumption on the
+continuum is used.
+
+Section~\ref{sec:filters} constructs the two ideals, the ultrafilter, the
+rank and the simultaneous reservations. Section~\ref{sec:local-bases}
+proves the interpolation properties within a single class.
+Section~\ref{sec:assembly} assembles the classes and proves all the
+matroid axioms, including maximal extension.
+Section~\ref{sec:counterexample} places $Q$ along the path and completes
+the ordinal descent argument. Section~\ref{sec:separate-consequences}
+derives the separate universal Covering and Packing consequences.

preprints/A-Counterexample-to-the-Infinite-Matroid-Packing-Covering-Conjecture-September-24-2026/build/sections/02-matroids.tex

@@ -0,0 +1,119 @@
+\section{Matroids and the self-dual reduction}\label{sec:matroids}
+
+We first record the infinite-matroid conventions and explain why, for a
+self-dual pair, it suffices to rule out an independent covering of the
+whole ground set. This reduction is essential: a mixed packing/covering
+partition need not be a packing or a covering on the whole set.
+
+\subsection{Independence, closure, and duality}
+
+An infinite matroid on a set $F$ is a family $\mathcal I\subseteq 2^F$
+satisfying the following axioms
+\cite[Section~1.1]{BDKPW2013}:
+\begin{description}
+\item[(I1)] $\varnothing\in\mathcal I$.
+\item[(I2)] Every subset of a member of $\mathcal I$ belongs to $\mathcal I$.
+\item[(I3)] If $I\in\mathcal I$ is not inclusion-maximal and
+  $B\in\mathcal I$ is inclusion-maximal, there is an element
+  $e\in B\setminus I$ such that $I\cup\{e\}\in\mathcal I$.
+\item[(IM)] For every $I\in\mathcal I$ and $I\subseteq Y\subseteq F$,
+  the family $\{J\in\mathcal I:I\subseteq J\subseteq Y\}$ has an
+  inclusion-maximal member.
+\end{description}
+The inclusion-maximal independent sets are the \emph{bases}. In
+particular, (IM) requires maximal extensions inside every subset, not
+only inside $F$.
+
+For a matroid $N=(F,\mathcal I)$ and $X\subseteq F$, its closure is
+\[
+ \operatorname{cl}_N(X)=X\cup
+ \{e\in F:\text{there is an independent }I\subseteq X
+                 \text{ with }I\cup\{e\}\text{ dependent}\}.
+\]
+This operator is extensive, monotone, and idempotent
+\cite[Theorem~4.2]{BDKPW2013}. A set $S$ is \emph{spanning} if
+$\operatorname{cl}_N(S)=F$.
+Restrictions are matroids, and the complements of the bases of $N$
+are the bases of its dual $N^*$
+\cite[Theorems~3.1 and~3.4]{BDKPW2013}. We use
+\[
+ N.X=N/(F\setminus X)=(N^*\!\upharpoonright X)^*
+\]
+for contraction onto $X$. Self-duality will always mean the equality
+$N=N^*$ on the same labeled ground set.
+
+\begin{lemma}\label{lem:closure-duality}
+Let $N$ be an infinite matroid on $F$.
+\begin{enumerate}
+\item If $S\subseteq X\subseteq F$, then
+\[
+ \operatorname{cl}_{N\upharpoonright X}(S)
+   =\operatorname{cl}_N(S)\cap X.
+\]
+\item A subset $S\subseteq F$ is spanning if and only if it contains
+  a basis of $N$.
+\item A subset $S\subseteq F$ is spanning in $N$ if and only if
+  $F\setminus S$ is independent in $N^*$.
+\end{enumerate}
+\end{lemma}
+\begin{proof}
+The first assertion follows directly from the closure definition:
+for $e\in X$, independence of sets contained in $S\cup\{e\}$ is
+the same in $N$ and in its restriction to $X$.
+
+For the second, use (IM) to take a maximal independent subset $B$ of
+$S$. For every $e\in S\setminus B$, the set $B\cup\{e\}$ is
+dependent, so $S\subseteq\operatorname{cl}_N(B)$. Monotonicity and
+idempotence therefore give
+\[
+ \operatorname{cl}_N(B)=\operatorname{cl}_N(S).
+\]
+If $S$ is spanning, so is $B$. An independent spanning set is a basis:
+an independent extension by an element of its closure would contradict
+the defining independent witness for closure. Conversely, a basis spans,
+because adjoining any element outside it gives a dependent set; any set
+containing a basis consequently spans.
+
+If $S$ contains a basis $B$, then $F\setminus S$ is contained in
+the dual basis $F\setminus B$. Conversely, extend an independent
+$F\setminus S$ to a basis $B^*$ of $N^*$ using (IM).
+The primal basis $F\setminus B^*$ is contained in $S$.
+\end{proof}
+
+\subsection{From a mixed partition to an independent covering}
+
+\begin{lemma}\label{lem:selfdual-reduction}
+Let $N_0,N_1$ be matroids on the same set $F$, with $N_i=N_i^*$
+on the same labels. Suppose there is a partition
+$F=P\mathbin{\dot\cup}C$, disjoint sets $S_0,S_1\subseteq P$
+spanning $N_0\!\upharpoonright P,N_1\!\upharpoonright P$,
+respectively, and independent sets $I_i$ of $N_i.C$ such that
+$I_0\cup I_1=C$. Then $F$ can be covered by one independent set
+of each of $N_0,N_1$.
+\end{lemma}
+\begin{proof}
+Self-duality gives
+\[
+ N_i.C=(N_i\!\upharpoonright C)^*.
+\]
+Thus $T_i=C\setminus I_i$ spans $N_i\!\upharpoonright C$ by
+Lemma~\ref{lem:closure-duality}. Also
+$T_0\cap T_1=C\setminus(I_0\cup I_1)=\varnothing$.
+Consequently the two sets
+\[
+ A_i=S_i\cup T_i\qquad(i=0,1)
+\]
+are disjoint. The restriction-closure identity shows that
+$P\subseteq\operatorname{cl}_{N_i}(S_i)$ and
+$C\subseteq\operatorname{cl}_{N_i}(T_i)$, so $A_i$ spans the
+full matroid $N_i$. Its complement $F\setminus A_i$ is independent
+in $N_i^*=N_i$. Finally,
+\[
+ (F\setminus A_0)\cup(F\setminus A_1)=F
+\]
+because $A_0\cap A_1=\varnothing$.
+\end{proof}
+
+We will construct a self-dual pair with no independent covering.
+Lemma~\ref{lem:selfdual-reduction} will then exclude every mixed
+packing/covering partition for that pair.

preprints/A-Counterexample-to-the-Infinite-Matroid-Packing-Covering-Conjecture-September-24-2026/build/sections/03-filters.tex

@@ -0,0 +1,278 @@
+\section{An ultrafilter rank and simultaneous reservations}
+\label{sec:filters}
+
+We construct two tools on a countable set. The first is an ordinal-valued
+function on an ultrafilter whose value can be increased while preserving
+a prescribed lower bound on the set. The second supplies two disjoint
+sets that meet each of fewer than continuum many prescribed sets in
+positive upper block density. These two tools will allow us to choose bases
+with an ordinal inequality while also separating them from all earlier
+choices.
+
+\subsection{Blocks, columns, and two ideals}
+
+Let $\mathfrak c=2^{\aleph_0}$, also regarded as its initial ordinal.
+For each integer $m\ge0$, let
+\[
+ D_m=\{m\}\times\{0,1\}^{\{0,1\}^m},\qquad
+ D=\coprod_{m\ge0}D_m,\qquad w_m=|D_m|=2^{2^m}.
+\]
+Thus $D$ is countably infinite. The tags $m$ make its finite blocks
+disjoint. Their sizes satisfy
+\begin{equation}\label{eq:block-growth}
+ w_m\ge2\sum_{j<m}w_j\qquad(m\ge1).
+\end{equation}
+Indeed, equality holds for $m=1$. If the inequality holds for $m$, then
+$2\sum_{j<m+1}w_j\le3w_m\le w_m^2=w_{m+1}$, since $w_m\ge4$.
+
+The density-zero ideal on $D$ is
+\[
+ \mathcal J=\left\{X\subseteq D:
+       \frac{|X\cap D_m|}{w_m}\longrightarrow0\right\}.
+\]
+An ideal is a family closed under taking subsets and finite unions;
+we call a set \emph{positive} for the ideal if it does not belong to it.
+The displayed family is a proper ideal containing all finite sets.
+
+Let $L$ and $R$ be labelled copies of $D$, and write
+$E_0=L\mathbin{\dot\cup}R$. For $X\subseteq E_0$, its two slices,
+viewed as subsets of $D$, are denoted by $X_L$ and $X_R$.
+When writing lifts, we identify $E_0$ with $D\times\{L,R\}$.
+Write $W_m$ for the union of the two copies of $D_m$.
+The ideal $\mathcal J_0$ on $E_0$ consists of the sets whose two slices
+belong to $\mathcal J$. Equivalently,
+\[
+ X\in\mathcal J_0
+ \quad\Longleftrightarrow\quad
+ \frac{|X\cap W_m|}{|W_m|}\longrightarrow0.
+\]
+This equivalence follows because the doubled-block density is the
+average of the two nonnegative slice densities. Complements are always
+taken in the relevant ground set, $D$ or $E_0$.
+
+\begin{lemma}[Independent columns]\label{lem:columns}
+There are subsets $A_t,B_t,G_t$ of $D$, indexed by $t<\mathfrak c$,
+with the following property. For any $r$ distinct columns from these
+three families and any prescribed membership pattern on them, the
+proportion of that pattern in $D_m$ is exactly $2^{-r}$ for all
+sufficiently large $m$.
+\end{lemma}
+
+\begin{proof}
+Assign distinct infinite binary strings to all the columns; this is
+possible because $3\mathfrak c=\mathfrak c$. For a column with string
+$h$, put $(m,f)$ in that column precisely when
+$f(h\upharpoonright m)=1$.
+The prefixes of any fixed finite list of distinct strings are distinct
+for all sufficiently large $m$. Prescribing the values of $f$ at these
+$r$ arguments leaves $2^{2^m-r}$ functions among the $2^{2^m}$ choices
+in $D_m$, proving the assertion.
+\end{proof}
+
+We need a larger ideal in addition to $\mathcal J$. Define
+\begin{equation}\label{eq:larger-ideal}
+ \mathcal K=\left\{X\subseteq D:
+ X\cap\bigcap_{t\in F}A_t\in\mathcal J
+ \text{ for some finite }F\subseteq\mathfrak c\right\},
+\end{equation}
+where an empty intersection is $D$. This is the ideal generated by
+$\mathcal J$ and the complements $A_t^c$. To see the equivalence, split
+$X$ into its intersection with the positive $A$-columns in
+\eqref{eq:larger-ideal} and its remaining part, which is covered by
+their complements. In particular $\mathcal J\subseteq\mathcal K$.
+Lemma~\ref{lem:columns} shows that $D\notin\mathcal K$, so this ideal
+is proper.
+
+We will repeatedly use the following consequence. A consistent finite
+pattern prescribing positive membership in $A$-columns and arbitrary
+signs in $G$-columns is $\mathcal K$-positive. Intersecting such a pattern
+with any further finitely many positive $A$-columns still gives a
+pattern with a fixed positive eventual density, so
+\eqref{eq:larger-ideal} excludes membership in $\mathcal K$.
+
+Write $X\subseteq_{\mathcal K}Y$ if $X\setminus Y\in\mathcal K$.
+On $E_0$, let $\mathcal K_0$ be the ideal of sets whose two slices
+belong to $\mathcal K$. Thus $\mathcal J_0\subseteq\mathcal K_0$.
+The distinction between these ideals matters: changes in
+$\mathcal K_0$ will preserve an ordinal condition, whereas differences
+positive for $\mathcal J_0$ will keep the resulting bases incomparable.
+
+\subsection{The ultrafilter and its ordinal rank}
+
+An ultrafilter $\mathcal U$ on $D$ is a proper filter that contains
+exactly one of $X$ and $D\setminus X$ for every $X\subseteq D$.
+
+\begin{lemma}\label{lem:ultrafilter}
+There is an ultrafilter $\mathcal U$ on $D$ disjoint from $\mathcal K$,
+containing every $G_t$, and satisfying
+\begin{equation}\label{eq:finitely-many-columns}
+ X\in\mathcal U\quad\Longrightarrow\quad
+ \{t<\mathfrak c:X\subseteq_{\mathcal K}G_t\}
+ \text{ is finite}.
+\end{equation}
+\end{lemma}
+
+\begin{proof}
+Call $Z\subseteq D$ forbidden if it is contained modulo $\mathcal K$
+in infinitely many of the $G_t$. Require a filter to contain all
+complements of members of $\mathcal K$, all $G_t$, and all complements
+of forbidden sets. We verify the finite-intersection property.
+
+Combine the finitely many $\mathcal K$-small exclusions into one set
+$N\in\mathcal K$. Let $F$ be the finite set of required positive
+$G$-indices, and let $Z_1,\ldots,Z_q$ be the forbidden sets to exclude.
+For each $j$, choose an index $t_j$ such that
+$Z_j\setminus G_{t_j}\in\mathcal K$, with the $t_j$ distinct and
+outside $F$. Each $Z_j$ has infinitely many witnessing indices, so
+these choices are possible. The pattern
+\[
+ P=\bigcap_{t\in F}G_t\cap\bigcap_{j=1}^qG_{t_j}^c
+\]
+is $\mathcal K$-positive. Moreover
+$P\cap Z_j\subseteq Z_j\setminus G_{t_j}\in\mathcal K$.
+It follows that
+\[
+ P\setminus\left(N\cup\bigcup_{j=1}^qZ_j\right)
+\]
+is still $\mathcal K$-positive: otherwise $P$ would be a finite union
+of members of $\mathcal K$. This nonempty set lies in the required
+finite intersection. Extend the resulting proper filter to an
+ultrafilter by the ultrafilter lemma. It contains no member of
+$\mathcal K$ and no forbidden set, as required.
+\end{proof}
+
+Fix such an ultrafilter. Call its members \emph{large} and its
+nonmembers \emph{small}. Small in this sense does not mean membership
+in either ideal. Every large set is $\mathcal J$-positive, and every
+complement of a $\mathcal K$-set is large.
+
+List all large subsets of $D$ as $(F_\xi)_{\xi<\mathfrak c}$,
+allowing repetitions if needed. For a large set $X$, define
+\begin{equation}\label{eq:ordinal-rank}
+ \rho(X)=\min\{\xi<\mathfrak c:
+                     F_\xi\subseteq_{\mathcal K}X\}.
+\end{equation}
+The set of indices is nonempty because $X$ occurs in the list.
+This is an auxiliary ordinal rank, distinct from matroid rank.
+
+\begin{lemma}[Rank invariance and monotonicity]\label{lem:rank}
+If $X\triangle Y\in\mathcal K$, then $X$ is large if and only if $Y$
+is large, and in that case $\rho(X)=\rho(Y)$.
+If $X,Y$ are large and $X\subseteq_{\mathcal K}Y$, then
+$\rho(X)\ge\rho(Y)$.
+\end{lemma}
+
+\begin{proof}
+The complement of $X\triangle Y$ belongs to $\mathcal U$.
+Intersecting it with either large set gives a large subset of the
+other, proving invariance of largeness. The two sets have exactly the
+same candidate indices in \eqref{eq:ordinal-rank}, proving rank
+invariance. Finally, $F_\xi\subseteq_{\mathcal K}X$ and
+$X\subseteq_{\mathcal K}Y$ imply
+$F_\xi\subseteq_{\mathcal K}Y$ by closure of the ideal under finite
+unions. Taking minima proves the last assertion.
+\end{proof}
+
+The next property supplies large sets of arbitrarily high rank inside
+any interval whose lower endpoint is small and upper endpoint is
+large.
+
+\begin{lemma}[Rank-raising thinning]\label{lem:thinning}
+Let $s\subseteq X\subseteq D$, with $s$ small and $X$ large, and let
+$\gamma<\mathfrak c$. There is a large set $u$ such that
+\[
+ s\subseteq u\subseteq X,\qquad \rho(u)>\gamma.
+\]
+\end{lemma}
+
+\begin{proof}
+For every $\xi\le\gamma$, the set
+$Y_\xi=F_\xi\cap(X\setminus s)$ is large. By
+\eqref{eq:finitely-many-columns}, it is contained modulo
+$\mathcal K$ in only finitely many $G_t$.
+The union of these finite sets of indices has cardinal at most
+$\max(\aleph_0,|\gamma+1|)<\mathfrak c$. Choose $t$ outside that union
+and put
+\[
+ u=s\cup(X\cap G_t).
+\]
+Then $u$ is large and lies between $s$ and $X$.
+If $F_\xi\subseteq_{\mathcal K}u$ for some $\xi\le\gamma$, then
+\[
+ Y_\xi\setminus G_t
+ =F_\xi\cap X\cap s^c\cap G_t^c
+ \subseteq F_\xi\setminus u\in\mathcal K,
+\]
+contrary to the choice of $t$. Thus $\rho(u)>\gamma$.
+\end{proof}
+
+This counting argument does not require regularity of
+$\mathfrak c$. For any fixed cardinal $\lambda<\mathfrak c$, a union
+of $\lambda$ finite sets has cardinal at most
+$\max(\lambda,\aleph_0)<\mathfrak c$. We will use this same estimate
+at each stage of a recursion of length $\mathfrak c$.
+
+\subsection{Two simultaneous reservations}
+
+The ordinal rank is unchanged by $\mathcal K$-small changes. We now
+show that such changes can nevertheless meet many prescribed sets
+in a way that remains positive for the smaller ideal $\mathcal J$.
+For $t<\mathfrak c$, define the following subsets of $E_0$, using the
+same labels in both copies:
+\[
+ H_{t,0}=(A_t^c\cap B_t^c)\times\{L,R\},\qquad
+ H_{t,1}=(A_t^c\cap B_t)\times\{L,R\}.
+\]
+They are disjoint, and both belong to $\mathcal K_0$ because their
+slices lie in $A_t^c$.
+
+\begin{lemma}[Simultaneous reservations]\label{lem:reservations}
+Let $\mathscr S$ be a family of fewer than $\mathfrak c$ subsets of
+$E_0$, each positive for $\mathcal J_0$. There is an index
+$t<\mathfrak c$ such that
+\[
+ S\cap H_{t,0}\notin\mathcal J_0
+ \quad\text{and}\quad
+ S\cap H_{t,1}\notin\mathcal J_0
+ \qquad(S\in\mathscr S).
+\]
+\end{lemma}
+
+\begin{proof}
+Fix first one $\mathcal J_0$-positive set $S$. Choose $\epsilon>0$
+such that the proportion of $S$ in $W_m$ is at least $\epsilon$ for
+infinitely many $m$. Call an index $t$ bad for $S$ if at least one
+of $S\cap H_{t,0}$ and $S\cap H_{t,1}$ belongs to $\mathcal J_0$.
+We prove that only finitely many indices are bad.
+
+Choose any finite list of distinct bad indices $t_1,\ldots,t_q$, and
+for each $i$ choose $k_i\in\{0,1\}$ with
+$S\cap H_{t_i,k_i}\in\mathcal J_0$. Put $H_i=H_{t_i,k_i}$.
+These cells use $2q$ distinct $A$- and $B$-columns. Avoiding $H_i$
+allows three of the four membership patterns on its two columns.
+Lemma~\ref{lem:columns} therefore gives $3^q$ avoiding patterns out of
+$4^q$ equally frequent patterns in every sufficiently late block.
+Duplicating the cells on the two copies does not change these proportions.
+In particular,
+\[
+ \frac{|W_m\setminus\bigcup_{i=1}^qH_i|}{|W_m|}
+ =\left(\frac34\right)^q
+\]
+for all sufficiently large $m$. Every point of $S\cap W_m$ either
+avoids all these cells or lies in one of the failing intersections, so
+\[
+ \frac{|S\cap W_m|}{|W_m|}
+ \le \left(\frac34\right)^q+
+       \sum_{i=1}^q\frac{|S\cap H_i\cap W_m|}{|W_m|}.
+\]
+Each summand tends to zero. Along the infinite subsequence on which
+$S$ has density at least $\epsilon$, this gives
+$\epsilon\le(3/4)^q$. Since $(3/4)^q\to0$, the size of any finite
+list of bad indices is bounded. Hence only finitely many indices are bad.
+
+For the family $\mathscr S$, take the union of these finite sets of
+bad indices. Its cardinal is less than $\mathfrak c$ by the estimate
+following Lemma~\ref{lem:thinning}. Any index outside that union
+works simultaneously for every $S\in\mathscr S$. If the family is
+empty, any index works.
+\end{proof}

preprints/A-Counterexample-to-the-Infinite-Matroid-Packing-Covering-Conjecture-September-24-2026/build/sections/04-local-bases.tex

@@ -0,0 +1,332 @@
+\section{Bases within a density-zero class}\label{sec:local-bases}
+
+We now construct families of candidate bases within one equivalence class
+modulo $\mathcal J_0$. The purpose is twofold: balanced finite changes
+must preserve the family, and intervals of subsets must contain or lie
+on one side of a member of the family. The latter property will supply
+maximal independent extensions in the eventual matroid.
+
+Fix $T\subseteq E_0$ such that $T$ and $T^c$ are
+$\mathcal J_0$-positive, and put
+\[
+ \mathcal C(T)=\{X\subseteq E_0:X\triangle T\in\mathcal J_0\}.
+\]
+We use $T$ as the chosen representative, or \emph{prototype}, of this class.
+Every member of $\mathcal C(T)$ and its complement are
+$\mathcal J_0$-positive. Indeed, an ideal-small member, together with its
+ideal-small symmetric difference from $T$, would make $T$ ideal-small;
+the same argument applies to complements.
+
+For $n\ge1$ and $m\ge0$, define
+\begin{align}
+ V_n&=\bigcup_{m<n}W_m,\qquad c_n=|V_n|,\notag\\
+ p_n^T(X)&=|X\cap V_n|-|T\cap V_n|,\notag\\
+ d_m^T(X)&=\frac{|X\cap W_m|-|T\cap W_m|}{|W_m|},\notag\\
+ f_n^T(X)&=p_n^T(X)+c_n\left(
+       \sup_{m\ge n}d_m^T(X)+\inf_{m\ge n}d_m^T(X)\right).
+       \label{eq:probes}
+\end{align}
+These quantities are defined for $X\in\mathcal C(T)$. The prefix count
+records a finite modification exactly once its affected blocks lie in
+$V_n$. The tail extrema allow later-block changes to influence earlier
+probes, and their sum changes sign when both $X$ and $T$ are complemented.
+Write
+\[
+ \ell_T(X)=\liminf_{n\to\infty}f_n^T(X),\qquad
+ h_T(X)=\limsup_{n\to\infty}f_n^T(X),
+\]
+allowing infinite values, and define two families
+\begin{align}
+ \mathcal B^-(T)&=\{X\in\mathcal C(T):-1<\ell_T(X)\le0\},\notag\\
+ \mathcal B^+(T)&=\{X\in\mathcal C(T):0\le h_T(X)<1\}.
+ \label{eq:local-bases}
+\end{align}
+We call these the lower-limit and upper-limit rules, respectively.
+In this section, a basis for a rule means a member of the indicated
+family; no matroid is asserted yet.
+
+\subsection{Probe estimates and continuity}
+
+\begin{lemma}[Probe properties]\label{lem:probe-properties}
+For $X,Y\in\mathcal C(T)$ the following statements hold.
+\begin{enumerate}
+\item $d_m^T(X)\to0$ and $p_n^T(X)/c_n\to0$.
+\item If $X\subseteq Y$, then
+\begin{equation}\label{eq:probe-monotone}
+ f_n^T(Y)-f_n^T(X)\ge |(Y\setminus X)\cap V_n|.
+\end{equation}
+\item Adding or deleting one element changes every probe by at most
+one. A finite modification with net change $b$ in cardinality satisfies
+$f_n^T(Y)=f_n^T(X)+b$ for all sufficiently large $n$.
+\item Complementation gives
+\[
+ f_n^{T^c}(X^c)=-f_n^T(X),\qquad
+ \mathcal B^+(T^c)=\{B^c:B\in\mathcal B^-(T)\}.
+\]
+\end{enumerate}
+\end{lemma}
+
+\begin{proof}
+The first density limit follows from $X\triangle T\in\mathcal J_0$.
+Also $p_n^T(X)=\sum_{m<n}|W_m|d_m^T(X)$. For any $\varepsilon>0$,
+the terms after a fixed index have absolute density at most
+$\varepsilon$; their contribution divided by $c_n$ is at most
+$\varepsilon$, and the finite initial contribution divided by $c_n$
+tends to zero.
+
+Inclusion increases each block count, hence both tail extrema; the
+prefix increases by $|(Y\setminus X)\cap V_n|$. This proves the second
+assertion. If an element changes in a block $W_k$ with $k<n$, only the
+prefix changes. If $k\ge n$, each tail extremum changes by at most
+$1/|W_k|$. By \eqref{eq:block-growth},
+\[
+ |W_k|\ge 2\sum_{m<k}|W_m|\ge 2c_n,
+\]
+so the total change is at most one. After every block affected by a
+finite modification is in the prefix, the tails agree and the prefix
+changes by exactly $b$.
+
+Finally, complementing both $X$ and $T$ negates every prefix and block
+density. The supremum of the negatives is the negative infimum, and
+vice versa. This proves the last assertion, including its half-open
+endpoint conventions.
+\end{proof}
+
+The next continuity statement is needed both to stop finite deletion
+procedures and to pass to their decreasing limit. Its common
+density-zero bound is essential to controlling the tail extrema.
+
+\begin{lemma}[Continuity within a fixed interval]\label{lem:block-continuity}
+Suppose $I\subseteq X_k\subseteq Y$ for $k\ge1$, where
+$I,Y\in\mathcal C(T)$, and membership of every element in $X_k$
+eventually agrees with its membership in a set $X$. Then
+$X\in\mathcal C(T)$ and, for every fixed $n\ge1$,
+\[
+ f_n^T(X_k)\longrightarrow f_n^T(X).
+\]
+\end{lemma}
+
+\begin{proof}
+We have $I\subseteq X\subseteq Y$ and $Y\setminus I\in\mathcal J_0$,
+so $X\in\mathcal C(T)$. On every fixed finite union of blocks, $X_k$
+eventually agrees with $X$. On the remaining blocks,
+\[
+ |d_m^T(X_k)-d_m^T(X)|
+ \le \frac{|(Y\setminus I)\cap W_m|}{|W_m|},
+\]
+uniformly in $k$, and the right-hand side tends to zero. Thus the
+sequences of block densities converge uniformly in $m$. Supremum and
+infimum over a fixed tail change by at most the uniform difference,
+while the fixed prefix eventually stabilizes. Formula
+\eqref{eq:probes} gives the result.
+\end{proof}
+
+\begin{lemma}[Antichains and finite changes]\label{lem:local-antichain}
+Each family $\mathcal B^-(T)$ and $\mathcal B^+(T)$ contains $T$, is
+an antichain under inclusion, and is preserved by finite modifications
+that add and delete equally many elements.
+\end{lemma}
+
+\begin{proof}
+The probes of $T$ vanish. Balanced finite changes preserve the class
+and eventually preserve every probe, so they preserve either rule.
+
+Suppose $X\subsetneq Y$ both satisfy the lower-limit rule. A finite
+difference of cardinality $k\ge1$ gives
+$\ell_T(Y)=\ell_T(X)+k>0$, a contradiction. For an infinite difference,
+$|(Y\setminus X)\cap V_n|\to\infty$. The probes of $X$ are eventually
+bounded below because their lower limit is finite, so
+Lemma~\ref{lem:probe-properties} implies $f_n^T(Y)\to\infty$, again
+a contradiction. The upper-limit case follows by complementing and
+using the lower-limit rule for $T^c$.
+\end{proof}
+
+\subsection{Modifications from a positive pool}
+
+To interpolate between subsets, we need to move the limiting probes
+in either direction while remaining in $\mathcal C(T)$. A positive
+pool of available elements suffices, even though we use only a
+density-zero subset of it.
+
+\begin{lemma}[Small modifications with divergent probes]\label{lem:positive-pool}
+Let $X\in\mathcal C(T)$.
+If $G\subseteq X^c$ is $\mathcal J_0$-positive, there is
+$P\subseteq G$ in $\mathcal J_0$ such that
+$f_n^T(X\cup P)\to+\infty$.
+If $G\subseteq X$ is $\mathcal J_0$-positive, there is
+$P\subseteq G$ in $\mathcal J_0$ such that
+$f_n^T(X\setminus P)\to-\infty$.
+\end{lemma}
+
+\begin{proof}
+For additions, choose $\delta>0$ such that $G$ has density at least
+$\delta$ in arbitrarily late blocks. Suppress the superscript $T$,
+put $a_n=\sup_{m\ge n}|d_m(X)|$, and choose strictly increasing
+positive integers $N_j$ such that, for $n\ge N_j$,
+\begin{equation}\label{eq:spike-cutoff}
+ \frac{|p_n(X)|}{c_n}+2a_n+\frac1{\sqrt{c_n}}\le 2^{-j}.
+\end{equation}
+This is possible by Lemma~\ref{lem:probe-properties}.
+
+For each sufficiently large $j$, put $b_j=2^{1-j}<\delta$ and choose
+an increasing sequence of blocks $m_j\ge N_{j+1}$ in which $G$ has
+density at least $\delta$, also requiring
+$1/|W_{m_j}|\le\delta-b_j$. From $G\cap W_{m_j}$ select exactly
+$\lceil b_j|W_{m_j}|\rceil$ elements. Rounding is possible because
+this number is at most $\delta|W_{m_j}|$. Let $P$ be the union of these
+selections. Its nonzero block densities lie between $b_j$ and
+$b_j+1/|W_{m_j}|$, and hence tend to zero. Thus $P\in\mathcal J_0$.
+
+If $N_j\le n<N_{j+1}$, the selected block $m_j$ belongs to the tail
+in \eqref{eq:probes}. The tail supremum for $X\cup P$ is at least
+$b_j-a_n$, and its tail infimum is at least $-a_n$. The prefix can only
+increase. Therefore \eqref{eq:spike-cutoff} gives
+\[
+ \frac{f_n(X\cup P)}{c_n}
+ \ge -\frac{|p_n(X)|}{c_n}+b_j-2a_n
+ \ge \frac1{\sqrt{c_n}}.
+\]
+These intervals contain every sufficiently large $n$, proving
+divergence to $+\infty$. Apply the addition statement to $X^c$ and
+$T^c$, and use complement symmetry, to obtain the deletion statement.
+\end{proof}
+
+\subsection{Interpolation and interval alternatives}
+
+The finite-change rule rounds any finite limiting value into the
+interval $(-1,0]$. The remaining case has opposite infinite limits at
+the two endpoints. We handle it by finite deletions whose effect on
+earlier probes is summable.
+
+\begin{lemma}[Interpolation]\label{lem:local-interpolation}
+If $I\subseteq Y$ are in $\mathcal C(T)$,
+$\ell_T(I)\le-1$, and $\ell_T(Y)>0$, then there is
+$B\in\mathcal B^-(T)$ with $I\subseteq B\subseteq Y$.
+\end{lemma}
+
+\begin{proof}
+Suppress $T$ from the notation. If $\ell(I)=a$ is finite, add
+$\lfloor-a\rfloor$ elements from $Y\setminus I$. There are enough:
+a smaller finite gap of size $h$ would give
+$\ell(Y)=a+h\le-1$. The resulting lower limit belongs to $(-1,0]$.
+If $\ell(Y)=b$ is finite, delete $\lceil b\rceil$ gap elements from
+$Y$. A smaller finite gap of size $h$ would instead give
+$\ell(I)=b-h>0$. Again the resulting lower limit belongs to $(-1,0]$.
+
+It remains to consider $\ell(I)=-\infty$ and $\ell(Y)=+\infty$.
+Set $X_0=Y$ and construct decreasing sets $X_j$ by finite deletions.
+At stage $j\ge1$, do nothing if some $n\ge j$ already satisfies
+$f_n(X_{j-1})\le0$. Otherwise choose $M_j\ge j$ so large that
+\begin{equation}\label{eq:stage-loss}
+ 2c_j\sup_{m\ge M_j}
+       \frac{|(Y\setminus I)\cap W_m|}{|W_m|}<2^{-j}.
+\end{equation}
+This is possible because $Y\setminus I\in\mathcal J_0$.
+Delete elements of $X_{j-1}\setminus I$ in blocks $m\ge M_j$, one
+at a time in increasing block order, stopping as soon as
+$f_n\le0$ for some $n\ge j$.
+
+This stopping time is finite. If every eligible element were deleted,
+the resulting set would differ from $I$ by finitely many elements,
+so it would have lower limit $-\infty$. Fix a strictly negative probe
+with index $n\ge j$ for that set. By
+Lemma~\ref{lem:block-continuity}, the same probe is negative after
+finitely many of the deletions. Just before the stopping deletion all
+probes with index at least $j$ were positive. Each drops by at most
+one, so
+\begin{equation}\label{eq:stage-reset}
+ f_n(X_j)\ge-1\qquad(n\ge j)
+\end{equation}
+at every stage that makes deletions. For $n<j$ the prefix is unchanged
+during the whole stage, and each tail extremum decreases by at most
+the density supremum in \eqref{eq:stage-loss}. Since $c_n\le c_j$,
+the entire stage decreases $f_n$ by less than $2^{-j}$.
+
+Let $X_\infty=\bigcap_{j\ge0}X_j$. It lies between $I$ and $Y$ and
+belongs to $\mathcal C(T)$; Lemma~\ref{lem:block-continuity} applies
+to this decreasing sequence. Every stage leaves a witness
+$f_n\le0$ with $n\ge j$. Such a witness persists under later
+deletions and in the limit, so $\ell(X_\infty)\le0$.
+
+There is a first stage $j_0$ that makes deletions. Otherwise the
+unchanged set $Y$ would have a nonpositive probe in every tail,
+contrary to $\ell(Y)=+\infty$. Fix $n\ge j_0$. After stage $j_0$ its
+probe is at least $-1$. Through stage $n$, every later stage either
+leaves it unchanged or restores this bound by \eqref{eq:stage-reset}.
+Stages $j>n$ then cost less than $2^{-j}$ each. Passing to the limit,
+\[
+ f_n(X_\infty)\ge-1-\sum_{j>n}2^{-j}=-1-2^{-n}.
+\]
+Consequently $-1\le\ell(X_\infty)\le0$. If the lower limit exceeds
+$-1$, take $B=X_\infty$. If it equals $-1$, add one point of
+$Y\setminus X_\infty$ to obtain lower limit zero. There are infinitely
+many such points: a finite difference from $Y$ would contradict
+$\ell(Y)=+\infty$ by the finite-change rule. This completes the
+interpolation.
+\end{proof}
+
+The local rules have the following interval property. Unless a basis
+lies below the lower endpoint, between the endpoints, or above the upper
+endpoint, the prototype has a positive part outside the lower endpoint
+and the upper endpoint has a positive part outside the prototype.
+These are the two differences needed to separate a new prototype class
+from this one in Section~\ref{sec:assembly}.
+
+\begin{proposition}[Interval alternatives for one class]
+\label{prop:local-intervals}
+Fix $\sigma\in\{-,+\}$ and let $I\subseteq Y\subseteq E_0$.
+If
+\[
+ T\setminus I\in\mathcal J_0
+ \quad\text{or}\quad
+ Y\setminus T\in\mathcal J_0,
+\]
+there is $B\in\mathcal B^\sigma(T)$ satisfying at least one of
+\begin{equation}\label{eq:local-alternatives}
+ B\subseteq I,\qquad I\subseteq B\subseteq Y,
+ \qquad Y\subseteq B.
+\end{equation}
+\end{proposition}
+
+\begin{proof}
+We first establish the interval alternatives when one or both endpoints
+belong to the class. For the lower-limit rule, if
+$X\in\mathcal C(T)$ has $\ell_T(X)\le-1$,
+Lemmas~\ref{lem:positive-pool} and~\ref{lem:local-interpolation}
+extend it to a basis by additions from any prescribed positive
+subset of $X^c$. If $\ell_T(X)>0$, their deletion versions give a
+basis contained in $X$, using any prescribed positive subset of $X$.
+Both unrestricted corrections are available because $X$ and $X^c$
+are positive.
+
+If both $I$ and $Y$ belong to $\mathcal C(T)$, a lower endpoint with
+positive lower limit supplies a basis below $I$, and an upper
+endpoint with lower limit at most $-1$ supplies a basis above $Y$.
+An endpoint that is itself a basis also gives an outcome in
+\eqref{eq:local-alternatives}. The only remaining case is
+$\ell_T(I)\le-1$ and $\ell_T(Y)>0$, when
+Lemma~\ref{lem:local-interpolation} gives a basis between them.
+
+If $I$ belongs to the class and $Y\setminus I$ is positive,
+correcting $I$ downwards, leaving it fixed when it is a basis, or
+correcting it upwards from the pool $Y\setminus I$ gives a basis
+below $I$ or between $I$ and $Y$.
+If $Y$ belongs to the class and the gap is positive, the
+corresponding corrections give a basis between $I$ and $Y$ or
+above $Y$.
+Complementation proves these same three conclusions for the
+upper-limit rule, exchanging the two one-endpoint conclusions.
+
+Now suppose $T\setminus I\in\mathcal J_0$.
+If $I\in\mathcal C(T)$, a $\mathcal J_0$-small gap puts $Y$ in the same class,
+whereas a positive gap permits the lower-endpoint conclusion just
+proved. In either case we obtain \eqref{eq:local-alternatives}.
+If $I\notin\mathcal C(T)$, then $T\cap I\in\mathcal C(T)$ and
+$I\setminus T$ is positive. Applying the lower-endpoint conclusion
+to $[T\cap I,I]$ gives a basis contained in $I$.
+
+Finally, if $Y\setminus T\in\mathcal J_0$, apply the result just
+proved to the prototype $T^c$ and the interval $[Y^c,I^c]$, with
+the opposite rule. Complementing its basis gives
+\eqref{eq:local-alternatives} for $T$ and the original rule.
+\end{proof}

preprints/A-Counterexample-to-the-Infinite-Matroid-Packing-Covering-Conjecture-September-24-2026/build/sections/05-assembly.tex

@@ -0,0 +1,342 @@
+\section{Assembling a self-dual matroid}\label{sec:assembly}
+
+We now choose the prototype classes to which the local rules will be applied.
+There are two requirements: every resulting basis must impose the desired
+strict inequality on ordinal ranks, and enough classes must be present to
+allow maximal independent extensions inside every subset of $E_0$.
+The interval criterion used for the second requirement is due to
+Bowler and Geschke~\cite[Theorem~6]{BowlerGeschke2016}; we include the full
+verification of the matroid axioms after constructing the bases.
+
+\subsection{Admissible sets and two monotone zones}
+
+For $X\subseteq E_0=L\dot\cup R$, put
+\[
+ u(X)=X_L,\qquad y(X)=D\setminus X_R.
+\]
+Call $X$ \emph{admissible} if either
+\begin{equation}\label{eq:admissible}
+ \begin{aligned}
+ &u(X),y(X)\in\mathcal U
+       &&\text{and }\rho(u(X))>\rho(y(X)),\\
+ \text{or }\quad
+ &u(X),y(X)\notin\mathcal U
+       &&\text{and }\rho(D\setminus u(X))>
+                         \rho(D\setminus y(X)).
+ \end{aligned}
+\end{equation}
+All ranks appearing here are evaluated on members of $\mathcal U$.
+Define the \emph{Upper zone} to consist of those $X$ for which $u(X)$ is
+large and either $y(X)$ is small, or $y(X)$ is large and
+$\rho(u(X))\le\rho(y(X))$.
+Define the \emph{Lower zone} by requiring $X^c$ to lie in the Upper zone.
+Complements of subsets of $E_0$ are taken in $E_0$.
+
+\begin{lemma}\label{lem:zones}
+The Upper zone, Lower zone, and admissible sets partition $2^{E_0}$.
+The Upper zone is upward closed, the Lower zone is downward closed,
+and all three statuses are invariant under symmetric differences in
+$\mathcal K_0$.
+Complementation exchanges the two zones and preserves admissibility.
+Moreover, $\varnothing$ is Lower, $E_0$ is Upper, and every admissible
+set and its complement are $\mathcal J_0$-positive.
+\end{lemma}
+
+\begin{proof}
+Write $u=u(X)$ and $y=y(X)$.
+When $u$ is large and $y$ small, $X$ is Upper; when $u$ is small and $y$
+large, it is Lower.
+If both are large, comparison of their ranks partitions the possibilities
+into Upper and admissible.
+If both are small, comparison of the ranks of their complements partitions
+the possibilities into Lower and admissible.
+Complementing $X$ replaces both $u$ and $y$ by their complements, proving
+the asserted symmetries.
+
+Suppose that $X\subseteq X'$ and $X$ is Upper.
+Then $u\subseteq u'$ and $y'\subseteq y$, with $u'$ large.
+If $y'$ is small, $X'$ is Upper by definition.
+Otherwise $y$ is large as well, and Lemma~\ref{lem:rank} gives
+\[
+ \rho(u')\le\rho(u)\le\rho(y)\le\rho(y'),
+\]
+so again $X'$ is Upper.
+Downward closure of the Lower zone follows by complementation.
+The invariance assertion follows from Lemma~\ref{lem:rank} in the two
+coordinates.
+The statuses of $\varnothing$ and $E_0$ follow directly from the definitions.
+
+If the first line of~\eqref{eq:admissible} holds, $X_L=u$ and
+$(X^c)_R=y$ are large, hence $\mathcal J$-positive.
+If the second line holds, $X_R=D\setminus y$ and
+$(X^c)_L=D\setminus u$ are large instead.
+In either case $X$ and $X^c$ are $\mathcal J_0$-positive.
+\end{proof}
+
+\begin{lemma}[Crossing the zones]\label{lem:zone-crossing}
+If $A\subseteq Z\subseteq E_0$, with $A$ Lower and $Z$ Upper, then
+there is an admissible $B$ satisfying $A\subseteq B\subseteq Z$.
+\end{lemma}
+
+\begin{proof}
+The sets $s=A_L$ and $X=Z_L$ are respectively small and large, and
+$s\subseteq X$.
+We keep $B_R=A_R$, so write $y=D\setminus A_R$.
+If $y$ is large, Lemma~\ref{lem:thinning} gives a large $u$ between
+$s$ and $X$ with $\rho(u)>\rho(y)$; take $B_L=u$.
+If $y$ is small, apply the same Lemma between the small set
+$D\setminus X$ and the large set $D\setminus s$ to obtain a large $v$
+with $\rho(v)>\rho(D\setminus y)$.
+Taking $B_L=D\setminus v$ gives the second line
+of~\eqref{eq:admissible}.
+Both constructions respect the two endpoint inclusions.
+\end{proof}
+
+\subsection{Reserving separation from earlier classes}
+
+Different prototypes must have positive differences in both directions.
+This will ensure that their local bases cannot contain one another.
+The next Lemma arranges these two inequalities while preserving the
+endpoint statuses needed in Lemma~\ref{lem:zone-crossing}.
+
+\begin{lemma}\label{lem:insert-prototype}
+Let $\mathcal T$ be a family of fewer than $\mathfrak c$ admissible subsets
+of $E_0$.
+Suppose $A\subseteq Z\subseteq E_0$, that $A$ is not Upper and $Z$ is
+not Lower, and that, for every $T\in\mathcal T$,
+\begin{equation}\label{eq:available-separation}
+ T\setminus A\notin\mathcal J_0,
+ \qquad Z\setminus T\notin\mathcal J_0.
+\end{equation}
+Then there is an admissible $B$ between $A$ and $Z$ such that
+$B\setminus T$ and $T\setminus B$ are $\mathcal J_0$-positive for every
+$T\in\mathcal T$.
+\end{lemma}
+
+\begin{proof}
+Put $G=Z\setminus A$.
+For every $T\in\mathcal T$ with $A\setminus T\in\mathcal J_0$,
+include the trace $G\setminus T$ in a list.
+It is positive by~\eqref{eq:available-separation}, since
+$Z\setminus T=(A\setminus T)\cup(G\setminus T)$.
+For every $T$ with $T\setminus Z\in\mathcal J_0$, also include the
+positive trace $T\cap G$, using
+$T\setminus A=(T\setminus Z)\cup(T\cap G)$.
+There are fewer than $\mathfrak c$ traces.
+Lemma~\ref{lem:reservations} supplies disjoint $\mathcal K_0$-small sets
+$H_0,H_1$ that meet every listed trace $\mathcal J_0$-positively.
+An empty list presents no restriction on their choice.
+
+Modify only the gap:
+\begin{equation}\label{eq:gap-reservations}
+ A'=A\cup(H_0\cap G),\qquad
+ Z'=Z\setminus(H_1\cap G).
+\end{equation}
+Because $H_0$ and $H_1$ are disjoint and $G$ is disjoint from $A$, we have
+$A'\subseteq Z'$.
+The endpoint changes are $\mathcal K_0$-small, so $A'$ is not Upper and
+$Z'$ is not Lower by Lemma~\ref{lem:zones}.
+If an endpoint is admissible, take it for $B$.
+Otherwise $A'$ is Lower and $Z'$ Upper, and
+Lemma~\ref{lem:zone-crossing} gives an admissible $B$ between them.
+
+For any $T\in\mathcal T$, if $A\setminus T$ is positive then so is
+$B\setminus T$.
+Otherwise the reserved positive set $H_0\cap(G\setminus T)$ lies in
+$B\setminus T$.
+Likewise, if $T\setminus Z$ is positive it is already contained in
+$T\setminus B$; otherwise the reserved positive set $H_1\cap T\cap G$
+is contained in $T\setminus B$.
+This proves both inequalities.
+\end{proof}
+
+\subsection{Meeting every interval}
+
+We choose admissible prototypes in complementary pairs.
+For each first prototype $T$ we retain the local bases
+$\mathcal B^-(T)$, and for $T^c$ we retain $\mathcal B^+(T^c)$.
+These two families are complements of one another by the local rules.
+Once a pair has been inserted, its rule and all its bases remain fixed.
+The aim is that, for every interval $I\subseteq Y\subseteq E_0$ with infinite gap,
+some basis $B$ satisfies the following alternative:
+\begin{equation}\label{eq:interval-alternative}
+ B\subseteq I,\qquad I\subseteq B\subseteq Y,
+ \qquad\text{or}\qquad Y\subseteq B.
+\end{equation}
+
+\begin{proposition}\label{prop:basis-family}
+There is a nonempty family $\mathcal B$ of admissible subsets of $E_0$
+with the following properties:
+\begin{enumerate}
+ \item $\mathcal B$ is an antichain and is closed under complements.
+ \item If $B\in\mathcal B$, $F\subseteq B$ and $H\subseteq E_0\setminus B$
+ are finite with $|F|=|H|$, then $(B\setminus F)\cup H\in\mathcal B$.
+ \item For every $I\subseteq Y\subseteq E_0$ with $Y\setminus I$ infinite,
+ the alternative~\eqref{eq:interval-alternative} holds.
+\end{enumerate}
+\end{proposition}
+
+\begin{proof}
+As $E_0$ is countable, list all intervals with infinite gap in a sequence
+of length $\mathfrak c$.
+Recursively retain complementary pairs of admissible prototypes, requiring
+the two directed differences of any distinct prototypes to be
+$\mathcal J_0$-positive.
+At any stage, call bases in the already retained local families
+\emph{old bases}.
+They are closed under complements.
+
+Consider the current interval $I\subseteq Y$.
+If~\eqref{eq:interval-alternative} already holds for an old basis,
+skip this stage. Otherwise Proposition~\ref{prop:local-intervals}
+implies that every old prototype $T$ satisfies
+\begin{equation}\label{eq:unmet-interval}
+ T\setminus I\notin\mathcal J_0,
+ \qquad Y\setminus T\notin\mathcal J_0.
+\end{equation}
+Thus an unmet interval has the positive differences needed to reserve
+separation from every old class.
+
+An admissible set cannot contain an Upper set or be contained in a Lower
+set, by Lemma~\ref{lem:zones}. Thus if $I$ is Upper, we must seek the
+allowed basis below $I$; if $Y$ is Lower, we must seek it above $Y$.
+Otherwise we seek a basis between the endpoints. Accordingly choose
+\[
+ [A,Z]=
+ \begin{cases}
+ [\varnothing,I],& I\text{ is Upper},\\
+ [Y,E_0],&Y\text{ is Lower},\\
+ [I,Y],&\text{otherwise}.
+ \end{cases}
+\]
+The first two cases cannot coincide, by monotonicity of the Upper zone.
+In every case $A$ is not Upper and $Z$ is not Lower.
+We verify~\eqref{eq:available-separation} for all old prototypes.
+In the third case it is~\eqref{eq:unmet-interval}.
+In the first case $T\setminus A=T$ is positive.
+If $I\setminus T$ were $\mathcal J_0$-small, $I\cap T$ would remain Upper and be
+contained in $T$, forcing the admissible $T$ to be Upper.
+Thus $Z\setminus T=I\setminus T$ is positive.
+In the second case $Z\setminus T=T^c$ is positive.
+If $T\setminus Y$ were $\mathcal J_0$-small, $T\cap Y$ would remain admissible and be
+contained in the Lower set $Y$, a contradiction.
+Thus $T\setminus A=T\setminus Y$ is positive as well.
+
+At a stage $\alpha<\mathfrak c$ there are at most two prototypes per
+earlier stage, hence fewer than $\mathfrak c$ old prototypes.
+Lemma~\ref{lem:insert-prototype} therefore supplies an admissible $B$
+between $A$ and $Z$ with both positive differences from every old
+prototype.
+Insert $B$ with the minus rule and $B^c$ with the plus rule.
+The complement has both positive differences from every old $T$, because
+$T^c$ is also old and
+\[
+ B^c\setminus T=T^c\setminus B,\qquad
+ T\setminus B^c=B\setminus T^c.
+\]
+The two new prototypes have positive differences from one another by
+Lemma~\ref{lem:zones}.
+They therefore define new classes.
+All probes of $B$ relative to itself vanish, so $B\in\mathcal B^-(B)$.
+Its placement in $[A,Z]$ supplies one of the three outcomes
+in~\eqref{eq:interval-alternative} for the original interval.
+
+This recursion uses no regularity assumption on $\mathfrak c$.
+For each individual $\alpha$ below its initial ordinal, the numbers of
+predecessors, prototypes, and requested traces are all bounded by a fixed
+cardinal less than $\mathfrak c$.
+The reservation Lemma applies with that bound.
+Choices can be made using a fixed well-order of $2^{E_0}$.
+
+Let $\mathcal B$ be the union of all retained local basis families.
+It is nonempty, since the interval $[\varnothing,E_0]$ receives an outcome.
+Every member remains admissible under its $\mathcal J_0$-small change
+from its prototype, and the family is closed under complements.
+Lemma~\ref{lem:local-antichain} gives the antichain and finite-change
+properties within each class.
+For different classes, the positive directed differences of their
+prototypes remain positive after $\mathcal J_0$-small changes.
+Thus their bases are incomparable as well.
+Finally every interval outcome persists, because no local family is
+subsequently altered.
+\end{proof}
+
+\subsection{The maximal-extension axiom}
+
+The interval construction is complete.
+We now check that it produces a matroid, including maximal extension
+inside arbitrary subsets; finite exchange alone does not supply this
+last property.
+
+\begin{theorem}\label{thm:local-matroid}
+There is a matroid $Q$ on $E_0=L\dot\cup R$ whose bases are precisely
+the family $\mathcal B$ in Proposition~\ref{prop:basis-family}.
+It is equal to its dual on the same labels, and all its bases are admissible.
+In particular, for every basis $B$,
+\begin{equation}\label{eq:basis-rank-decrease}
+ B_L\in\mathcal U
+ \quad\Longrightarrow\quad
+ D\setminus B_R\in\mathcal U
+ \quad\text{and}\quad
+ \rho(B_L)>\rho(D\setminus B_R).
+\end{equation}
+Every independent set of $Q$ that is not a basis can be extended by any
+element outside it.
+\end{theorem}
+
+\begin{proof}
+Define the independent sets to be all subsets of members of $\mathcal B$.
+Since $\mathcal B$ is nonempty, this gives (I1), and heredity (I2) is
+immediate.
+The maximal independent sets are exactly $\mathcal B$:
+each independent set is contained in a member, and the antichain property
+prevents any such member from having a proper independent extension.
+
+Suppose $I$ is independent but not maximal, and choose $B_0\in\mathcal B$
+properly containing it.
+Let $e\in E_0\setminus I$.
+If $e\in B_0$, then $I\cup\{e\}$ is independent.
+Otherwise choose $f\in B_0\setminus I$; the balanced swap
+$(B_0\setminus\{f\})\cup\{e\}$ is a basis containing $I\cup\{e\}$.
+This proves the final assertion of the theorem.
+For any maximal independent $B_1$, its difference $B_1\setminus I$ is
+nonempty, since $B_1\subseteq I$ would contradict its maximality.
+Taking $e$ from this difference proves (I3).
+
+For (IM), fix an independent $I\subseteq Y\subseteq E_0$.
+If $Y\setminus I$ is finite, choose an independent extension of $I$
+containing a maximum number of elements of that finite gap.
+It is maximal within $Y$.
+If the gap is infinite, use~\eqref{eq:interval-alternative}.
+A basis $B\subseteq I$ forces $B=I$: put $I$ inside a basis $B'$ and
+apply the antichain property to $B\subseteq I\subseteq B'$.
+Thus in this case $I$ itself is maximal within $Y$.
+A basis between $I$ and $Y$ is the required maximal extension.
+A basis containing $Y$ makes $Y$ independent, so $Y$ itself is the
+required extension.
+This proves all four independent-set axioms.
+
+The bases of the dual matroid are the complements of the bases of $Q$.
+Since $\mathcal B$ is complement-closed, $Q^*=Q$ on $E_0$.
+Admissibility was established in Proposition~\ref{prop:basis-family},
+and~\eqref{eq:basis-rank-decrease} is its first case.
+\end{proof}
+
+\begin{remark}\label{rem:uniform-local}
+The matroid $Q$ is uniform in the sense that every subset is independent
+or spanning~\cite{BowlerGeschke2016}.
+Indeed, if $X$ is dependent, maximize an independent subset $I$ inside
+$X$ using (IM).
+If $I$ were not a basis of $Q$, the last assertion of
+Theorem~\ref{thm:local-matroid} would allow another element of $X$ to be
+added.
+Thus $X$ contains a basis and is spanning.
+Every basis and its complement are infinite by admissibility, so $Q$
+has infinite rank and corank.
+Every finite $F\subseteq E_0$ is independent: starting from any basis,
+insert the elements of $F$ that are missing and delete equally many
+elements outside $F$, using its infinitude and balanced finite changes.
+Yet $E_0$ is dependent, because every basis has a nonempty complement.
+Thus $Q$ is not finitary, and self-duality shows that it is not cofinitary
+either.
+\end{remark}

preprints/A-Counterexample-to-the-Infinite-Matroid-Packing-Covering-Conjecture-September-24-2026/build/sections/06-counterexample.tex

@@ -0,0 +1,153 @@
+\section{Two matroids on a double ray}\label{sec:counterexample}
+
+We now use the local matroid $Q$ from
+Theorem~\ref{thm:local-matroid} to construct the required pair.
+Recall that $Q=Q^*$ on $E_0=L\mathbin{\dot\cup}R$, where each
+of $L,R$ is a labeled copy of the countable set $D$. The ultrafilter
+$\mathcal U$ and ordinal rank $\rho$ have the following consequence
+for every basis $B$ of $Q$:
+\begin{equation}\label{eq:outward-rank}
+ B_L\in\mathcal U
+ \quad\Longrightarrow\quad
+ D\setminus B_R\in\mathcal U
+ \quad\text{and}\quad
+ \rho(B_L)>\rho(D\setminus B_R).
+\end{equation}
+The rank is antitone under inclusion of members of $\mathcal U$.
+An independent covering will propagate a member of $\mathcal U$
+along a ray, strictly decreasing this rank at every step.
+
+\subsection{Direct sums and the common ground set}
+
+For completeness, the direct sums we use satisfy all the infinite-matroid
+axioms. Let $(Q_j:j\in J)$ be matroids on pairwise disjoint sets $F_j$,
+where $J$ is countable. Declare $I\subseteq\bigcup_{j\in J}F_j$
+independent if $I\cap F_j$ is independent in every $Q_j$.
+Axioms (I1) and (I2) follow componentwise. The maximal independent
+sets are exactly the unions of one basis from each component: a
+nonmaximal component permits an extension of the union, and if all
+components are maximal no element can be adjoined.
+
+For (I3), if $I$ is not maximal, some $I\cap F_j$ is not maximal
+in $Q_j$. Given a global basis $B$, its part $B\cap F_j$ is a
+component basis. Apply (I3) there and adjoin the resulting element to
+$I$. For (IM), given independent $I\subseteq Y$, choose in each
+component a maximal independent extension of $I\cap F_j$ inside
+$Y\cap F_j$. Their union is a maximal independent extension of $I$
+inside $Y$. These choices are available in ZFC. This proves that the
+direct sum is a matroid, and also proves its stated description of
+bases. If every component is self-dual on its own labels, then the
+complement of each global basis is again a global basis, so the direct
+sum is self-dual on the same labels.
+
+Take vertices $v_n$ indexed by $n\in\mathbb Z$. Between $v_n$ and
+$v_{n+1}$ put the bundle
+\[
+ e_n=\{n\}\times D,
+ \qquad E=\mathop{\dot\bigcup}_{n\in\mathbb Z}e_n.
+\]
+Thus every bundle has the fixed label map $(n,d)\mapsto d$ onto $D$.
+The ground set $E$ is countably infinite: $D$ is a countably infinite
+union of nonempty finite blocks, and $E=\mathbb Z\times D$.
+Designate $e_{-1}$, joining $v_{-1}$ and $v_0$, as the central bundle.
+At every vertex put a copy $Q_n$ of $Q$ on $e_{n-1}\cup e_n$.
+The inward bundle is its $L$-coordinate, and the outward bundle its
+$R$-coordinate; explicitly,
+\[
+\begin{array}{c|cc}
+ &L&R\\\hline
+ n\le -1&e_n&e_{n-1}\\
+ n\ge 0&e_{n-1}&e_n.
+\end{array}
+\]
+Both middle vertices therefore use the central bundle as $L$.
+All copies use the same label maps, ultrafilter, and rank.
+
+
+
+Define
+\[
+ M_0=\bigoplus_{n\in 2\mathbb Z}Q_n,
+ \qquad
+ M_1=\bigoplus_{n\in 2\mathbb Z+1}Q_n.
+\]
+For either parity, the component supports partition $E$, since each
+bundle has exactly one endpoint of that parity. Hence these are
+matroids on the same countably infinite set $E$. The preceding
+direct-sum verification and $Q=Q^*$ give $M_i=M_i^*$ on the same
+labels for $i=0,1$.
+
+The parity placement and inward/outward coordinates are shown in
+Figure~\ref{fig:double-ray}.
+
+\begin{figure}[htbp]
+\centering
+\input{figures/double-ray.tex}
+\caption{Each segment represents a copy of $D$. A copy of $Q$ uses
+the two bundles incident with its vertex. Filled vertices contribute
+to $M_0$, open vertices to $M_1$. The central bundle is $e_{-1}$;
+from either endpoint, the $L$-coordinate points toward the center and
+the $R$-coordinate points outward.}
+\label{fig:double-ray}
+\end{figure}
+
+Every finite subset of $E$ is independent in each $M_i$, because
+all its component parts are finite and every finite subset of $Q$ is
+independent by Remark~\ref{rem:uniform-local}. But $E$ is dependent
+in each $M_i$: its intersection with any component support is the
+dependent whole ground set of that copy of $Q$. Thus neither $M_i$
+is finitary. Since $M_i=M_i^*$, neither is cofinitary either.
+
+In fact, neither matroid has a nonempty finitary or cofinitary direct
+summand. Suppose a nonempty set $F$ supports a finitary summand.
+Every subset of $F$ is independent, since all its finite subsets are
+independent, so $F$ is the summand's only basis. Thus every basis of
+$M_i$ contains $F$. But complements of bases are also bases of $M_i$,
+a contradiction. Dualizing the direct-sum basis description gives the
+same contradiction for a cofinitary summand.
+
+\subsection{The descending ranks}
+
+Suppose that independent sets $J_i$ of $M_i$ cover $E$.
+Replacing $J_1$ by $E\setminus J_0\subseteq J_1$, we obtain an
+independent partition $E=J_0\mathbin{\dot\cup}J_1$.
+Assign each element of a bundle to the endpoint whose parity agrees
+with its part in this partition. At every vertex, the assigned elements
+on its two incident bundles form an independent set of its copy of $Q$.
+
+The central bundle is partitioned between $v_{-1}$ and $v_0$.
+Exactly one of the two label sets belongs to $\mathcal U$.
+Start at that endpoint and move outward forever. Write $x_k\subseteq D$
+for the labels received by the $k$th vertex on its inward bundle.
+Thus $x_0\in\mathcal U$. If the current vertex receives $z_k$
+on its outward bundle, the next vertex receives exactly
+\[
+ x_{k+1}=D\setminus z_k
+\]
+there, because the assignment is a partition and the label maps agree.
+
+Assume inductively that $x_k\in\mathcal U$. The independent
+assignment at this vertex is contained in a basis $B_k$ of its copy
+of $Q$. Put
+\[
+ u_k=(B_k)_L,
+ \qquad y_k=D\setminus(B_k)_R.
+\]
+We have $x_k\subseteq u_k$, so $u_k\in\mathcal U$.
+Equation~\eqref{eq:outward-rank} gives $y_k\in\mathcal U$ and
+$\rho(u_k)>\rho(y_k)$. Since $z_k\subseteq(B_k)_R$,
+we have $y_k\subseteq x_{k+1}$; hence $x_{k+1}\in\mathcal U$.
+Antitonicity now yields
+\begin{equation}\label{eq:ordinal-descent}
+ \rho(x_k)\ \ge\ \rho(u_k)\ >\ \rho(y_k)\ \ge\
+ \rho(x_{k+1}).
+\end{equation}
+This produces an infinite strictly descending sequence of ordinals,
+which is impossible: its set of values has a least element, and the
+next value would be smaller.
+
+Thus $M_0,M_1$ have no independent covering of $E$.
+By Lemma~\ref{lem:selfdual-reduction}, they have no packing/covering
+partition either. Their common ground set is countably infinite and
+both matroids are self-dual on the same labels, completing the proof
+of Theorem~\ref{thm:main}.

preprints/A-Counterexample-to-the-Infinite-Matroid-Packing-Covering-Conjecture-September-24-2026/build/sections/07-consequences.tex

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+\section{The separate Covering and Packing conjectures}
+\label{sec:separate-consequences}
+
+Bowler and Carmesin also formulated separate Covering and Packing
+conjectures. For a set-indexed family $\mathcal M=(M_i:i\in\Theta)$ on $E$,
+write $\mathcal M\!\upharpoonright Y=(M_i\!\upharpoonright Y:i\in\Theta)$
+and $\mathcal M.Y=(M_i.Y:i\in\Theta)$. Their Covering Conjecture asserts,
+for every such family, the equivalence
+\begin{equation}\label{eq:covering-conjecture}
+\begin{aligned}
+ &\mathcal M\text{ has a covering}\\
+ &\quad\Longleftrightarrow\quad
+ \text{for every }Y\subseteq E,\\
+ &\qquad\bigl[\mathcal M\!\upharpoonright Y\text{ has a packing}
+ \Longrightarrow
+ \mathcal M\!\upharpoonright Y\text{ has a covering}\bigr].
+\end{aligned}
+\end{equation}
+Their Packing Conjecture asserts, again for every such family,
+\begin{equation}\label{eq:packing-conjecture}
+\begin{aligned}
+ &\mathcal M\text{ has a packing}\\
+ &\quad\Longleftrightarrow\quad
+ \text{for every }Y\subseteq E,\\
+ &\qquad\bigl[\mathcal M.Y\text{ has a covering}
+ \Longrightarrow
+ \mathcal M.Y\text{ has a packing}\bigr].
+\end{aligned}
+\end{equation}
+These are the conjectures of~\cite[Conjectures~5.2 and~6.3]{BowlerCarmesin2015}.
+
+\begin{corollary}\label{cor:separate-covering-packing}
+The unrestricted separate Covering and Packing Conjectures of Bowler and
+Carmesin are both false in ZFC.
+\end{corollary}
+
+\begin{proof}
+Bowler and Carmesin's universal equivalences identify intersection for
+every pair, packing/covering for every pair, packing/covering for every
+set-indexed family, \eqref{eq:covering-conjecture} for every set-indexed
+family, and \eqref{eq:packing-conjecture} for every set-indexed family
+\cite[Section~7]{BowlerCarmesin2015}. Theorem~\ref{thm:main} falsifies the
+pairwise packing/covering assertion, so both separate universal conjectures
+are false. These last two failures are consequences of equivalence between
+universal statements; this argument does not identify their witnessing
+families with the pair $(M_0,M_1)$.
+\end{proof}
+

preprints/A-Counterexample-to-the-Infinite-Matroid-Packing-Covering-Conjecture-September-24-2026/paper.pdf

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