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Lemma 7. If \( G \) is the internal direct product of \( {N}_{1},\ldots ,{N}_{n} \) then \( G \) is isomorphic to \( \times {}_{i}{N}_{i} \) . On the other hand \( { \times }_{i}{N}_{i} \) is the internal direct product of the standard embeddings of the \( {N}_{i} \) .
Proof: Exercise.
No
Lemma 9. In a modular lattice, the map \( u \mapsto u \sqcap y \) is a lattice isomorphism from \( \left\lbrack {x \sqcup y/x}\right\rbrack \) to \( \left\lbrack {y/x \sqcap y}\right\rbrack \) , with inverse \( v \mapsto x \sqcup v \) .
Proof: In any lattice, \( u \mapsto u \sqcap y \) is readily verified to be a meet preserving map between the two sublattices, and \( v \mapsto x \sqcup v \) a join preserving map in the opposite direction. It therefore suffices to show that the maps are inverse to each other (since being order preserving they then pre...
Yes
Lemma 10. In a modular lattice, suppose \( a \leq A, b \leq B \) . Then \( a \sqcup \left( {A \cap B}\right) /a \sqcup \left( {A \cap b}\right) \) and \( A \cap B/\left( {A \cap b}\right) \sqcup \left( {a \cap B}\right) \) are perspective.
Proof: The proof of lemma 2 goes through; that \( \alpha \cap \beta = \left( {a \cap B}\right) \sqcup \left( {A \cap b}\right) \) follows by modularity.
No
Lemma 12. If \( y \leq x \) then \( l\left( {x/y}\right) = l\left( x\right) - l\left( y\right) \) . Also \( l\left( {x \sqcup y}\right) = l\left( x\right) + l\left( y\right) - l\left( {x \sqcap y}\right) \) .
Proof: There is a maximal chain from \( y \) to 0 containing \( x \) by corollary 6, and the first claim follows. The second claim follows because \( l\left( {x \sqcup u/x}\right) = l\left( {y/x \sqcap y}\right) \) by lemma 9 .
No
Lemma 24. If \( R \) is Artinian then \( \operatorname{Rad}\left( R\right) \) is nilpotent.
Proof: Write \( I \) for \( \operatorname{Rad}\left( R\right) \) ; the sequence \( I,\left\lbrack {I}^{2}\right\rbrack ,\ldots \) is eventually constant, say \( \left\lbrack {I}^{j}\right\rbrack = \left\lbrack {I}^{n}\right\rbrack \) for \( j \geq n \) . Write \( J \) for \( \left\lbrack {I}^{n}\right\rbrack \), and su...
Yes
Theorem 26. \( R \) is semisimple iff \( R \) is Artinian and \( \operatorname{Rad}\left( R\right) = 0 \) .
Proof: One direction was already observed. Conversely suppose \( R \) is Artinian and \( \operatorname{Rad}\left( R\right) = 0 \) . If \( L \) is a left ideal then it is not nilpotent, by the hypothesis that \( \operatorname{Rad}\left( R\right) = 0 \) . By lemma \( {25L} \) contains an idempotent \( e \) . If \( L \) i...
No
Theorem 5. Suppose \( \left\{ {A}_{j}\right\} \) is a set of subsets of some set \( X \), each equipped with a topology. Suppose \( \left\{ {A}_{j}\right\} \) is closed under intersection, \( {A}_{j} \cap {A}_{k} \) has the topology inherited from either \( {A}_{j} \) or \( {A}_{k} \), and \( {A}_{j} \cap {A}_{k} \) is...
Proof: Suppose \( K \subseteq {A}_{j} \) is closed in \( {A}_{j} \) ; then by the hypotheses \( K \cap {A}_{k} \) is closed in \( {A}_{k} \) for all \( k \), whence \( K \) is closed in \( X \) . In particular \( {A}_{j} \) is closed in \( X \) . The argument for open intersections is similar.
Yes
Theorem 6 (Lindelof’s theorem). A second countable space is a Lindelof space.
Proof: Let \( B \) be a countable base and \( C \) a cover of the space \( X \) . For each \( x \in X \) choose a set \( S \in B \) containing \( x \) and a subset of a set of \( C \) . For each such \( S \), of which there are only countably many, choose a superset in \( C \) . The resulting subcollection of \( C \) i...
Yes
Corollary 7. In a second countable space any uncountable set \( S \) has a limit point.
Proof: If \( S \) has no limit point then for each \( x \in S \) we may choose a basic open set containing \( x \) but no other point of \( S \) .
No
Theorem 8 (Urysohn’s lemma). \( X \) is normal iff, given disjoint closed sets \( A, B \) there is a continuous function \( f : X \mapsto \left\lbrack {0,1}\right\rbrack \) such that \( f\left( w\right) = 0 \) for \( w \in A \) and \( f\left( w\right) = 1 \) for \( w \in B \) .
Proof: Suppose \( X \) is normal; for each rational \( r \) of the form \( p/{2}^{n} \) for \( n \geq 0 \) and \( 0 \leq p \leq {2}^{n} \), a set \( {A}_{r} \) will be defined, open if \( r \neq 0 \) . Let \( {A}_{0} = A \) and \( {A}_{1} = {B}^{c} \) . Inductively, if \( t = \left( {{2p} + 1}\right) /{2}^{n + 1} \) le...
Yes
Corollary 9 (Tietze extension theorem). If \( X \) is a normal space, \( K \) is a closed subset, and \( f : K \mapsto \left\lbrack {-1,1}\right\rbrack \) be a continuous function. Then there is a continuous function \( g : X \mapsto \left\lbrack {-1,1}\right\rbrack \) such that \( g \upharpoonright K = f \) .
Proof: Let \( {K}_{0} = \{ x : f\left( x\right) \leq - 1/3\} ,{L}_{0} = \{ x : f\left( x\right) \geq 1/3\} \) . Then \( {K}_{0},{L}_{0} \) are closed and disjoint, so by Urysohn’s lemma there is a continuous function \( {g}_{0} : X \mapsto \left\lbrack {-1/3,1/3}\right\rbrack \) which is \( - 1/3 \) on \( {K}_{0} \) an...
No
Lemma 10 (shrinking lemma). Suppose \( \left\{ {{U}_{j} : j \in J}\right\} \) is a point finite open cover of a normal space \( X \) . Then there is an open cover \( \left\{ {V}_{j}\right\} \) where \( {V}_{j}^{\mathrm{{cl}}} \subseteq {U}_{j} \) .
Proof: Say that a partial function \( \phi \) on \( J \) is a partial shrinking if \( \phi \left( j\right) \) is an open set whose closure is contained in \( {U}_{j} \), and the sets \( \phi \left( j\right) \) for \( j \in \operatorname{Dom}\left( \phi \right) \), together with the sets \( {U}_{j} \) for \( j \notin \o...
Yes
Corollary 11. Under the same hypotheses, there are functions \( {f}_{j} : {U}_{j} \mapsto \left\lbrack {0,1}\right\rbrack \), such that \( {f}_{j}\left( x\right) = 0 \) if \( x \notin {U}_{j} \), and for each \( x \in X,\mathop{\sum }\limits_{j}{f}_{j}\left( x\right) = 1 \) (where by hypothesis the sum is finite).
Proof: By the shrinking lemma and Urysohn’s lemma, choose a function \( {g}_{j} \) which is 1 on a closed subset of \( {U}_{j} \) and 0 outside \( {U}_{j} \) . Then set \( {f}_{j}\left( x\right) = {g}_{j}\left( x\right) /\mathop{\sum }\limits_{j}{g}_{j}\left( x\right) \) .
Yes
Lemma 16. If \( C \) is a locally finite cover of a connected space then \( C \) is countable (or finite).
Proof: Let \( {D}_{0} \) be any member of \( C \), and let \( {D}_{n + 1} \) be \( {D}_{n} \) with the sets of \( C \) which intersect sets of \( {D}_{n} \) added. Let \( D = \mathop{\bigcup }\limits_{n}{D}_{n} \), and let \( Y = \bigcup D \) . \( Y \) is certainly open. If \( U \in C \) intersects \( Y \) then it is i...
No
Theorem 17. Suppose \( X \) is a locally compact Hausdorff space.\na. If \( X \) is second countable then \( X \) is \( \sigma \) -compact.\nb. If \( X \) is \( \sigma \) -compact \( X \) is paracompact.\nc. If \( X \) is connected and paracompact then \( X \) is \( \sigma \) -compact.
Proof: For part a, let \( {U}_{i} \) be a countable base; we claim that the \( {U}_{i} \) for which \( {U}_{i}^{\text{cl }} \) is compact are still a base, and taking the closures of these yields the theorem. Let \( V \) be open and \( x \in V \) . Then since \( X \) is locally compact Hausdorff there is a compact neig...
Yes
Theorem 23 (Tychanoff’s theorem). The product of a collection of compact spaces is compact.
The sets \( K \times \left( {{ \times }_{k \neq j}{X}_{j}}\right), K \subseteq {X}_{j}, K \) closed, form a subbase for the closed sets in the product topology. Let \( C \) be a collection of these with the finite intersection property. Let \( {C}_{i} = \left\{ {{\pi }_{i}\left( K\right) : K \in C}\right\} \) where \( ...
Yes
Lemma 28. Suppose \( \left\langle {{X}_{i},{d}_{i}}\right\rangle, i \geq 1 \), is a countable family of pseudo-metric spaces. The binary function on \( { \times }_{i}{X}_{i} \) given by\n\n\[ d\left( {\left\langle {x}_{i}\right\rangle ,\left\langle {y}_{i}\right\rangle }\right) = \mathop{\sum }\limits_{i}\min \left( {1...
Proof: First note that \( \min \left( {1, d\left( {x, y}\right) }\right) \) is a pseudo-metric (exercise 10), and has the same topology since the open balls of radius less than 1 are a base for the topology of either function. We leave the verification that \( d \) is a pseudo-metric to exercise 10 . Write \( x \) for ...
No
a. The map \( j : X \mapsto {X}^{\text{cmpl }} \) induced by \( {j}_{1} \) is an isometry.
For part a, clearly \( d\left( {{j}_{1}\left( x\right) ,{j}_{1}\left( y\right) }\right) = d\left( {x, y}\right) \) .
No
Lemma 1. If, for a family \( {\alpha }_{i} : c \mapsto F\left( i\right) \), for each \( a{\operatorname{Hom}}^{a} \) maps the family to a limit cone, then \( \langle c,\alpha \rangle \) is a limit for \( F \) . Dually if for a family \( {\alpha }_{i} : F\left( i\right) \mapsto c \), for each \( b{\operatorname{Hom}}_{b...
Proof: Suppose \( \left\{ {{\alpha }_{i} : i \in J}\right\} \) is a family and \( \mu : F\left( i\right) \mapsto F\left( j\right) \) . By hypothesis (in fact using only that \( {\operatorname{Hom}}^{c} \) maps the family to a cone), \( {\mu }^{L}{\alpha }_{i}^{L} = {\alpha }_{j}^{L} \), so \( \mu {\alpha }_{i} = {\mu }...
Yes
Lemma 3. Suppose \( \left\langle {{F}_{i},{G}_{i},{\psi }_{i}}\right\rangle, i = 1,2 \), are adjunctions from \( A \) to \( C \), and \( \alpha \) is a natural transformation from \( {G}_{1} \) to \( {G}_{2} \) . Then there is a unique natural transformation \( \beta \) from \( {F}_{2} \) to \( {F}_{1} \) such that the...
Proof: The function\n\n\[ \n{\beta }_{ac}^{\prime } = {\psi }_{2ac}^{-1}{\alpha }_{c}^{L}{\psi }_{1ac} \n\]\n\nis the unique one making the diagram commutative. One verifies that the \( {\beta }_{ac}^{\prime } \) are the components of a natural transformation from \( \operatorname{Hom}\left( {{F}_{1}\left( a\right), c}...
Yes
Theorem 4. Suppose \( G \) is a functor from \( {B}^{\mathrm{{op}}} \times C \) to \( A \), and each functor \( G\left( {b, - }\right) \) has a left adjoint \( {F}_{b} \) . Then there is a unique functor \( F : A \times B \mapsto C \) agreeing with the \( {F}_{b} \), such that the equivalence (1) holds. The dual statem...
Proof: The object function of \( F \) is clear; the arrow function must be specified. Suppose \( f : {b}_{2} \mapsto {b}_{1} \), and let \( {F}_{i} = {F}_{{b}_{i}},{G}_{i} = G\left( {{b}_{i}, - }\right) \), and \( {\alpha }_{c} = G\left( {f,{\iota }_{c}}\right) \) . By (1),(2) commutes; so by lemma 3 natural transforma...
Yes
Lemma 2. If \( {g}_{i} : a \mapsto {b}_{i} \) is a pullback for \( {f}_{i} : {b}_{i} \mapsto c \) and \( {f}_{1} \) is epic, then \( {g}_{2} \) is epic and \( {f}_{1}^{K} \equiv {g}_{1}{g}_{2}^{K} \) . Dually if \( {g}_{i} : {b}_{i} \mapsto a \) is a pushout for \( {f}_{i} : c \mapsto {b}_{i} \) and \( {f}_{1} \) is mo...
Proof: Let \( {p}_{i} \) and \( {m}_{i} \) be the projections and injections for \( {b}_{1} \times {b}_{2} \) . Let \( \phi = {f}_{1}{p}_{1} - {f}_{2}{p}_{2} \), and \( k = \operatorname{Ker}\left( \phi \right) \) , so that \( {g}_{i} \) may be taken as \( {p}_{i}k \) . Now, \( \phi {m}_{1} = {f}_{1} \), so \( \phi \) ...
Yes
Lemma 3. \( \operatorname{Im}\left( f\right) \equiv \operatorname{Ker}\left( g\right) \) iff \( {gf} = 0 \), and if \( {gx} = 0 \) then \( {x\epsilon } = {ft} \) for some \( t \) and epic \( \epsilon \) . Dually \( \operatorname{Coim}\left( f\right) \equiv \operatorname{Coker}\left( g\right) \) iff \( {fg} = 0 \), and ...
Proof: Write \( f \) as a coimage-image pair \( {me} \) . If \( \operatorname{Ker}\left( g\right) = m \) then \( {gm} = 0 \), so \( {gf} = 0 \) ; and if \( {gx} = 0 \) then \( x = {m\theta } \) for some \( \theta \), so by pullback \( {x\epsilon } = {m\theta \epsilon } = {met} = {ft} \) for some \( t \) and epic \( \ep...
Yes
Theorem 7 (NINE LEMMA). Given a diagram in an Abelian category\n\n![9108176a-2e3c-43c0-9470-80fd40d2a9ad_224_2.jpg](images/9108176a-2e3c-43c0-9470-80fd40d2a9ad_224_2.jpg)\n\nwith exact columns and bottom two rows, there exist unique \( {m}_{0} \) and \( {e}_{0} \) making the diagram commutative; further the top row ext...
Proof: Exercise 5.
No
Theorem 8 (Middle nine lemma). Suppose that, in the diagram of theorem 7 short exact sequences are given in the first and third rows, and arrows \( m, e \) in the middle row. Then the middle row extends to a short exact sequence.
Proof: Exercise 6.
No
Lemma 16. If given a left ideal \( I \subseteq R \) and a linear map \( f : I \mapsto A \), there is a linear map \( {f}^{\prime } : R \mapsto A \) extending \( f \), then \( A \) is injective.
Proof: Suppose \( C \) is a submodule of \( B \), and \( f : C \mapsto A \) . By a typical application of Zorn’s lemma (see for example theorem 15.4) it suffices to show that, for a submodule \( D \) with \( C \subseteq D \), and \( x \in B - D \), a function \( g : D \mapsto A \) can be extended to the submodule gener...
Yes
Theorem 26. Suppose \( R \) is an integral domain, and \( M \) an \( R \) -module. Let \( Q \) denote \( R/{R}_{ \neq } \), and let \( J \) denote \( Q/R \) . Then \( \operatorname{Tor}\left( M\right) \cong {\operatorname{Tor}}_{1}\left( {M, J}\right) \) .
Proof: \( R \) and \( Q \) are both flat, so from the long exact sequence for \( 0 \rightarrow R \rightarrow Q \rightarrow J \rightarrow 0,{\operatorname{Tor}}_{n}\left( {M, J}\right) = 0 \) for \( n \geq 2 \) . If \( M \) is torsion then \( M \otimes Q = 0 \) (if \( {rm} = 0 \) for \( r \neq 0 \) then \( m \otimes {r}...
No
Corollary 3. Any two algebraic closures of a field \( F \) are isomorphic, by an isomorphism fixing \( F \) .
Proof: Let \( {K}_{1} \) and \( {K}_{2} \) be two algebraic closures. By the theorem there is an embedding \( \sigma : {K}_{1} \mapsto {K}_{2} \) fixing \( F \) . Further \( \sigma \left\lbrack {K}_{1}\right\rbrack \) is algebraically closed and \( {K}_{2} \) is algebraic over it, so \( \sigma \left\lbrack {K}_{1}\righ...
Yes
Lemma 4. Suppose \( E \supseteq F \) is an extension of fields. Let \( S \subseteq E \) be an algebraically independent subset of size \( k > 0 \), and let \( T \) be a subset of size \( k + 1 \) . If \( S \cup \{ t\} \) is algebraically dependent for each \( t \in T \) then \( T \) is algebraically dependent.
Proof: Suppose \( \{ 1,\ldots, k\} \) is the disjoint union of \( {I}_{s} \) and \( {I}_{t} \) ; we claim that every \( {s}_{i} \) and \( {t}_{i} \) is algebraic over \( F\left( {\left\{ {{s}_{i} : i \in {I}_{s}}\right\} \cup \left\{ {{s}_{i} : i \in {I}_{t}}\right\} }\right) \) . The claim is proved by induction on \(...
No
Lemma 8. Suppose \( \\left| x\\right| \) satisfies the restrictions of lemma 7. Then \( \\left| x\\right| \) satisfies the triangle inequality iff \( \\left| x\\right| \\leq 1 \) implies \( \\left| {x + 1}\\right| \\leq 2 \) .
Proof: If the triangle inequality holds and \( \\left| x\\right| \\leq 1 \) then \( \\left| {x + 1}\\right| \\leq 2 \) . For the converse, supposing \( \\left| x\\right| \\leq \\left| y\\right| \), \( \\left| {x + y}\\right| = \\left| y\\right| \\left| {x/y + 1}\\right| \\leq 2\\left| y\\right| = 2\\max \\left( {\\left...
Yes
Lemma 12. An absolute value \( X \) is non-Archimedean iff \( \left| x\right| \leq 1 \) for all \( x \) in the prime field of \( K \) .
Proof: One direction follows by lemma 11. The proof in the other direction proceeds similarly to that of the converse direction of lemma 10; using the binomial theorem and the hypothesis one obtains \( {\left| x + y\right| }^{n} \leq \) \( \left( {n + 1}\right) {\left( \max \left( \left| x\right| ,\left| y\right| \righ...
Yes
Corollary 17. Suppose \( A, B, C, D \) are commutative rings.\na. If \( B \) is a finitely generated \( A \) -module then \( B \supseteq A \) is integral.\nb. If \( B \supseteq A \) is integral and \( B \) is finitely generated over \( A \) as a ring then \( B \) is a finitely generated \( A \) -module.\nc. If \( C \su...
Proof: For part a, let \( M = B \) in the theorem. For part b, first note that if \( B \) is a finitely generated \( A \) -module, say by \( {u}_{1},\ldots ,{u}_{m} \), and \( C \) is a finitely generated \( B \) -module, say by \( {v}_{1},\ldots ,{v}_{n} \), then \( C \) is a finitely generated \( A \) -module, by \( ...
Yes
Lemma 29. Suppose \( {N}_{i} \) is a \( P \) -primary submodule for \( 1 \leq i \leq r \) and \( N = { \cap }_{1 \leq i \leq r}{N}_{i} \) ; then \( N \) is \( P \) -primary.
Proof: By hypothesis \( \operatorname{Rad}\left( {\operatorname{Ann}\left( {M/{N}_{i}}\right) }\right) = P \) for all \( i \), so by theorem 26 and the preceding paragraph \( \operatorname{Rad}\left( {\operatorname{Ann}\left( {M/N}\right) }\right) = P \) . Suppose \( {ax} \in N \) and \( x \notin N \) ; then for some \...
Yes
Lemma 30. Suppose \( N \) is a \( P \) -primary submodule, and \( x \in M \) . Then \( \operatorname{Rad}\left( \left( {N : x}\right) \right) = R \) if \( x \in N \), else \( P \) .
Proof: If \( x \in N \) then clearly \( \left( {N : x}\right) = R \) ; suppose \( x \notin N \) . If \( a \in P \) then \( {a}^{n} \in \operatorname{Ann}\left( {M/N}\right) \) for some \( n \), so \( {a}^{n} \in \left( {N : x}\right) \), so \( a \in \operatorname{Rad}\left( \left( {N : x}\right) \right) \) . If \( a \n...
Yes
Corollary 35. If \( F \) is algebraically closed an ideal \( I \) in \( F\left\lbrack \mathbf{x}\right\rbrack \) is maximal iff it is the ideal generated by \( \left\{ {{x}_{1} - {a}_{1},\ldots ,{x}_{n} - {a}_{n}}\right\} \) for some \( {a}_{1},\ldots ,{a}_{n} \in F \) .
Proof: One direction is proved above. Let \( I \) be maximal, so that \( F\left\lbrack \mathbf{x}\right\rbrack /I \) is a field. By the theorem, this field is an algebraic extension of \( F \), and since \( F \) is algebraically closed it equals \( F \) . Lemma 33 then shows that it has the specified form.
Yes
Corollary 36 (weak Nullstellensatz). If \( F \) is algebraically closed and \( I \) is a proper ideal in \( F\left\lbrack \mathbf{x}\right\rbrack \) then \( V\left( I\right) \) is nonempty.
Proof: Choose a maximal ideal \( J \) containing \( I \), and suppose it is generated by \( \left\{ {{x}_{1} - {a}_{1},\ldots ,{x}_{n} - {a}_{n}}\right\} \) . Then \( \mathbf{a} \in V\left( I\right) \) where \( \mathbf{a} = \left\langle {{a}_{1},\ldots ,{a}_{n}}\right\rangle \) .
Yes
Theorem 37 (Nullstellensatz). Suppose \( F \) is algebraically closed, and \( I \) is an ideal in \( F\\left\\lbrack \\mathbf{x}\\right\\rbrack \) . Suppose \( p \\in F\\left\\lbrack \\mathbf{x}\\right\\rbrack \) vanishes on \( V\\left( I\\right) \) . Then \( {p}^{m} \\in I \) for some integer \( m \), that is, \( p \\...
Proof: Suppose \( {p}_{1},\\ldots ,{p}_{t} \) generate \( I \), and let \( J \) be the ideal in \( F\\left\\lbrack {\\mathbf{x}, y}\\right\\rbrack \) generated by \( {p}_{1},\\ldots ,{p}_{t},1 - {yp} \) (this is called Rabinowitsch’s trick). \( J \) is readily verified to have an empty zero set, whence it is the entire...
Yes
Lemma 47. Suppose \( x, d \in {\mathcal{H}}_{h} \) and \( d \neq 0 \) . Then there are \( q, r \in \mathcal{H} \) with \( x = {qd} + r \) and \( Q\left( r\right) < Q\left( d\right) \) .
Proof: First we show the theorem when \( d \) is an integer \( m \) . Let \( \zeta = \left( {1 + i + j + k}\right) /2 \) . Every \( x \in {\mathcal{H}}_{h} \) can be written uniquely as \( {x}_{0}^{\prime }\zeta + {x}_{1}^{\prime }i + {x}_{2}^{\prime }j + {x}_{3}^{\prime }k \), indeed \( {x}_{0}^{\prime } = 2{x}_{0} \)...
Yes
Lemma 48. If \( p \) if an odd prime then for some \( r, s \) with \( 0 \leq r, s < p,1 + {r}^{2} + {s}^{2} \equiv 0{\;\operatorname{mod}\;p} \) .
Proof: The numbers \( {x}^{2} \) with \( 0 \leq x \leq \left( {p + 1}\right) /2 \) are incongruent \( {\;\operatorname{mod}\;p} \), and likewise the numbers \( 1 - {y}^{2} \) with \( 0 \leq y \leq \left( {p + 1}\right) /2 \) . Therefore \( {x}^{2} \equiv 1 - {y}^{2}{\;\operatorname{mod}\;p} \) for some such \( x, y \) ...
No
Corollary 4. Suppose \( D \) is a complete lattice and \( G \) is an order preserving map from \( D \) to a poset \( C \) . Then \( G \) has a left adjoint iff it preserves meets.
Proof: It has already been shown that a right adjoint preserves meets. For the converse, let \( F\left( c\right) = \) \( \sqcap \{ d \in D : c \leq G\left( d\right) \} \) . Since \( G \) preserves meets \( {GF}\left( c\right) = \sqcap \{ G\left( d\right) : c \leq G\left( d\right) \} \), whence \( {GF}\left( c\right) \g...
Yes
Lemma 11. Suppose \( F \) is a functor from \( A \) to \( B, G \) is a functor from \( B \) to \( A \), and \( {FG} \) and \( {GF} \) are naturally equivalent to the identity functor. Then \( F \) and \( G \) are faithful, full, and both left and right adjoints to each other.
Proof: Let \( \alpha \) be the natural equivalence in \( A \) from \( {GF} \) to the identity. Suppose \( f : {a}_{1} \mapsto {a}_{2} \) is an arrow in \( A \) ; the equation \( f{\alpha }_{{a}_{1}} = {\alpha }_{{a}_{2}}{GF}\left( f\right) \) shows that \( f \) is determined by \( F\left( f\right) \), i.e., \( F \) is ...
Yes
Lemma 12. Suppose \( A, B \in {\operatorname{Lat}}_{B} \) and \( f : A \mapsto B \) is a morphism. Let \( {\beta }_{a}^{A} \) be as above.\na. For \( a, b \in A,{\beta }_{a \sqcap b} = {\beta }_{a} \cap {\beta }_{b} \) .
Proof: For part \( \mathrm{a}, f \in {\beta }_{a \sqcap b} \) iff \( f\left( {a \sqcap b}\right) = 1 \) iff \( f\left( a\right) \sqcap f\left( b\right) = 1 \) iff \( f\left( a\right) = 1 \) and \( f\left( b\right) = 1 \) iff \( f \in {\beta }_{a} \cap {\beta }_{b} \) .
Yes
Lemma 13. If \( A \in {\operatorname{DLat}}_{B} \), an open set in \( \operatorname{Spec}\left( A\right) \) is compact iff it is \( {\beta }_{a} \) for some \( a \in A \) .
Proof: Suppose \( {\beta }_{a} \subseteq { \cup }_{b \in S}{\beta }_{b} \), i.e., if \( f\left( a\right) = 1 \) then for some \( b \in {Sf}\left( b\right) = 1 \) . Let \( I \) be the ideal generated by \( S \) and let \( F \) be the filter \( {a}^{ \geq } \) . By lemma 11.16, if \( I \cap F = \varnothing \) then there ...
Yes
Lemma 25. Let \( L \) be a complete lattice, and \( T \subseteq \operatorname{Pt}\left( L\right) \) . Then \( T \) is compact iff \( {\sigma }_{T} \) is Scott open.
Proof: First, note that in a lattice, a subset \( U \) is Scott open iff it is \( \geq \) -closed, and whenever \( \sqcup S \in U \) then for some finite \( F \subseteq S, \sqcup F \in U \) . Second, recall that \( { \cup }_{x \in S}{\beta }_{x} = {\beta }_{\sqcup S} \) ; whence \( T \subseteq { \cup }_{x \in S}{\beta ...
Yes
Lemma 29. Let \( L \) be a complete lattice, and let \( \prec \) be \( { \prec }_{T} \) where \( T \) is the Scott topology.\na. If \( x \prec y \) then \( x \ll y \).
Proof: Suppose \( y \in U \) where \( U \) is Scott open and \( U \subseteq {x}^{ \geq } \), and suppose \( y \leq \sqcup S \) . Then \( \sqcup S \in U \), so \( \sqcup F \in U \) for some finite \( F \subseteq S \), and \( x \leq \sqcup F \) . This proves part a.
Yes
Lemma 30. A topology \( T \) on \( \operatorname{Hom}\left( {Y, Z}\right) \) is in \( {\mathcal{T}}_{2} \) iff it makes \( {\varepsilon }_{YZ} \) continuous.
Proof: For any topology \( T\bar{\varepsilon } \), being the identity, is continuous from \( \operatorname{Hom}\left( {Y, Z}\right) \) to \( \operatorname{Hom}\left( {Y, Z}\right) \), both equipped with \( T \) . If \( T \in {\mathcal{T}}_{2} \) it follows that \( \varepsilon \) is continuous. Conversely suppose \( T \...
No
Lemma 31. \( {\mathcal{T}}_{1} \) is closed down, \( {\mathcal{T}}_{2} \) is closed up, and any topology in \( {\mathcal{T}}_{1} \) is weaker than any topology in \( {\mathcal{T}}_{2} \) . In particular, there is at most one exponential topology.
Proof: If \( \bar{g} \) is continuous when \( \operatorname{Hom}\left( {Y, Z}\right) \) is equipped with \( T \), and \( {T}^{\prime } \subseteq T \), then \( \operatorname{Hom}\left( {Y, Z}\right) \) is continuous when equipped with \( {T}^{\prime } \) ; thus, \( {\mathcal{T}}_{1} \) is closed down. If \( \bar{g} \) i...
Yes
Lemma 33. Let \( S \) denote the Scott topology on \( \Omega \left( Y\right) \), and \( T \) any other topology.\na. \( S \in {\mathcal{T}}_{1} \).
Proof: Suppose \( W \subseteq \times Y \) is open, \( x \in W \), and \( O \) is a Scott open neighborhood of \( {W}_{x} \) . To prove part a, we will find an open neighborhood \( U \) of \( x \) such that \( {W}_{u} \in O \) for all \( u \in U \) . For each \( y \) with \( \langle x, y\rangle \in {W}_{x} \) choose ope...
Yes
Corollary 35. A sober space is exponentiable iff it is strongly locally compact. A Hausdorff space is exponentiable iff it is locally compact. For a sober exponentiable space \( Y \) the exponential topology on \( \operatorname{Hom}\left( {Y, Z}\right) \) is the compact-open topology.
Proof: The first two claims follows by lemma 28. For the third, as observed above the compact-open topology is \( {T}^{ * } \) where \( T \) has the set of \( {K}^{u} \) for \( K \) compact as a subbase; so it suffices to show that these are a base for the Scott topology. Given a set \( O \) open in the Scott topology ...
Yes
Lemma 1. Suppose \( X \) is a real Hilbert space, \( S \subseteq X \) is a nonempty closed convex subset, and \( x \in X - S \) . Then there is a unique point \( y \in S \) such that \( \left| {y - x}\right| = \inf \{ \left| {w - x}\right| : w \in S\} \) .
Proof: By translating \( S \) we may suppose that \( x = 0 \) . Let \( d = \inf \{ \left| w\right| : w \in S\} \), and let \( {w}_{n} \in S \) be such that \( \left| {w}_{n}\right| \) converges to \( d \) . Since \( S \) is convex, \( \left| {\left( {{w}_{n} + {w}_{m}}\right) /2}\right| \in S \), so \( \left| {{w}_{n} ...
Yes
Lemma 2. Suppose \( X \) is a real Hilbert space, \( S \subseteq X \) is a nonempty closed convex subset and \( x \in X - S \) . Then \( z \in S \) is the unique point which minimizes \( \left| {y - x}\right| \) iff for all \( y \in S,\left( {x - z}\right) \cdot \left( {y - z}\right) \leq 0 \) .
Proof: By the convexity of \( S, z \) is the closest point iff for each \( y \in S \), for each \( t \in \left\lbrack {0,1}\right\rbrack {\left| x - \left( z + t\left( y - z\right) \right) \right| }^{2} \geq \) \( {\left| x - z\right| }^{2} \) . By algebra the requirement for \( y \) holds iff for each \( t \in \left\l...
Yes
Corollary 8. A set \( S \subseteq {\mathcal{R}}^{n} \) is a polyhedral cone iff \( S = \operatorname{Ccone}\left( {{a}_{1},\ldots ,{a}_{m}}\right) \) for some \( {a}_{1},\ldots ,{a}_{m} \).
Proof: Suppose \( S = \operatorname{Ccone}\left( {{a}_{1},\ldots ,{a}_{m}}\right) \). By restricting to a subspace we can assume \( A \) has full rank. Let \( T \) be the set of halfspaces containing the \( {a}_{i} \), which are a halfspace of the hyperplane spanned by \( n - 1 \) linearly independent \( {a}_{i} \); cl...
Yes
Corollary 11. A polyhedron is bounded iff it is the convex hull of a finite set of points.
Proof: If a polyhedron \( S \) is bounded, the cone as in the theorem must be \( \varnothing \) . Conversely by the theorem the convex hull of a finite set of points is a polyhedron, and it is clearly bounded.
Yes
Lemma 4. Let \( S \subseteq {\mathcal{R}}^{n} \), and let \( \left\{ {S}_{i}\right\} \) be a (finite or) countable disjoint collection of sets with \( S = \cup {S}_{i} \) . Then\n\na. \( \sum {\mu }_{I}\left( {S}_{i}\right) \leq {\mu }_{I}\left( S\right) \),\n\nb. \( \sum {\mu }^{ * }\left( {S}_{i}\right) \geq {\mu }^{...
Proof: For part a it suffices to show that if \( c < {\mu }_{1} = \sum {\mu }_{I}\left( {S}_{i}\right) \) then \( c < {\mu }_{I}\left( S\right) \) . If \( {\mu }_{1} \) is finite there are packings \( {P}_{i} \) of \( {S}_{i} \) with \( \nu \left( {P}_{i}\right) \geq \left( {c/{\mu }_{1}}\right) {\mu }_{I}\left( {S}_{i...
Yes
Theorem 6. Suppose \( T \) is a linear transformation of determinant \( \lambda \neq 0 \), and \( S \subseteq {\mathcal{R}}^{n} \) is Lebesgue measurable. Then \( T\left\lbrack S\right\rbrack \) is Lebesgue measurable, and \( \mu \left( {T\left\lbrack S\right\rbrack }\right) = \left| \lambda \right| \mu \left( S\right)...
Proof: The theorem is clear for \( T \) a diagonal or permutation matrix and \( S \) a cell, whence for any measurable \( S \) . It thus suffices to show the theorem for a matrix of an elementary operation. For such a matrix in \( {\mathcal{R}}^{2} \), a cell \( C \) may be dissected into rectangles and triangles whose...
No
Corollary 10. Suppose \( S \subseteq {\mathcal{R}}^{n} \) is bounded and convex, \( S = - S, L \) is a full rank lattice, and \( \mu \left( S\right) > \) \( {2}^{n}\det \left( L\right) \) ; then \( S \) contains a nonzero point of \( L \) .
Proof: Let \( S = G\left\lbrack T\right\rbrack \) where \( G \) is a generator matrix for \( L \) . Then \( T \) satisfies the hypotheses of the theorem, and the image of a nonzero point of \( {\mathcal{Z}}^{n} \) is a nonzero point of \( L \) .
No
Corollary 11. Let \( G \) be an \( n \times n \) real matrix with entries \( {g}_{ij} \) and \( \det \left( G\right) \neq 0 \), and let \( {\zeta }_{i}\left( x\right) \) be the linear form \( \mathop{\sum }\limits_{j}{g}_{ij}{x}_{j} \) . Suppose \( {\lambda }_{i} \) for \( 1 \leq i \leq n \) are positive real numbers w...
Proof: Let \( S = \left\{ {x : \left| {x}_{i}\right| < {\lambda }_{i}}\right. \) for \( \left. {1 \leq i \leq n}\right\} \) and apply corollary 10 .
Yes
Lemma 23. For a Coxeter group \( G \), and \( w \in G \), if \( l\left( {w{\rho }_{r}}\right) = l\left( w\right) + 1 \) then \( \bar{w}\left( {e}_{r}\right) \in \) Plc.
Proof: Suppose \( w, r \) is a counterexample with \( l\left( w\right) \) least. Since \( l\left( {e}_{r}\right) = 1, l\left( w\right) = 1 \) . For some \( {w}_{1} \) and \( s, w = {w}_{1}{\rho }_{s} \) and \( l\left( {w}_{1}\right) = l\left( w\right) - 1 \) . Inductively, if \( l\left( {{w}_{i}{\rho }_{t}}\right) = l\...
Yes
Theorem 24. For a Coxeter group \( G \), the homomorphism \( w \mapsto \bar{w} \) from \( G \) to \( \bar{G} \) is an isomorphism.
Proof: It is surjective by definition. If \( \bar{w} = 1 \) and \( w \neq 1 \) then for some \( r, w = {w}^{\prime }{\rho }_{r} \) and \( l\left( w\right) = l\left( {w}^{\prime }\right) + 1 \).\n\nBy lemma \( {23},{\bar{w}}^{\prime }\left( {e}_{r}\right) \in \) Plc. But \( - {\bar{w}}^{\prime }\left( {e}_{r}\right) = {...
Yes
Lemma 9. Let \( f : X \mapsto Y \) be a vector space homomorphism spaces in \( F \) -NLS. Then \( f \) is continuous iff there is a real number \( r > 0 \) such that \( \left| {f\left( x\right) }\right| \leq r\left| x\right| \) .
Proof: It was observed in section 3 that \( f \) is continuous iff it is continuous at 0 . If \( r \) exists then \( f \) is continuous at 0 ; indeed, given \( \epsilon \) if \( \left| x\right| < \epsilon /r \) then \( \left| {f\left( x\right) }\right| < \epsilon \) (nontriviality is not required for this direction). F...
Yes
Corollary 10. Two norms \( {\left| x\right| }_{1} \) and \( {\left| x\right| }_{2} \) in a vector space \( X \) induce the same topology iff there are real numbers \( {r}_{1} > 0 \) and \( {r}_{2} > 0 \) such that \( {\left| x\right| }_{1} < {r}_{1}{\left| x\right| }_{2} \) and \( {\left| x\right| }_{2} < {r}_{2}{\left...
Proof: The identity function is continuous from norm 1 to norm 2 iff \( {r}_{1} \) exists, and from norm 2 to norm 1 iff \( {r}_{2} \) exists.
No
Lemma 11. Suppose \( Y \subseteq X \) is a subspace. Let \( \left| {x + Y}\right| = \inf \{ \left| {x + y}\right| : y \in Y\} \) . a. \( \left| {x + Y}\right| \) is a pseudo-norm on \( X/Y \) .
Proof: Letting \( y \) vary over \( Y \) , \[ \left| {{ax} + Y}\right| = \inf \{ \left| {{ax} + y}\right| \} = \inf \{ \left| {{ax} + {ay}}\right| \} = \left| a\right| \inf \{ \left| {x + y}\right| \} = \left| a\right| \left| {x + Y}\right| . \] Also \[ \left| {{x}_{1} + {x}_{2} + Y}\right| = \inf \left\{ \left| {{x}_{...
Yes
Lemma 13. Suppose \( X \) is a normed linear space and \( f : X \mapsto F \) is a linear functional whose kernel is not all of \( X \) . Then \( f \) is continuous iff \( \operatorname{Ker}\left( f\right) \) is closed. Further in this case \( X \) is homeomorphic to the product of \( \operatorname{Ker}\left( f\right) \...
Proof: The first claim follows by lemma 5. In the proof of of b implies c, open balls may be used rather than balanced sets, noting that \( f\left\lbrack {B}_{0, r}\right\rbrack \) is bounded if it is not all of \( F \) ; also lemma 9 may be appealed to. Now, if \( w \notin \operatorname{Ker}\left( f\right) \) then \( ...
Yes
Corollary 27. Suppose \( \langle L, R,\alpha \rangle \) is an adjunction. If \( R \) is a functor to groups then \( L \) is a functor to cogroups, and conversely.
Proof: If \( R \) is a functor to groups then given \( H,\operatorname{Hom}\left( {H, R\left( -\right) }\right) \) is a group, and so \( \operatorname{Hom}\left( {L\left( H\right) , - }\right) \) is. Remaining details are left to the reader.
No
Lemma 2. Let \( X \) be a nonempty topological space and \( S \) a nonempty subset.\na. If \( X \) is irreducible and \( S \) is open then \( S \) is irreducible.
Proof: For part a, if \( {U}_{1},{U}_{2} \) are disjoint open subsets of \( S \) they are disjoint open subsets of \( X \) .
No
Lemma 9. Suppose \( F \) is a field, \( p \in F\left\lbrack x\right\rbrack \), and \( e \) is greater than the degree of any variable in \( p \) . For \( 1 \leq i < n \) let \( {y}_{i} = {x}_{i} - {x}_{n}^{{e}^{i}} \) . Then \( p = q\left( {{y}_{1},\ldots ,{y}_{n - 1},{x}_{n}}\right) \) where \( q \) is monic as a poly...
Proof: If \( {x}_{1}^{{d}_{1}}\cdots {x}_{n}^{{d}_{n}} \) is a monomial of \( p \), when written in terms of \( {y}_{1},\ldots ,{y}_{n - 1},{x}_{n} \) its highest degree term is \( {x}_{n}^{f} \) where \( f = {d}_{n} + {d}_{1}e + \cdots + {d}_{n - 1}{e}^{n - 1} \) . The \( {d}_{i} \) are just the digits in the \( e \) ...
Yes
Lemma 11. Suppose \( A \) is a finitely generated \( F \) -algebra over a field \( F \), and an integral domain. If \( P \) is a nontrivial prime ideal of \( A \) then \( \operatorname{Trdeg}\left( {A/P}\right) < \operatorname{Trdeg}\left( A\right) \) .
Proof: Suppose \( {a}_{1} + P,\ldots ,{a}_{n} + P \) are algebraically independent, and \( b \in P \) is nonzero. Suppose \( p\left( {{a}_{1},\ldots ,{a}_{n}, b}\right) = 0 \) ; then \( p\left( {{a}_{1}, + P\ldots ,{a}_{n} + P,0}\right) = 0 \) in \( A/P \), so \( p\left( {{x}_{1},\ldots ,{x}_{n},0}\right) \) must be th...
No
Lemma 12. Suppose \( F \) is a field; then \( \operatorname{Kdim}\left( {F\left\lbrack x\right\rbrack }\right) = n \) .
Proof: For \( 0 \leq i \leq n \) let \( {P}_{i} \) be the ideal generated by \( {x}_{1},\ldots ,{x}_{i}.{P}_{i} \) is a prime ideal, because \( p \in {P}_{i} \) iff, as a polynomial with coefficients in \( F\left\lbrack {{x}_{i + 1},\ldots ,{x}_{n}}\right\rbrack \), the constant term is 0 . Further these ideals form a ...
Yes
Lemma 14. Suppose \( R \) is a Noetherian ring. Then there are only finitely many minimal prime ideals in \( R \) .
Proof: Let \( S \) be the set of ideals \( I \) such that \( R/I \) has infinitely many minimal prime ideals. Since \( R \) is Noetherian, if \( S \) is nonempty then it contains a maximal element \( I \) . We may thus assume that \( R \) is such that for any ideal \( I \) in \( R \), there are only finitely many prime...
No
Corollary 16. Suppose \( A \) is a finitely generated \( F \) -algebra over a field \( F \), and an integral domain. Then all maximal chains of prime ideals have length \( \operatorname{Trdeg}\left( A\right) \) .
Proof: The proof is by induction on \( d = \operatorname{Trdeg}\left( A\right) \) . If \( d = 0 \), then \( A \) is an integral extension of \( F \), so by theorem 20.21.e \( A \) is a field, and the only prime ideal is \( \{ 0\} \) . Suppose \( {P}_{0} \subset \cdots {P}_{e} \) is a chain of prime ideals. It follows b...
No
Lemma 24. Suppose \( R \) is a \( \mathcal{N} \) -graded commutative ring and \( M \) is a \( \mathcal{Z} \) -graded \( R \) -module. Suppose \( \operatorname{Ann}\left( x\right) \) is a prime ideal \( P \) . Then \( P \) is homogeneous.
Proof: Let \( x = {x}_{1} + \cdots + {x}_{s} \) where the \( {x}_{i} \) are homogeneous and \( {x}_{1} \) is of least degree. We prove the lemma by induction on \( s \) . If \( r \in P \), write \( r = {r}_{1} + \cdots + {r}_{t} \) where the \( {r}_{i} \) are homogeneous and of increasing degree. Then \( {r}_{1}{x}_{1}...
Yes
Lemma 25. Suppose \( R \) is a Noetherian commutative ring and \( M \) is a finitely generated \( R \)-module. Then there is a chain \( 0 = {M}_{0} \subset \cdots \subset {M}_{m} = M \) of submodules, such that for \( i > 0{M}_{i}/{M}_{i - 1} \) is isomorphic to \( R/{P}_{i} \) for some prime ideal \( {P}_{i} \). If \(...
Proof: As noted in section 8.5, \( M \) is Noetherian. By lemma 23 there is an \( x \in M \) such that \( \operatorname{Ann}\left( x\right) \) is a prime ideal \( P \). As noted in section 8.1, the submodule \( {M}_{1} = {Rx} \) is isomorphic to \( R/P \). Continuing inductively with \( M/{M}_{1} \), using the correspo...
No
Lemma 32. For \( S \subseteq {\mathcal{N}}^{n}, S \) has only finitely many minimal elements in the product order.
Proof: Since the minimal elements of \( S \) and those of \( {S}^{ \geq } \) are the same, we may assume that \( S \) is \( \geq \) -closed. If \( n = 1 \) the claim follows because \( \mathcal{N} \) is well-ordered. For \( n > 1 \), let \( {S}_{i} = \left\{ {\left\langle {{u}_{1},\ldots ,{u}_{n - 1}}\right\rangle : \l...
Yes
Lemma 2. Suppose \( A \) is an integrally closed integral domain, \( K \) is its field of fractions, \( L \) is a finite separable extension of \( K \), and \( B \) is the integral closure of \( A \) in \( L \). a. For \( b \in B \), the coefficients of the irreducible polynomial of \( b \) over \( K \), and the norm a...
Proof: For part a, since the quantities in question are in the ring generated over \( A \) by the conjugates of \( b \) they are all integral over \( A \), and in \( K \). Since \( A \) is integrally closed, they are in \( A \). For part b, choose a basis \( {e}_{1},\ldots ,{e}_{n} \) for \( L \) over \( K \). By multi...
Yes
Lemma 4. Suppose \( L \supseteq K \) is a finite extension of dimension \( n \) . If \( {l}^{\prime } = {Tl} \) for an \( n \times n \) matrix \( T \) over \( K \) and column vectors \( l,{l}^{\prime } \) then \( \operatorname{Disc}\left( {{l}_{1}^{\prime },\ldots ,{l}_{n}^{\prime }}\right) = \det {\left( T\right) }^{2...
Proof: \( \;\operatorname{Disc}\left( {{l}_{1}^{\prime },\ldots ,{l}_{n}^{\prime }}\right) = \det \left( N\right) \) where\n\n\[ \n{N}_{ij} = \operatorname{Tr}\left( {{l}_{i}^{\prime }{l}_{j}^{\prime }}\right) = \operatorname{Tr}\left( {\mathop{\sum }\limits_{{rs}}{T}_{ir}{T}_{js}{l}_{r}{l}_{s}}\right) = \mathop{\sum }...
Yes
Lemma 15. Suppose \( R \) is a Dedekind domain, \( F \) is the field of fractions, \( O \) is a valuation ring in \( F \) containing \( R, M \) is the maximal ideal of \( O \), and \( P = M \cap R \). Then \( O = {R}_{P} \).
Proof: Let \( U = O - M \). If \( s \in R - P \) then \( s \in O - M = U \); this shows that \( {R}_{P} \subseteq O \). If \( P = \{ 0\} \) then \( F = O \) follows. Otherwise, \( {P}_{P} \subseteq \left\lbrack {MO}\right\rbrack = M \). That is, we have valuation rings \( {O}_{2} \subseteq {O}_{1} \), with maximal idea...
Yes
Lemma 21. Suppose \( A \) is a Dedekind domain, \( P \) is a prime ideal, and \( e > 1 \) . Then the \( A \) -module \( {P}^{e - 1}/{P}^{e} \) is isomorphic to \( A/P \) .
Proof: Since \( {P}^{e - 1} \supset {P}^{e} \), there is an element \( a \in {P}^{e - 1} - {P}^{e} \) . The map \( x \mapsto {ax} + {P}^{e} \) from \( A \) to \( {P}^{e - 1}/{P}^{e} \) is an \( A \) -module homomorphism. It is surjective, because \( {Aa} + {P}^{e} \) contains \( {P}^{e} \) and is contained in \( {P}^{e...
Yes
Corollary 24. Given an a.i.r. \( A \), let \( {\beta }_{A} \) be as in theorem 23. Then every ideal class of \( A \) contains an ideal \( I \) with \( \operatorname{Inrm}\left( I\right) \leq {\beta }_{A} \) .
Proof: Let \( C \) be the class, let \( J \) be an ideal in the inverse class, and choose \( x \in J \) such that \( \mathrm{N}\left( x\right) \leq \) \( {\beta }_{A}\operatorname{Inrm}\left( J\right) \) . Then \( J \supseteq {Ax} \), so \( \left\lbrack {IJ}\right\rbrack = {Ax} \) for some \( I \), and \( I \in C \) . ...
Yes
Lemma 28. If \( p \) is a polynomial whose coefficients are algebraic numbers then there is an a.n.f. \( K \) containing the coefficients, such that \( p = c{p}^{\prime } \) where \( c \in K \) and \( {p}^{\prime } \in {O}_{K} \) is primitive. In any such \( {O}_{K}, c \) and \( {p}^{\prime } \) are unique up to units.
Proof: Suppose \( p = \mathop{\sum }\limits_{{i = 0}}^{n}\left( {{a}_{i}/{b}_{i}}\right) {x}^{i} \), and let \( F \) be an a.n.f. containing the coefficients. Let \( e = {b}_{1}\cdots {b}_{n} \) ; then \( {ep} \in {O}_{F}\left\lbrack x\right\rbrack \) . Let \( I \subseteq {O}_{F} \) be the ideal generated by the coeffi...
No
Lemma 5. Suppose \( R \) is a field, and \( {p}_{1} = q{p}_{2} + {p}_{3} \), where \( {n}_{1} \geq {n}_{2} > {n}_{3} \) and \( {p}_{3} \neq 0 \). a. For \( 0 \leq i < {n}_{3},{S}_{i}\left( {{p}_{1},{p}_{2}}\right) = {\left( -1\right) }^{\left( {{n}_{1} - i}\right) \left( {{n}_{2} - i}\right) }{\pi }_{2}^{{n}_{1} - {n}_...
Proof: The argument is a generalization of the argument for \( i = 0 \) given above. Let \( N \) be the copy of Sylvester’s matrix for \( {p}_{2} \) and \( {p}_{3} \) in the lower right corner of \( {M}^{\prime \prime } \). Then \( {N}_{i} \) for \( 0 \leq i < {n}_{3} \) is obtained from \( N \) by deleting the rows of...
Yes
Corollary 7. In Collins’ reduced method, \( {p}_{i} = {\rho }_{i}{S}_{{n}_{i - 1} - 1}\left( {{p}_{1},{p}_{2}}\right) \) for \( 3 \leq i \leq k \), where \( {\rho }_{i} \in R \) . In fact,
\[ {\rho }_{i} = {\left( -1\right) }^{{n}_{i - 2} - {n}_{i - 1} + 1}{\Pi }_{l = 3}^{i - 1}{\left( -1\right) }^{\left( {{n}_{l - 1} - {n}_{i - 1} + 1}\right) \left( {{n}_{l - 2} - {n}_{i - 1} + 1}\right) }{\pi }_{l - 1}^{\left( {{n}_{l - 2} - {n}_{l - 1} + 1}\right) \left( {{n}_{l - 1} - {n}_{l} - 1}\right) } \] Proof: ...
Yes
Lemma 10. Suppose \( a, b \in F, a < b, p \in F\left\lbrack x\right\rbrack \), and \( p\left( a\right) p\left( b\right) < 0 \) . Then \( p\left( c\right) = 0 \) for some \( c \) with \( a < c < b \) .
Proof: Assume \( p \) is monic, and factors into irreducible factors as \( \left( {x - {r}_{1}}\right) \cdots \left( {x - {r}_{k}}\right) {q}_{1}\left( x\right) \cdots {q}_{l}\left( x\right) \) where \( {r}_{1} \leq \cdots \leq {r}_{k} \) and \( {q}_{j} \) is an irreducible quadratic. It is readily shown that \( {q}_{j...
No
Lemma 16. Suppose \( L \subseteq {\mathcal{R}}^{n} \) is a lattice, \( {b}_{1},\ldots ,{b}_{n} \) is an LLL-reduced basis, and \( {x}_{1},\ldots ,{x}_{t} \in L \) are linearly independent. Then for \( 1 \leq j \leq t,\left| {b}_{j}\right| \leq {2}^{\left( {n - 1}\right) /2}\max \left\{ {{x}_{1},\ldots ,{x}_{t}}\right\}...
Proof: For \( 1 \leq i \leq n \) and \( 1 \leq j \leq t \) there are \( {r}_{ij} \in \mathcal{Z}\left( {{\widehat{r}}_{ij} \in \mathcal{R}}\right) \) such that \( {x}_{j} = \mathop{\sum }\limits_{i}{r}_{ij}{b}_{i}\left( {{\widehat{x}}_{j} = \mathop{\sum }\limits_{i}{\widehat{r}}_{ij}{\widehat{b}}_{i}}\right) \) , where...
Yes
a. If the rows or columns of a matrix are permuted then the determinant is unchanged if the permutation is even, and multiplied by -1 if the permutation is odd.
Proof: Part a follows by the remarks above concerning permutation matrices.
No
Theorem 1.1.6. Let \( p\left( t\right) \) be a given polynomial of degree \( k \) . If \( \lambda, x \) is an eigenvalue-eigenvector pair of \( A \in {M}_{n} \), then \( p\left( \lambda \right), x \) is an eigenvalue-eigenvector pair of \( p\left( A\right) \) . Conversely, if \( k \geq 1 \) and if \( \mu \) is an eigen...
Proof. We have\n\n\[ p\left( A\right) x = {a}_{k}{A}^{k}x + {a}_{k - 1}{A}^{k - 1}x + \cdots + {a}_{1}{Ax} + {a}_{0}x,\;{a}_{k} \neq 0 \]\n\nand \( {A}^{j}x = {A}^{j - 1}{Ax} = {A}^{j - 1}{\lambda x} = \lambda {A}^{j - 1}x = \cdots = {\lambda }^{j}x \) by repeated application of the eigenvalue-eigenvector equation. Thu...
Yes
Theorem 1.1.9. Let \( A \in {M}_{n} \) be given. Then \( A \) has an eigenvalue. In fact, for each given nonzero \( y \in {\mathbf{C}}^{n} \), there is a polynomial \( g\left( t\right) \) of degree at most \( n - 1 \) such that \( g\left( A\right) y \) is an eigenvector of \( A \) .
Proof. Let \( m \) be the least integer \( k \) such that the vectors \( y,{Ay},{A}^{2}y,\ldots ,{A}^{k}y \) are linearly dependent. Then \( m \geq 1 \) since \( y \neq 0 \), and \( m \leq n \) since any \( n + 1 \) vectors in \( {\mathbf{C}}^{n} \) are linearly dependent. Let \( {a}_{0},{a}_{1},\ldots ,{a}_{m} \) be s...
Yes
What are the eigenvalues and determinant of \( I + x{y}^{ * } \) ?
Using (0.8.5.11) and the fact that \( \operatorname{adj}\left( {\alpha I}\right) = {\alpha }^{n - 1}I \), we compute\n\n\[ \n{p}_{I + x{y}^{ * }}\left( t\right) = \det \left( {{tI} - \left( {I + x{y}^{ * }}\right) }\right) = \det \left( {\left( {t - 1}\right) I - x{y}^{ * }}\right) \n\]\n\n\[ \n= \det \left( {\left( {t...
Yes
Let \( x, y \in {\mathbf{C}}^{n}, x \neq 0 \), and \( A \in {M}_{n} \) . Suppose that \( {Ax} = {\lambda x} \) and let the eigenvalues of \( A \) be \( \lambda ,{\lambda }_{2},\ldots ,{\lambda }_{n} \) . What are the eigenvalues of \( A + x{y}^{ * } \) ?
First observe that \( \left( {t - \lambda }\right) x = \left( {{tI} - A}\right) x \) implies that \( \left( {t - \lambda }\right) \operatorname{adj}\left( {{tI} - A}\right) x = \operatorname{adj}\left( {{tI} - A}\right) \left( {{tI} - A}\right) x = \det \left( {{tI} - A}\right) x \), that is,\n\n\[ \left( {t - \lambda ...
Yes
Theorem 1.2.17. Let \( A \in {M}_{n} \) . There is some \( \delta > 0 \) such that \( A + {\varepsilon I} \) is nonsingular whenever \( \varepsilon \in \mathbf{C} \) and \( 0 < \left| \varepsilon \right| < \delta \) .
Proof. Observation 1.1.8 ensures that \( \lambda \in \sigma \left( A\right) \) if and only if \( \lambda + \varepsilon \in \sigma \left( {A + {\varepsilon I}}\right) \) . Therefore, \( 0 \in \sigma \left( {A + {\varepsilon I}}\right) \) if and only if \( \lambda + \varepsilon = 0 \) for some \( \lambda \in \sigma \left...
Yes
Theorem 1.2.18. Let \( A \in {M}_{n} \) and suppose that \( \lambda \in \sigma \left( A\right) \) has algebraic multiplicity \( k \) . Then \( \operatorname{rank}\left( {A - {\lambda I}}\right) \geq n - k \) with equality for \( k = 1 \) .
Proof. Apply the preceding observation to the characteristic polynomial \( {p}_{A}\left( t\right) \) of a matrix \( A \in {M}_{n} \) that has an eigenvalue \( \lambda \) with multiplicity \( k \geq 1 \) . If we let \( B = A - \) \( {\lambda I} \), then zero is an eigenvalue of \( B \) with multiplicity \( k \) and henc...
Yes
Theorem 1.3.3. Let \( A, B \in {M}_{n} \) . If \( B \) is similar to \( A \), then \( A \) and \( B \) have the same characteristic polynomial.
Proof. Compute\n\n\[ \n{p}_{B}\left( t\right) = \det \left( {{tI} - B}\right) \]\n\n\[ \n= \det \left( {t{S}^{-1}S - {S}^{-1}{AS}}\right) = \det \left( {{S}^{-1}\left( {{tI} - A}\right) S}\right) \]\n\n\[ \n= \det {S}^{-1}\det \left( {{tI} - A}\right) \det S = {\left( \det S\right) }^{-1}\left( {\det S}\right) \det \le...
Yes
Corollary 1.3.4. Let \( A, B \in {M}_{n} \) and suppose that \( A \) is similar to \( B \) . Then\n\n(a) A and B have the same eigenvalues.\n\n(b) If \( B \) is a diagonal matrix, its main diagonal entries are the eigenvalues of \( A \) .\n\n(c) \( B = 0 \) (a diagonal matrix) if and only if \( A = 0 \) .\n\n(d) \( B =...
Exercise. Verify the assertions in the preceding corollary.
No
Theorem 1.3.7. Let \( A \in {M}_{n} \) be given. Then \( A \) is similar to a block matrix of the form\n\n\[ \left\lbrack \begin{matrix} \Lambda & C \\ 0 & D \end{matrix}\right\rbrack ,\;\Lambda = \operatorname{diag}\left( {{\lambda }_{1},\ldots ,{\lambda }_{k}}\right), D \in {M}_{n - k},1 \leq k < n \]\n\nif and only ...
Proof. Suppose that \( k < n \), the \( n \) -vectors \( {x}^{\left( 1\right) },\ldots ,{x}^{\left( k\right) } \) are linearly independent, and \( A{x}^{\left( i\right) } = {\lambda }_{i}{x}^{\left( i\right) } \) for each \( i = 1,\ldots, k \) . Let \( \Lambda = \operatorname{diag}\left( {{\lambda }_{1},\ldots ,{\lambd...
Yes
Lemma 1.3.8. Let \( {\lambda }_{1},\ldots ,{\lambda }_{k} \) be \( k \geq 2 \) distinct eigenvalues of \( A \in {M}_{n} \) (that is, \( {\lambda }_{i} \neq {\lambda }_{j} \) if \( i \neq j \) and \( 1 \leq i, j \leq k \) ), and suppose that \( {x}^{\left( i\right) } \) is an eigenvector associated with \( {\lambda }_{i...
Proof. Suppose that there are complex scalars \( {\alpha }_{1},\ldots {\alpha }_{k} \) such that \( {\alpha }_{1}{x}^{\left( 1\right) } + {\alpha }_{2}{x}^{\left( 2\right) } + \cdots + {\alpha }_{r}{x}^{\left( r\right) } = 0 \) . Let \( {B}_{1} = \left( {A - {\lambda }_{2}I}\right) \left( {A - {\lambda }_{3}I}\right) \...
Yes
Theorem 1.3.9. If \( A \in {M}_{n} \) has \( n \) distinct eigenvalues, then \( A \) is diagonalizable.
Proof. Let \( {x}^{\left( i\right) } \) be an eigenvector associated with the eigenvalue \( {\lambda }_{i} \) for each \( i = \) \( 1,\ldots, n \) . Since all the eigenvalues are distinct, Lemma 1.3.8 ensures that the vectors \( {x}^{\left( 1\right) },\ldots ,{x}^{\left( n\right) } \) are linearly independent. Theorem ...
Yes
Lemma 1.3.10. Let \( {B}_{1} \in {M}_{{n}_{1}},\ldots ,{B}_{d} \in {M}_{{n}_{d}} \) be given and let \( B \) be the direct sum\n\n\[ B = \left\lbrack \begin{matrix} {B}_{1} & & 0 \\ & \ddots & \\ 0 & & {B}_{d} \end{matrix}\right\rbrack = {B}_{1} \oplus \cdots \oplus {B}_{d} \]\n\nThen \( B \) is diagonalizable if and o...
Proof. If for each \( i = 1,\ldots, d \) there is a nonsingular \( {S}_{i} \in {M}_{{n}_{i}} \) such that \( {S}_{i}^{-1}{B}_{i}{S}_{i} \) is diagonal, and if we define \( S = {S}_{1} \oplus \cdots \oplus {S}_{d} \), then one checks that \( {S}^{-1}{BS} \) is diagonal.\n\nFor the converse, we proceed by induction. Ther...
Yes
Theorem 1.3.12. Let \( A, B \in {M}_{n} \) be diagonalizable. Then \( A \) and \( B \) commute if and only if they are simultaneously diagonalizable.
Proof. Assume that \( A \) and \( B \) commute, perform a similarity transformation on both \( A \) and \( B \) that diagonalizes \( A \) (but not necessarily \( B \) ) and groups together any repeated eigenvalues of \( A \) . If \( {\mu }_{1},\ldots ,{\mu }_{d} \) are the distinct eigenvalues of \( A \) and \( {n}_{1}...
Yes
Lemma 1.3.19. Let \( \mathcal{F} \subset {M}_{n} \) be a commuting family. Then some nonzero vector in \( {\mathbf{C}}^{n} \) is an eigenvector of every \( A \in \mathcal{F} \) .
Proof. There is always a nonzero \( \mathcal{F} \) -invariant subspace, namely, \( {\mathbf{C}}^{n} \) . Let \( m = \) \( \min \left\{ {\dim V : V}\right. \) is a nonzero \( \mathcal{F} \) -invariant subspace of \( \left. {\mathbf{C}}^{n}\right\} \) and let \( W \) be any given \( \mathcal{F} \) -invariant subspace suc...
Yes
Theorem 1.3.21. Let \( \mathcal{F} \subset {M}_{n} \) be a family of diagonalizable matrices. Then \( \mathcal{F} \) is a commuting family if and only if it is a simultaneously diagonalizable family. Moreover, for any given \( {A}_{0} \in \mathcal{F} \) and for any given ordering \( {\lambda }_{1},\ldots ,{\lambda }_{n...
Proof. If \( \mathcal{F} \) is simultaneously diagonalizable, then it is a commuting family by a previous exercise. We prove the converse by induction on \( n \) . If \( n = 1 \), there is nothing to prove since every family is both commuting and diagonal. Let us suppose that \( n \geq 2 \) and that, for each \( k = 1,...
Yes
Theorem 1.3.22. Suppose that \( A \in {M}_{m, n} \) and \( B \in {M}_{n, m} \) with \( m \leq n \) . Then the \( n \) eigenvalues of \( {BA} \) are the \( m \) eigenvalues of \( {AB} \) together with \( n - m \) zeroes; that is, \( {p}_{BA}\left( t\right) = {t}^{n - m}{p}_{AB}\left( t\right) \) . If \( m = n \) and at ...
Proof. A computation reveals that\n\n\[ \left\lbrack \begin{matrix} {I}_{m} & - A \\ 0 & {I}_{n} \end{matrix}\right\rbrack \left\lbrack \begin{matrix} {AB} & 0 \\ B & {0}_{n} \end{matrix}\right\rbrack \left\lbrack \begin{matrix} {I}_{m} & A \\ 0 & {I}_{n} \end{matrix}\right\rbrack = \left\lbrack \begin{matrix} {0}_{m} ...
Yes