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Proposition 3.6. Let \( S \) be a semigroup. Then\n\n\[ \n{S}^{I}l = {Sl} \n\] | Proof. Since \( S \subset {S}^{I} \) we need only show that \( {S}^{I}l \leq {Sl} \) .\n\nSuppose \( T \subset {S}^{I} \) is a type I semigroup. We show that \( T - I \) is a type I semigroup. If \( T - I = T \) we are done, so assume \( I \in T \) . Let\n\n\[ \n\varphi : T - I \rightarrow A,\;A\text{ aperiodic } \n\]\... | Yes |
Proposition 3.7. Let \( {S}_{2} * {S}_{1} \) be a semidirect product of semigroups \( {S}_{1} \) and \( {S}_{2} \) . Then\n\n\[ \left( {{S}_{2} * {S}_{1}}\right) l \leq {S}_{2}l + {S}_{1}l \] | Proof. Consider the projection morphism\n\n\[ \pi : {S}_{2} * {S}_{1} \rightarrow {S}_{1} \]\n\nProposition 3.1 yields\n\n\[ \left( {{S}_{2} * {S}_{1}}\right) l \leq {\pi l} + {S}_{1}l \]\n\nso it suffices to show that \( {\pi l} \leq {S}_{2}l \) .\n\nEvery member of ker \( \pi \) is in the form \( {S}_{2} * W \), wher... | Yes |
Corollary 3.8. (A2) is satisfied by \( l : \) TS \( \rightarrow N \) . That is\n\n\[ \left( {X \circ Y}\right) l \leq {Xl} + {Yl} \] | Proof. Since \( X \circ Y \prec {X}^{ \circ } \circ {Y}^{ \circ } \) and \( {Xl} = {X}^{ \circ }l \), it suffices to assume that \( X = \left( {Q, S}\right) \) and \( Y = \left( {P, T}\right) \) are complete. In this case, the semigroup of \( X \circ Y \) is \( {S}^{\left( n\right) } * T \), where \( n = \operatorname{... | Yes |
Proposition 3.9. Let \( X \) be a ts. Then \[ {Xl} = \sup \left\{ {{X}_{e}l \mid e\text{ an idempotent of }X}\right\} \] | Proof. Let \( X = \left( {Q, S}\right) \) . Since the semigroup of \( {X}_{e} \) is \( {eSe} \), it suffices to prove \[ {Sl} = \sup \left\{ {\left( {eSe}\right) l \mid e \in S,{e}^{2} = e}\right\} \] Let \( E \) be the set of idempotents in \( S \), and let \( T = {ESE} \) . The reduction theorem (Theorem XI, 4.1) imp... | Yes |
Lemma 3.10. Let \( d : \) TS \( \rightarrow N \) be a function that satisfies (A1)-(A7). Let \( X = \left( {Q, S}\right) \) be a ts. Then\n\n\[ \n{Xd} \leq \sup \left\{ {{Td} \mid T \in {S}_{1}}\right\} \n\]\n\n(3.6)\n\n\[ \n{Xd} \leq {S}_{\text{II }}d + 1 \n\]\n\n(3.7) | Proof. By Corollary I, 9.4,\n\n\[ \nX \prec S \times A \n\]\n\nfor some aperiodic ts \( A \) . Thus, using (A1),(A3), and (A5’), we obtain \( {Xd} \leq {Sd} \) . Therefore, it suffices to prove (3.6) and (3.7) with \( X \) replaced by \( S \) .\n\nLet \( \mathbf{V} \) be an \( \mathbf{S} \) -variety and let\n\n\[ \n\va... | No |
Theorem 3.11. \( l : \) TS \( \rightarrow N \) is the largest function that satisfies (A1)-(A7). | Proof. We first establish\n\nIf \( S \) is a nonaperiodic type I semigroup then \( {Sl} \geq {S}_{\text{II }}l + 1 \)\n\n(3.11)\n\nLet \( {S}_{\text{II }}l = n \) and let\n\n\[ \n{S}_{\mathrm{{II}}} = {U}_{0} \supset {T}_{1} \supset \cdots \supset {T}_{n} \supset {U}_{n} \n\]\n\nbe a series of length \( n \) for \( {S}... | Yes |
Lemma 1.2. Let \( M \) be a left \( S \) -module and define a right action of \( S \) on \( {M}^{ * } = {\operatorname{Hom}}_{R}\left( {M, R}\right) \) by\n\n\[ \left( {ft}\right) \left( m\right) = f\left( {tm}\right) - D\left( {f\left( m\right) }\right) ,\;f\text{ in }{M}^{ * },\;m\text{ in }M, \]\n\n\[ \left( {fr}\ri... | Proof. The endomorphisms of the abelian group \( {M}^{ * } \) defined by \( t \) and any element \( r \) of \( R \) are readily verified to satisfy \( {tr} = {rt} + D\left( r\right) \) . Thus \( {M}^{ * } \) is indeed a right \( S \) -module. A routine computation verifies the commutativity of 1.3. | Yes |
Corollary 1.7. Let \( M \) be a left \( S \) -module.\n\n(a) \( {\operatorname{w.dim}}_{R}M \leq {\operatorname{w.dim}}_{S}M \leq {\operatorname{w.dim}}_{R}M + 1 \) . Thus, w.gl.dim \( S \leq \) w.gl.dim \( R + 1 \).\n\n(b) If \( M \neq 0 \) is finitely generated as an \( R \) -module, \( R \) is left noetherian and \(... | Proof. The first inequality in (a) is a consequence of the fact that any flat \( S \) -resolution of \( M \) is also a flat \( R \) -resolution. Combining the long exact sequence for Tor arising from 1.1 with the isomorphisms\n\n\[ \n{\operatorname{Tor}}^{S}\left( {N, S{ \otimes }_{R}M}\right) \cong {\operatorname{Tor}... | Yes |
Lemma 2.1. w.gl.dim \( R + n \leq \) w.gl.dim \( {A}_{n}\left( R\right) \leq \) w.gl.dim \( R + {2n} \) . | Proof. Since \( {A}_{n}\left( R\right) = {A}_{1}\left( {{A}_{n - 1}\left( R\right) }\right) \), it suffices to prove the lemma for \( n = 1 \) . Now \( {A}_{1}\left( R\right) = R\left\lbrack x\right\rbrack \left\lbrack {t, d/{dx}}\right\rbrack \) and w.gl.dim \( R\left\lbrack x\right\rbrack = \) w.gl.dim \( R + 1\lbrac... | Yes |
Theorem 2.2. Let \( R \) be a left and right noetherian ring of finite global dimension \( d \) . Then gl.dim \( {A}_{1}\left( R\right) = d + 2 \) if and only if there exists a left \( {A}_{1}\left( R\right) \) -module \( M \) that is a finitely generated \( R \) -module with \( {\operatorname{ldim}}_{R}M = d \) . | Proof. If such an \( M \) exists, then applying Corollary 1.7(b) twice (noting \( R\left\lbrack x\right\rbrack = R\left\lbrack {x,0}\right\rbrack \) and \( \left. {{A}_{1}\left( R\right) = R\left\lbrack x\right\rbrack \left\lbrack {t, d/{dx}}\right\rbrack }\right) \) we first find \( {\operatorname{l.dim}}_{R\left\lbra... | Yes |
Lemma 2.3. Let \( R \) be a left and right noetherian ring that does not contain the rational numbers.\n\n(a) There are cyclic \( R \) -modules that are abelian torsion groups and for every left \( R \) -module \( M \) that is an abelian torsion group we have w. \( {\dim }_{R}M \leq m\left( R\right) \) . | Proof. Since \( R \) does not contain the rational numbers, the image of some natural number \( k \) is not a unit in \( R \) . Thus \( R/{Rk} \) is a cyclic left \( R \) -module that is an abelian torsion group. Next, let \( M \) denote a finitely generated left \( R \) -module that is an abelian torsion group. Then t... | Yes |
Lemma 2.4. Let \( R \) be a commutative noetherian ring and let \( M \neq 0 \) be a left \( {A}_{1}\left( R\right) \) -module that is finitely generated as a left \( R \) -module, then \( M \) is an abelian torsion group. | Proof. Let \( I \) be the annihilator of \( M \) in \( {A}_{1}\left( R\right) \) . Then \( I \) is a two sided ideal in \( {A}_{1}\left( R\right) \) that is the kernel of the ring homomorphism \( {A}_{1}\left( R\right) \rightarrow \) \( {\operatorname{End}}_{R}\left( M\right) \) given by the left \( {A}_{1}\left( R\rig... | Yes |
Lemma 2.5. Let \( R \) be a commutative noetherian ring and \( x \) an indeterminate over \( R \) . Let \( R\left( x\right) \) denote the ring of fractions of \( R\left\lbrack x\right\rbrack \) with regard to the multiplicative set of monic polynomials. Then gl.dim \( R\left( x\right) = \) gl.dim \( R \) . | Proof. The multiplicative set of monic polynomials contains no zero divisors. Moreover, the usual division algorithm holds in \( R\left\lbrack x\right\rbrack \) if the divisor is monic. Hence an element of \( R\left( x\right) \) may be written as \( q + r/m \), where \( q, r, m \) lie in \( R\left\lbrack x\right\rbrack... | Yes |
Corollary 2.8. If \( R \) is a noetherian commutative ring of finite global dimension \( d \), gl.dim \( {A}_{1}\left( R\right) = d + 2 \) if and only if \( R \) has a residue class field \( F \) of finite characteristic with \( {\operatorname{ldim}}_{R}F = d \) . | Proof. From Theorem 2.6 it is clear that gl.dim \( {A}_{1}\left( R\right) = d + 2 \) can only occur if \( m\left( R\right) = d \) . But it is well known that \( d = \sup \) gl.dim \( {R}_{P} \), where \( P \) runs through the maximal ideals of \( R \) and gl.dim \( {R}_{P} = {\operatorname{l.dim}}_{R}R/P \) . | No |
Example 3.1. By [5, p. 503], \( {A}_{s}\left( Q\right) \) has a division ring of quotients which we denote by \( R \) . We relabel the \( x \) ’s and \( t \) ’s in \( R \) as \( u \) ’s and \( v \) ’s. If \( n \leq s \) the additive group of \( R \) has a structure as left \( {A}_{n}\left( R\right) \) -module given by ... | On the other hand, it is clear that for all \( n,{A}_{n}\left( R\right) \) can be obtained from \( {A}_{n + s}\left( Q\right) \) as a ring of fractions. Thus \( {A}_{n}\left( R\right) \) is flat as both a right and left \( {A}_{n + s}\left( Q\right) \) -module and \( {A}_{n}\left( R\right) { \otimes }_{{A}_{n + s}\left... | Yes |
Again let \( R \) be the division ring of Example 3.1 with \( s > 1 \) and let \( {R}^{\prime } = R \oplus Q\left\lbrack {{w}_{1},\ldots ,{w}_{k}}\right\rbrack \) with \( k < s \) . Then | \[ \text{gl.dim}{A}_{n}\left( {R}^{\prime }\right) = \begin{array}{ll} \mid n + k, & 0 \leq n \leq k \\ {2n}, & k \leq n \leq s \\ \mid n + s, & s \leq n. \end{array} \] Thus \( {\Delta }_{n}\left( {R}^{\prime }\right) = 1,0 \leq n \leq k;{\Delta }_{n}\left( {R}^{\prime }\right) = 2, k < n \leq s;{\Delta }_{n}\left( {R... | Yes |
Lemma 2.2. Let \( R \) denote a right noetherian ring with w.gl.dim \( R = \) \( d < \infty, N \) a fixed arbitrary right \( R \) -module and \( F \) the functor defined by \( F\left( X\right) = {\operatorname{Tor}}_{d}^{R}\left( {N, X}\right) \) for any left \( R \) -module \( X \) . Then\n\n(i) \( F \) is left exact ... | Proof. Part (i) is immediate from the definition of w.gl.dim and the appropriate long exact homology sequence. As for Part (ii), by [3, II, Ex. 2, pp. 31-32], for any right \( R \) -module \( Y \), there is a natural homomorphism \( Y{ \otimes }_{R}\Pi {X}_{i} \rightarrow \Pi \left( {Y{ \otimes }_{R}{X}_{i}}\right) \) ... | Yes |
Lemma 2.3. Let \( F \) denote a functor from the category of left \( R \) -modules to the category of abelian groups satisfying Parts (i) and (ii) of Lemma 2.2. Then for any \( R \) -module \( M \) and a family \( \left\{ {M}_{i}\right\} \) of submodules of \( M \) , \[ \cap F\left( {M}_{i}\right) = F\left( {\cap {M}_{... | Proof. Denote the projections \( {\Pi M}/{M}_{i} \rightarrow M/{M}_{i} \) and \( M \rightarrow M/{M}_{i} \) by \( {p}_{i} \) and \( {q}_{i} \), respectively. The commutative diagram with exact top row, yields in view of the left exactness of \( F \), a commutative diagram with exact rows: Now, if \( A, B, C \) denote a... | No |
Lemma 3.3. Let \( R \) denote a ring with w.gl.dim \( R = d < \infty \) . Then \( V \), the kernel of the endomorphism \( {\operatorname{Tor}}_{d}^{R}\left( \alpha \right) \), is isomorphic to \( {\operatorname{Tor}}_{d + 1}^{S}\left( {N, M}\right) \) . | Proof. The long exact sequence of Tor \( {}^{S} \) induced by (3.1) together with the isomorphism \( {\operatorname{Tor}}^{S}\left( {N, S{ \otimes }_{R}M}\right) \cong {\operatorname{Tor}}^{R}\left( {N, M}\right) \), shows that\n\n\[\n\operatorname{Ker}\left( {{\operatorname{Tor}}_{d}^{S}\left( {N,{a}_{l}\left( M\right... | No |
Lemma 3.6. Let \( R \) be a ring with w.gl.dim \( R = d < \infty \) . Then, using the convention of Lemma 2.2(i), for any \( v \) in \( V \cap {\operatorname{Tor}}_{d}^{R}\left( {N,{K}_{i}}\right) \), we have \( {\operatorname{Tor}}_{d}^{R}\left( {N,{\pi }_{i}}\right) \left( v\right) = 0. \) | Proof. As before, let \( Z \) be an \( S \) -projective resolution of \( N \) . Let \( f \) be a cocycle representing the cohomology class \( v \) . By Lemma 2.2(i), we may assume\n\n\[ f = \sum {z}_{h} \otimes {k}_{h},\;\text{ with }{k}_{h}\text{ in }{K}_{i}. \]\n\nSince \( v \) is in \( V \), the cohomology class of ... | Yes |
Theorem 3.8. Let \( R \) be a left and right noetherian ring, gl.dim. \( R = \) w.gl.dim \( R = d < \infty \), and \( S = R\left\lbrack t\right\rbrack \) the Ore extension of \( R \) with regard to a derivation \( D \) . If \( M \) is a left \( S \) -module with w.dim \( {}_{S}M = d + 1 \), then \( M \) contains an \( ... | Proof. By hypothesis, there exists a right \( S \) -module \( N \) such that \( {\operatorname{Tor}}_{d + 1}^{S}\left( {N, M}\right) \neq 0 \) . Let \( {M}^{\prime } \) be an arbitrary finitely generated \( R \) -submodule of \( M \) . Since \( R \) is left noetherian, \( K\left( {M}^{\prime }\right) = K \) as defined ... | Yes |
Lemma 4.1. Let \( R \) denote a commutative noetherian ring with gl.dim \( R = d < \infty \) , \( \mathfrak{M} \) the set of maximal ideals of \( R \), and \( M \) a finitely generated left \( R \) -module with \( {\dim }_{R}M = d \) . Then there exists an element \( {\mathfrak{m}}_{0} \) of \( \mathfrak{M} \) such tha... | Proof. By [3, VII, Ex. 9-11, pp. 141-2], \[ d = \mathop{\sup }\limits_{{\mathfrak{m}\text{ in }\mathfrak{M}}}{\dim }_{{R}_{\mathfrak{m}}}{M}_{\mathfrak{m}} = \mathop{\sup }\limits_{{\mathfrak{m}\text{ in }\mathfrak{M}}}\text{ gl.dim }{R}_{\mathfrak{m}}, \] and for any \( {R}_{\mathfrak{m}} \) -module \( P \), we have \... | Yes |
Theorem 4.2 (cf. [5, Th. 22]). Let \( R \) be a commutative noetherian ring, gl.dim \( R = d < \infty \), and \( S = R\left\lbrack t\right\rbrack \) the Ore extension of \( R \) with regard to a derivation \( D \) . If gl.dim \( S = d + 1 \), then there exists a maximal ideal \( {\mathfrak{m}}_{0} \) of \( R \) such th... | Proof. By Theorem 3.8, there exists a left \( S \) -module \( M \) that is finitely generated as an \( R \) -module with \( {\operatorname{w.dim}}_{R}M = {\dim }_{R}M = d \) . Let \( {\mathfrak{m}}_{0} \) be a maximal ideal of \( R \) satisfying the conditions of Lemma 4.1 and \( x \) an element of \( M \) such that \(... | Yes |
Proposition 1.1 (Diamond). Suppose that \( \pi : {D}_{p} \rightarrow {\mathrm{{GL}}}_{2}\left( A\right) \) is a continuous representation where \( A \) is an Artinian local ring with residue field \( k \), a finite field of characteristic p. Suppose \( \pi \approx \left( \begin{matrix} {\chi }_{1} & \varepsilon \\ 0 & ... | Proof (taken from [Dia, Prop. 6.1]). We may replace \( \pi \) by \( \pi \otimes {\chi }_{2}^{-1} \) and we let \( \varphi = {\chi }_{1}{\chi }_{2}^{-1} \) . Then \( \pi \cong \left( \begin{matrix} \varphi & \varepsilon & t \\ 0 & 1 & \end{matrix}\right) \) determines a cocycle \( t : {D}_{p} \rightarrow M\left( 1\right... | Yes |
Proposition 1.6.\n\n\[ \n\\# {H}_{L}^{1}\left( {{\\mathbf{Q}}_{\\sum }/\\mathbf{Q}, X}\\right) /\\# {H}_{{L}^{ * }}^{1}\left( {{\\mathbf{Q}}_{\\sum }/\\mathbf{Q},{X}^{ * }}\\right) = {h}_{\\infty }\\mathop{\\prod }\\limits_{{q \\in \\sum }}{h}_{q} \n\]\n\nwhere\n\[ \n\\begin{cases} {h}_{q} & = \\# {H}^{0}\left( {{\\mat... | Proof. Adapting the exact sequence of Poitou and Tate (cf. [Mi2, Th. 4.20]) we get a seven term exact sequence\n\n\[ \n0 \\rightarrow {H}_{L}^{1}\left( {{\\mathbf{Q}}_{\\sum }/\\mathbf{Q}, X}\\right) \\rightarrow {H}^{1}\left( {{\\mathbf{Q}}_{\\sum }/\\mathbf{Q}, X}\\right) \\rightarrow \\mathop{\\prod }\\limits_{{q \\... | Yes |
Proposition 1.7. If \( q \notin \sum \), and \( X \) is an arbitrary finite \( \operatorname{Gal}\left( {{\mathbf{Q}}_{\sum }/\mathbf{Q}}\right) \) - module of p-power order, | Proof. Consider the short exact sequence of inflation-restriction: \n\nThe proposition follows when we note that\n\n\[ \# {H}^{0}\left( {{\mathbf{Q}}_{q},{X}^{ * }}\right) = \# {H}^{1}{\left( {\mathbf{Q}}_{q}^{\mathrm{... | No |
Proposition 1.8. If \( q \in \mathcal{M}\;\left( {q \neq p}\right) \) and \( X = {V}_{{\lambda }^{n}} \) then \( {h}_{q} = 1 \) . | Proof. This is a straightforward calculation. For example if \( q \) is of type (A) then we have\n\n\[ \n{L}_{n, q} = \ker \left\{ {{H}^{1}\left( {{\mathbf{Q}}_{q},{V}_{{\lambda }^{n}}}\right) \rightarrow {H}^{1}\left( {{\mathbf{Q}}_{q},{W}_{{\lambda }^{n}}/{W}_{{\lambda }^{n}}^{0}}\right) \oplus {H}^{1}\left( {{\mathb... | Yes |
Proposition 1.9. (i) If \( X = {V}_{{\lambda }^{n}} \) then | \[ {h}_{p}{h}_{\infty } = \# {\left( \mathcal{O}/\lambda \right) }^{3n}\# {H}^{0}\left( {{\mathbf{Q}}_{p},{V}_{{\lambda }^{n}}^{ * }}\right) /\# {H}^{0}\left( {\mathbf{Q},{V}_{{\lambda }^{n}}^{ * }}\right) \] in the unrestricted case.\n\nProof. Case (i) is trivial. | No |
Corollary 1. In case (i), \( {J}_{H}{\widehat{\left( N)(\overline{\mathbf{Q}}\right) }}_{\mathfrak{m}} \simeq {\mathbf{T}}_{\mathfrak{m}}^{2}\; \) and \( \;{\mathrm{{Ta}}}_{\mathfrak{m}}\left( {{J}_{H}\left( N\right) \left( \overline{\mathbf{Q}}\right) }\right) \simeq \) \( {\mathbf{T}}_{\mathrm{m}}^{2} \) . In case (i... | In each case the first isomorphisms of Corollary 1 follow from the theorem together with the rank 2 result alluded to previously. Corollary 2 and the second isomorphisms of corollary 1 then follow on applying duality (2.4). (In the proof and in all applications we will only use the notion of a Gorenstein \( {\mathbf{Z}... | Yes |
Proposition 2.7. Suppose that \( \mathfrak{m} \) is a maximal ideal of \( \mathbf{T} = {\mathbf{T}}_{H}\left( N\right) \) associated to an irreducible representation. Suppose that \( q + {Np} \) . Then\n\n\[ \left( {\Delta }_{q}\right) = {\left( q - 1\right) }^{2}\left( {{T}_{q}^{2}-\langle q\rangle {\left( 1 + q\right... | The proof is a trivial generalization of that of Proposition 2.6. | No |
Proposition 4.1. There is an isomorphism\n\n\[ \n{H}_{\mathrm{{unr}}}^{1}\left( {{\mathbf{Q}}_{\sum }/\mathbf{Q},{Y}^{ * }}\right) \overset{ \sim }{ \rightarrow }\operatorname{Hom}{\left( \operatorname{Gal}\left( {M}_{\infty }/L\left( \nu \right) \right) ,\left( K/\mathcal{O}\right) \left( \nu \right) \right) }^{\opera... | Proof. The sequence is obtained from the inflation-restriction sequence as follows. First we can replace \( {H}^{1}\left( {{\mathbf{Q}}_{\sum }/\mathbf{Q},{Y}^{ * }}\right) \) by \n\n\[ \n{\left\{ {H}^{1}\left( {\mathbf{Q}}_{\sum }/L,\;\left( K/\mathcal{O}\right) \left( \nu \right) \right) \oplus {H}^{1}\left( {\mathbf... | Yes |
Proposition 4.6.\n\n\\[ \left\\langle {{f}_{\\varphi },{f}_{\\varphi }}\\right\\rangle = \\frac{1}{{16}{\\pi }^{3}}{N}^{2}\\left\\{ {\\mathop{\\prod }\\limits_{\\substack{{q \\mid N} \\ {q \\notin {S}_{\\varphi }} }}\\left( {1 - \\frac{1}{q}}\\right) }\\right\\} {L}_{N}\\left( {2,{\\varphi }^{2}\\overline{\\widehat{\\c... | Proof. One begins with a formula of Petersson that for an eigenform of weight 2 on \\( {\\Gamma }_{1}\\left( N\\right) \\) says\n\n\\[ \\langle f, f\\rangle = {\\left( 4\\pi \\right) }^{-2}\\Gamma \\left( 2\\right) \\left( \\frac{1}{3}\\right) \\pi \\left\\lbrack {{\\mathrm{{SL}}}_{2}\\left( \\mathbf{Z}\\right) : {\\Ga... | Yes |
Proposition 1. Suppose that \( \mathcal{O} \) is a complete discrete valuation ring and that \( \varphi : S \rightarrow T \) is a surjective local \( \mathcal{O} \) -algebra homomorphism between complete local Noetherian \( \mathcal{O} \) -algebras. Suppose further that \( {\mathfrak{p}}_{T} \) is a prime ideal of \( T... | Proof. First we consider the case where \( u = 0 \) . We may assume that the generators \( {x}_{1},\ldots ,{x}_{r} \) lie in \( {\mathfrak{p}}_{T} \) by subtracting their residues in \( T/{\mathfrak{p}}_{T} \rightarrow \mathcal{O} \) . By (ii) we may also write\n\n\[ S \simeq \mathcal{O}\llbracket {x}_{1},\ldots ,{x}_{... | Yes |
Lemma 4.1 \( {T}_{1},\cdots ,{T}_{s - 1}, S \) 全体线性无关。 | Proof: 如前所述,“沾光代表” \( S \) 是因为弟子担任了第 \( \left( {s + 1}\right) \) 代代表而毫不费力当选的。 这些弟子,也就是 \( {R}_{s + 1} \) 的成员 \( {r}_{1},\cdots ,{r}_{q} \) ,具有一项关键的性质 - 弟子本身当然是第 \( \left( {s + 1}\right) \) 代成员,不仅如此,他们的线性组合也一定是第 \( \left( {s + 1}\right) \) 代成员 (除了 \( o \) 以外)。换句话说,无论选择什么样的数 \( {c}_{1},\cdots ,{c}_{q} \) (除了 \( {c}_{1} =... | Yes |
Lemma 4.2 令 \( V \) 为线性空间, \( U \) 为 \( V \) 的线性子空间。设 \( \left( {{v}_{1},\cdots ,{v}_{n}}\right) \) 为 \( \mathrm{V} \) 的基底,并且其中前 \( m \) 项 \( \left( {{v}_{1},\cdots ,{v}_{m}}\right) \) 构成 \( U \) 的基底。这时,如果 \( \left( {{v}_{1}^{\prime },\cdots ,{v}_{m}^{\prime }}\right) \) 是 \( U \) 的另一组基底,那么 \( \left( {{\mathbf{v}}_{1}^... | Proof: 让我们来逐条确认构成基底的条件。首先是 “是否能够表示 \( V \) 内的任何向量 \( x \) ”。 因为 \( \left( {{v}_{1},\cdots ,{v}_{n}}\right) \) 是 \( V \) 的基底,所以可以选择合适的数 \( {c}_{1},\cdots ,{c}_{n} \) ,使得\n\n\[ \mathbf{x} = {c}_{1}{\mathbf{v}}_{1} + \cdots + {c}_{m}{\mathbf{v}}_{m} + {c}_{m + 1}{\mathbf{v}}_{m + 1} + \cdots + {c}_{n}{\mathbf{v}}_{n} \]\n... | "Yes" |
Problem 2.6. \( p, q \) 为大于一的整数, \( \gcd \left( {p,{6q}}\right) = 1 \) . 证明: | \[ \mathop{\sum }\limits_{{k = 1}}^{{q - 1}}{\left\lfloor \frac{pk}{q}\right\rfloor }^{2} \equiv {2p}\mathop{\sum }\limits_{{k = 1}}^{{q - 1}}k\left\lfloor \frac{pk}{q}\right\rfloor \left( {{\;\operatorname{mod}\;q} - 1}\right) \] | Yes |
Proposition 1. \( e\left( {\xi }_{hj}\right) = \left( {h + j}\right) \mathbf{u} \) and \( {p}_{1}\left( {\xi }_{hj}\right) = 2\left( {h - j}\right) \mathbf{u} \), where \( \mathbf{u} \mathrel{\text{:=}} {\mathbf{u}}_{1} \) . | Proof. As noted above, this is a special case of Hopf quaternion bundles with \( n = 1 \) . In this case, \( e\left( {\gamma }_{\mathbb{R}}\right) = \mathbf{u} \) and \[ {p}_{1}\left( {\gamma }_{\mathbb{R}}\right) = - {c}_{2}\left( {{\gamma }_{\mathbb{C}} \otimes \mathbb{C}}\right) = - {c}_{2}\left( {{\gamma }_{\mathbb... | Yes |
Lemma 1. Suppose \( M \) is an open subset of \( {\mathbb{R}}^{m} \) . Let \( X \subset M \) be a closed subset in \( M \) and \( K \) a compact subset of \( M \) . Suppose \( f : M \rightarrow {\mathbb{R}}^{n} \) is a smooth map and \( f{ \pitchfork }_{X}\{ 0\} \) . Fix a compact \( {K}^{\prime } \subset M \) with \( ... | Proof. Let \( \lambda : M \rightarrow \left\lbrack {0,1}\right\rbrack \) be a smooth cut-off function such that \( {\left. \lambda \right| }_{K} \equiv 1 \) and \( \lambda { \mid }_{c{K}^{\prime }} \equiv 0 \) . According to the fact mentioned above, we can take \( y \in {\mathbb{R}}^{n} \) with \( \left| y\right| < \v... | Yes |
Lemma 2. If \( k \geq n \) and \( p \geq n \), then the homomorphism \( \tau : \pi \left( {\mathbf{T}\left( {\widetilde{\gamma }}_{p}^{k}\right) }\right) \rightarrow {\Omega }_{n} \) is surjective. | Proof. Let \( {M}^{n} \) be a compact smooth manifold of dimension \( n \) . By Whitney embedding theorem, one can embed \( {M}^{n} \) into \( {\mathbb{R}}^{n + k} \) for some \( k \) . Let \( T{N}^{k} \) be the normal vector bundle of \( {M}^{n} \) in \( {\mathbb{R}}^{n + k} \) (the superscript indicates that \( {TN} ... | Yes |
Theorem 4. (Thom)\n\n\( \Omega \otimes \mathbb{Q} \cong \mathbb{Q}\left\lbrack {{\mathbb{{CP}}}^{2},{\mathbb{{CP}}}^{4},{\mathbb{{CP}}}^{6},\cdots }\right\rbrack . \) | Proof. By lemma 2, we know that \( {\Omega }_{n} \) is a homomorphic image of \( {\pi }_{n + k}\left( {\mathbf{T}\left( {\widehat{\gamma }}_{p}^{k}\right) }\right) \) . By theorem 3 , we have\n\n\[ \n{\pi }_{n + k}\left( {\mathbf{T}\left( {\widetilde{\gamma }}_{p}^{k}\right) }\right) \otimes \mathbb{Q} \cong {H}_{n + k... | Yes |
Theorem 5. (Poincaré Duality)\n\nIf \( M \) is a \( n \) -dimensional oriented compact manifold, then\n\n\[ \n{H}^{k}\left( {M;R}\right) = {H}_{n - k}\left( {M;R}\right) \n\] \n\nfor \( k = 0,1,\cdots, n \), where \( R \) can be any coefficient ring. | We consider the pairing\n\n\[ \n{H}_{i}\left( {M;F}\right) \times {H}_{n - i}\left( {M;F}\right) \rightarrow F. \n\] \n\nSince \( \dim {H}^{i}\left( {M;F}\right) = \dim {H}_{n - i}\left( {M;F}\right) = \dim {H}^{n - i}\left( {M;F}\right) \) by Poincaré duality and natural isomorphism of vector spaces, we can view \( {H... | Yes |
Theorem 6. (Thom)\n\n\( \sigma : \Omega \rightarrow \mathbb{Z} \) is a ring homomorphism, i.e., \( \sigma \) satisfies\n\n(a) \( \sigma \left( {M + N}\right) = \sigma \left( M\right) + \sigma \left( N\right) \) .\n\n(b) \( \sigma \left( {M \times N}\right) = \sigma \left( M\right) \times \sigma \left( N\right) \) .\n\n... | Proof. Let \( \dim M = m,\dim N = n \) and \( \dim W = m + 1 \) .\n\n(a) This is obvious.\n\n(b) Let \( V \mathrel{\text{:=}} M \times N \) . If \( 4 \nmid \dim V \), then \( 4 \nmid m \) or \( 4 \nmid n \) . Thus \( \sigma \left( {M \times N}\right) = 0 \) and\n\n\( \sigma \left( M\right) \times \sigma \left( N\right)... | Yes |
Proposition 2. Let \( A \) be the graded polynomial ring \( R\left\lbrack t\right\rbrack \) where \( t \) is a variable of degree 1. Given \( f\left( t\right) \in 1 + {\lambda }_{1}t + {\lambda }_{2}{t}^{2} + \cdots \in R\left\lbrack \left\lbrack t\right\rbrack \right\rbrack \), there is an unique multiplicative sequen... | Proof. By definition, we want to find \( \left\{ {K}_{n}\right\} \) satisfying\n\n\[ K\left( {1 + t}\right) = 1 + {K}_{1}\left( t\right) + {K}_{2}\left( {t,0}\right) + {K}_{3}\left( {t,0,0}\right) + \cdots = 1 + {\lambda }_{1}t + {\lambda }_{2}{t}^{2} + {\lambda }_{3}{t}^{3}\cdots . \]\n\nThat is, we want to find \( {K... | Yes |
Theorem 7. (Hirzebruch's Signature Formula)\n\nLet \( \\left\\{ {L}_{n}\\right\\} \) be the multiplicative sequence belonging to\n\n\[ \n\\frac{\\sqrt{t}}{\\tanh \\sqrt{t}} = 1 + \\frac{1}{3}t - \\frac{1}{45}{t}^{2} + \\cdots + \\frac{{\\left( -1\\right) }^{n - 1}{2}^{2n}{B}_{n}}{\\left( {2n}\\right) !}{t}^{n} + \\cdot... | Proof. By Thom’s cobordism theorem, we only need to check that \( \\sigma \\left( {\\mathbb{C}{\\mathbb{P}}^{2k}}\\right) = \) \( L\\left( {\\mathbb{{CP}}}^{2k}\\right) \) for each \( k \\in \\mathbb{N} \) . We have already computed that \( \\sigma \\left( {\\mathbb{{CP}}}^{2k}\\right) = 1 \) . To compute \( L\\left( {... | Yes |
Lemma 2. For any \( 1 \leq i, j \leq m \), arcs \( {S}^{ - }\left( {E}_{i}\right) \) and \( {S}^{ + }\left( {E}_{j}\right) \) intersect if and only if the midpoints \( {Q}_{i} \) and \( {Q}_{j} \) of edges \( {E}_{i} \) and \( {E}_{j} \) are antipodal. | Proof of lemma 2. Again, by properties (a, b) above, the endpoints of arc \( {S}^{ - }\left( {E}_{i}\right) \) cannot belong to \( {S}^{ + }\left( {E}_{j}\right) \) and vice versa. The two arcs are either disjoint or intersecting.\n\nAssume that arcs \( {S}^{ - }\left( {E}_{i}\right) \) and \( {S}^{ + }\left( {E}_{j}\r... | Yes |
Lemma 2. Let \( U \) and \( V \) be two antipodal points and assume that plane \( P \), passing through 0, separates the interiors of \( \Gamma - U \) and \( \Gamma - V \) . Let \( {\Psi }_{1} = \left( {\Gamma - U}\right) \cap P \) and \( {\Psi }_{2} = \left( {\Gamma - V}\right) \cap P \) . Then \( \Delta \cap \left( {... | Proof. The sets \( \Gamma - U \) and \( - \Gamma + V \) lie in the same closed half space bounded by \( P \) . Therefore, for any points \( X \in \left( {\Gamma - U}\right) \) and \( Y \in \left( {-\Gamma + V}\right) \), we have \( X + Y \in P \) if and only if \( X, Y \in P \) . Then\n\n\[ \left( {\Delta - \left( {U -... | Yes |
Lemma 1. If \( \left( {x, y}\right) \) a representation of a local champion \( c \) then \( {xy} < 0 \) . | Proof. Suppose indirectly that \( x \geq 0 \) and \( y \geq 0 \) and consider the values \( w\left( c\right) \) and \( w\left( {c + a}\right) \) . All representations of the numbers \( c \) and \( c + a \) in the form \( {au} + {bv} \) can be written as\n\n\[ c = a\left( {x - {kb}}\right) + b\left( {y + {ka}}\right) ,\... | Yes |
Lemma 2. Let \( c = {ax} - {by} \) where \( \left| x\right| + \left| y\right| \) is minimal and \( x, y \) have the same sign. The number \( c \) is a local champion if and only if \( \left| x\right| < b \) and \( \left| x\right| + \left| y\right| = \left\lfloor \frac{a + b}{2}\right\rfloor \) . | Proof. Without loss of generality we may assume \( x, y > 0 \) . The numbers \( c - a \) and \( c + b \) can be written as \[ c - a = a\left( {x - 1}\right) - {by}\;\text{ and }\;c + b = {ax} - b\left( {y - 1}\right) \] and trivially \( w\left( {c - a}\right) \leq \left( {x - 1}\right) + y < w\left( c\right) \) and \( ... | Yes |
Lemma 3. Let \( c = {ax} - {by} \) and assume that \( x \) and \( y \) have the same sign, \( \left| x\right| < b,\left| y\right| < a \) and \( \left| x\right| + \left| y\right| = \left\lfloor \frac{a + b}{2}\right\rfloor \) . Then \( w\left( c\right) = x + y \) . | Proof. By definition \( w\left( c\right) = \min \{ \left| {x - {kb}}\right| + \left| {y - {ka}}\right| : k \in \mathbb{Z}\} \) . If \( k \leq 0 \) then obviously \( \left| {x - {kb}}\right| + \left| {y - {ka}}\right| \geq x + y \) . If \( k \geq 1 \) then\n\n\[ \left| {x - {kb}}\right| + \left| {y - {ka}}\right| = \lef... | Yes |
Lemma 1. \( \left( {a,0,0}\right) \rightarrow \left( {0,{2}^{a},0}\right) \) for every \( a \geq 1 \) . | Proof. We prove by induction that \( \left( {a,0,0}\right) \rightarrow \left( {a - k,{2}^{k},0}\right) \) for every \( 1 \leq k \leq a \) . For \( k = 1 \) , apply Move 1 to the first stack:\n\n\[ \left( {a,0,0}\right) \rightarrow \left( {a - 1,2,0}\right) = \left( {a - 1,{2}^{1},0}\right) . \]\n\nNow assume that \( k ... | Yes |
For every positive integer \( n \), let \( {P}_{n} = \underset{n}{\underbrace{{2}^{{2}^{{.}^{{.}^{{.}^{2}}}}}}} \) (e.g. \( {P}_{3} = {2}^{{2}^{2}} = {16} \) ). Then \( \left( {a,0,0,0}\right) \rightarrow \left( {0,{P}_{a},0,0}\right) \) for every \( a \geq 1 \) . | Similarly to Lemma 1, we prove that \( \left( {a,0,0,0}\right) \rightarrow \left( {a - k,{P}_{k},0,0}\right) \) for every \( 1 \leq k \leq a \). For \( k = 1 \), apply Move 1 to the first stack: \[ \left( {a,0,0,0}\right) \rightarrow \left( {a - 1,2,0,0}\right) = \left( {a - 1,{P}_{1},0,0}\right) . \] Now assume that t... | Yes |
Lemma 1. For every \( 1 \leq i < j \leq 3 \), consider those circles \( \Omega \left( {P, r}\right) \) in the half-plane \( H \) which are tangent to \( {h}_{i} \) and \( {h}_{j} \) . (a) The locus of the centers of these circles is the angle bisector \( {\beta }_{ij} \) between \( {h}_{i} \) and \( {h}_{j} \) . (b) Th... | Proof. Part (a) is obvious. To prove part (b), notice that the circles which are tangent to \( {h}_{i} \) and \( {h}_{j} \) are homothetic with the common homothety center \( B \) (see Fig. 2). Then part (b) also becomes trivial. | No |
Lemma 3. The curved quadrilateral \( {\mathcal{Q}}_{ij} = \widehat{{V}_{i, j}{V}_{i + 1, j}}{V}_{i + 1, j + 1}\widehat{{V}_{i, j + 1}} \) is circumscribed if and only if \( {u}_{i, i + 1} = {v}_{j, j + 1} \) . | Proof. First suppose that the curved quadrilateral \( {\mathcal{Q}}_{ij} \) is circumscribed and \( \Omega \left( {P, r}\right) \) is its inscribed circle. By Lemma 1 and Lemma 2 we have \( r = {u}_{i, i + 1} \cdot d\left( P\right) \) and \( r = {v}_{j, j + 1} \cdot d\left( P\right) \) as well. Hence, \( {u}_{i, i + 1}... | Yes |
Lemma 1. For every Fibonacci-type sequence \( {a}_{0},{a}_{1},{a}_{2},\ldots \), and every \( k \geq 0 \), we have \( {a}_{k} = \) \( {F}_{k - 1}{a}_{0} + {F}_{k}{a}_{1} \) | Proof. Apply induction on \( k \) . The base cases \( k = 0,1 \) are trivial. For the step, from the induction hypothesis we get\n\n\[ \n{a}_{k + 1} = {a}_{k} + {a}_{k - 1} = \left( {{F}_{k - 1}{a}_{0} + {F}_{k}{a}_{1}}\right) + \left( {{F}_{k - 2}{a}_{0} + {F}_{k - 1}{a}_{1}}\right) = {F}_{k}{a}_{0} + {F}_{k + 1}{a}_{... | Yes |
For any distinct integers \( x \) and \( y \), the function \( {\Delta }_{\operatorname{lcm}\left( {f\left( x\right), f\left( y\right) }\right) }f \) is \( 2\left( {y - x}\right) \) -periodic. | Proof. Denote \( L = \operatorname{lcm}\left( {f\left( x\right), f\left( y\right) }\right) \) . Applying (3) twice, we obtain\n\n\[ \n{\Delta }_{L}f\left( b\right) = {\Delta }_{L}f\left( {{2x} - b - L}\right) = {\Delta }_{L}f\left( {{2y} - \left( {b + 2\left( {y - x}\right) }\right) - L}\right) = {\Delta }_{L}f\left( {... | Yes |
Lemma 2. Let \( g \) be a function. If \( t \) and \( s \) are nonzero integers such that \( {\Delta }_{ts}g = 0 \) and \( {\Delta }_{t}{\Delta }_{t}g = 0 \), then \( {\Delta }_{t}g = 0 \) . | Proof. Assume, without loss of generality, that \( s \) is positive. Let \( a \) be an arbitrary integer. Since \( {\Delta }_{t}{\Delta }_{t}g = 0 \), we have\n\n\[ \n{\Delta }_{t}g\left( a\right) = {\Delta }_{t}g\left( {a + t}\right) = \cdots = {\Delta }_{t}g\left( {a + \left( {s - 1}\right) t}\right) .\n\]\n\nThe sum... | Yes |
Property 2. The elements \( {a}_{1} \) and \( {a}_{2} \) cannot be simultaneously partitioning. Also, \( {a}_{n - 2} \) and \( \overline{{a}_{n - 1}} \) cannot be simultaneously partitioning | Proof. Assume that \( {a}_{1} \) and \( {a}_{2} \) are partitioning. By the claim, it follows that \( {a}_{2} = {g}_{1} = {\ell }_{1} = \) \( \operatorname{lcm}\left( {a}_{1}\right) = {a}_{1} \), a contradiction.\n\nSimilarly, assume that \( {a}_{n - 2} \) and \( {a}_{n - 1} \) are partitioning. The claim yields that \... | Yes |
Property 1. Any positive integer \( n \) has at most one odd and at most one even representation. | Proof. We first show that every integer \( n \) has at most one even representation. Since \( S \) is infinite, there exists \( x \in S \) such that \( x > \max \{ n, N\} \) . Then, the number \( n + x \) must be clean, and \( x \) does not appear in any even representation of \( n \) . If \( n \) has more than one eve... | No |
Property 2. Fix \( s \in S \) . Suppose that a number \( n > N \) has no even representation. Then \( n + {2as} \) has an even representation containing \( s \) for all integers \( a \geq 1 \) . | Proof. It is sufficient to prove the following statement: If \( n \) has no even representation without \( s \) , then \( n + {2s} \) has an even representation containing \( s \) (and hence no even representation without \( s \) by Property 1).\n\nNotice that the odd representation of \( n + s \) does not contain \( s... | Yes |
Property 3. Every sufficiently large integer has an even representation. | Proof. Fix any \( s \in S \), and let \( r \) be an arbitrary element in \( \{ 1,2,\ldots ,{2s}\} \) . Then, Property 2 implies that the set \( {Z}_{r} = \{ r + {2as} : a \geq 0\} \) contains at most one number exceeding \( N \) with no even representation. Therefore, \( {Z}_{r} \) contains finitely many positive integ... | No |
Property 4. For any \( s, t \in S \) with \( N < s < t \), the even representation of \( t \) contains \( s \) . | Proof. Suppose the contrary. Then, \( s + t \) has at least two odd representations: one obtained by adding \( s \) to the even representation of \( t \) and one obtained by adding \( t \) to the even representation of \( s \) . Since the latter does not contain \( s \), these two odd representations of \( s + t \) are... | Yes |
Lemma 1. Suppose that \( G = \left( {V, E}\right) \) is a graph with odd chromatic number \( k \geq 3 \), and let (1) be one of its leximinimal colorings. Then \( G \) contains an odd cycle which visits all color classes \( {V}_{1},{V}_{2},\ldots ,{V}_{k} \) | Proof of Lemma 1. Let us call a cycle colorful if it visits all color classes.\n\nDue to the definition of the chromatic number, \( {V}_{1} \) is nonempty. Choose an arbitrary vertex \( v \in {V}_{1} \) . We construct a colorful odd cycle that has only one vertex in \( {V}_{1} \), and this vertex is \( v \) .\n\nWe dra... | Yes |
Lemma 1. Let \( {EFGH} \) be a circumscribed quadrilateral, and let \( M \) be its incenter. Then\n\n\[ \frac{{EF} \cdot {FG}}{{GH} \cdot {HE}} = \frac{F{M}^{2}}{H{M}^{2}} \] | Proof. Notice that \( \angle {EMH} + \angle {GMF} = \angle {FME} + \angle {HMG} = {180}^{ \circ },\angle {FGM} = \angle {MGH} \), and \( \angle {HEM} = \angle {MEF} \) (see Figure 3). By the law of sines, we get\n\n\[ \frac{EF}{FM} \cdot \frac{FG}{FM} = \frac{\sin \angle {FME} \cdot \sin \angle {GMF}}{\sin \angle {MEF}... | Yes |
Lemma 2. Let \( {EFGH} \) and \( {E}^{\prime }{F}^{\prime }{G}^{\prime }{H}^{\prime } \) be circumscribed quadrilaterals such that \( \angle E + \angle {E}^{\prime } = \) \( \angle F + \angle {F}^{\prime } = \angle G + \angle {G}^{\prime } = \angle H + \angle {H}^{\prime } = {180}^{ \circ } \) . Then \[ \frac{{EF} \cdo... | Proof. Let \( M \) and \( {M}^{\prime } \) be the incenters of \( {EFGH} \) and \( {E}^{\prime }{F}^{\prime }{G}^{\prime }{H}^{\prime } \), respectively. We use the notation \( \left\lbrack {XYZ}\right\rbrack \) for the area of a triangle \( {XYZ} \) . Taking into account the relation \( \angle {FME} + \angle {F}^{\pri... | Yes |
Lemma 1. A triangulation of a convex polygon \( \Pi \) cannot contain two parallelograms. | Proof. Arguing indirectly, assume that \( {P}_{1} \) and \( {P}_{2} \) are two parallelograms contained in some triangulation \( \mathcal{T} \) . If they have a common triangle in \( \mathcal{T} \), then we may assume that \( {P}_{1} \) consists of triangles \( {ABC} \) and \( {ADC} \) of \( \mathcal{T} \), while \( {P... | Yes |
Lemma 2. Every triangle in a Thaiangulation \( \mathcal{T} \) of \( \Pi \) contains a side of \( \Pi \) . | Proof. Let \( {ABC} \) be a triangle in \( \mathcal{T} \) . Apply an affine transform such that \( {ABC} \) maps to an equilateral triangle; let \( {A}^{\prime }{B}^{\prime }{C}^{\prime } \) be the image of this triangle, and \( {\Pi }^{\prime } \) be the image of \( \Pi \) . Clearly, \( \mathcal{T} \) maps into a Thai... | Yes |
Lemma 2. Assume that \( n > 1 \) is bad. Then there exists a \( j \in \{ 1,2,3\} \) such that \( {a}_{n + j} \geq \) \( {a}_{n - 1} + j + 1 \), and \( {a}_{n + i} \geq {a}_{n - 1} + i \) for all \( 1 \leq i < j \) . | Proof. Recall that \( {b}_{n - 1} = 1 \) . Set\n\n\[ m = \inf \left\{ {i > 0 : {b}_{n + i - 1} > 1}\right\} \]\n\n(possibly \( m = + \infty \) ). We claim that \( j = \min \{ m,3\} \) works. Again, we distinguish several cases, according to the value of \( m \) ; in each of them we use Lemma 1 without reference.\n\nCas... | Yes |
Lemma 1. Assume that \( X < {2x} \). Then on the interval \( (X - x;x\rbrack \) the function \( g \) attains at most two values - namely, \( X \) and, possibly, some \( Y > X \). Similarly, if \( X > {2x} \), then \( g \) attains at most two values on \( \lbrack x;X - x) \) -namely, \( X \) and, possibly, some \( Y < X... | Proof. We start with the first claim of the lemma. Notice that \( X - x < x \), so the considered interval is nonempty.\n\nTake any \( a \in \left( {X - x;x}\right) \) with \( g\left( a\right) \neq X \) (if it exists). If \( g\left( a\right) < X \), then (*) yields \( g\left( a\right) \leq \) \( a + x \leq g\left( x\ri... | Yes |
Lemma 2. If \( X < {2x} \), then \( g \) is constant on \( \left( {X - x;x}\right) \) . Similarly, if \( X > {2x} \), then \( g \) is constant on \( \left( {x;X - x}\right) \) . | Proof. Again, it suffices to prove the first claim only. Assume, for the sake of contradiction, that there exist \( a, b \in \left( {X - x;x}\right) \) with \( g\left( a\right) \neq g\left( b\right) \) ; by Lemma 1, we may assume that \( g\left( a\right) = X \) and \( Y = g\left( b\right) > X \) . Notice that \( \min \... | Yes |
Lemma 3. If \( X < {2x} \), then \( g\left( a\right) = X \) for all \( a \in \left( {X - x;x}\right) \). Similarly, if \( X > {2x} \), then \( g\left( a\right) = X \) for all \( a \in \left( {x;X - x}\right) \). | Proof. Again, we only prove the first claim.\n\nBy Lemmas 1 and 2, this claim may be violated only if \( g \) takes on a constant value \( Y > X \) on \( \left( {X - x, x}\right) \). Choose any \( a, b \in \left( {X - x;x}\right) \) with \( a < b \). By \( \left( *\right) \), we have\n\n\[ Y \geq b + x \geq X \]\n\n(2)... | Yes |
Lemma 1. Let \( X \) and \( Y \) be finite sets of positive integers. The functions \( {f}_{X * Y} \) and \( {f}_{X} \circ {f}_{Y} \) are equal. | Proof. We have\n\n\[ \n{f}_{X * Y}\left( {\mathbb{Z}}_{ > 0}\right) = {\mathbb{Z}}_{ > 0} \smallsetminus \left( {X * Y}\right) = \left( {{\mathbb{Z}}_{ > 0} \smallsetminus X}\right) \smallsetminus {f}_{X}\left( Y\right) = {f}_{X}\left( {\mathbb{Z}}_{ > 0}\right) \smallsetminus {f}_{X}\left( Y\right) = {f}_{X}\left( {{\... | Yes |
Lemma 2. Suppose that \( X \) and \( Y \) are finite sets of positive integers satisfying \( X * Y = Y * X \) and \( \left| X\right| = \left| Y\right| \) . Then, we must have \( X = Y \) . | Proof. Assume that \( X \) and \( Y \) are not equal. Let \( s \) be the largest number in exactly one of \( X \) and \( Y \) . Without loss of generality, say that \( s \in X \smallsetminus Y \) . The number \( {f}_{X}\left( s\right) \) counts the \( {s}^{th} \) number not in \( X \), which implies that\n\n\[ \n{f}_{X... | Yes |
Lemma 1. Let two circles in the configuration cross at \( x \) and \( y \) . Then \( x \) and \( y \) are either both yellow or both non-yellow. | Proof. This is because the numbers of interior vertices on the four arcs \( x \) and \( y \) determine on the two circles have like parities. | No |
Lemma 2. If \( \overset{⏜}{xy},\overset{⏜}{yz} \), and \( \overset{⏜}{zx} \) are circular arcs of three pairwise distinct circles in the configuration, then the number of yellow vertices in the set \( \{ x, y, z\} \) is odd. | Proof. Let \( {C}_{1},{C}_{2},{C}_{3} \) be the three circles under consideration. Assume, without loss of generality, that \( {C}_{2} \) and \( {C}_{3} \) cross at \( x,{C}_{3} \) and \( {C}_{1} \) cross at \( y \), and \( {C}_{1} \) and \( {C}_{2} \) cross at \( z \) . Let \( {k}_{1} \) , \( {k}_{2},{k}_{3} \) be the... | Yes |
Lemma 3. Each face of \( G \) has equally many red and blue vertices. In particular, each face has an even number of non-yellow vertices. | Proof. Trace the boundary of a face once in circular order, and consider the colours each vertex is assigned in the colouring of the two circles that cross at that vertex, to infer that colours of non-yellow vertices alternate. | No |
Lemma 1. If \( m \geq 0 \) is an integer, then \( {4}^{m} \) is representable if and only if either of \( {2m} + 1 \) and \( {2m} + 2 \) is good. | Proof. The case \( m = 0 \) is obvious, so we may assume that \( m \geq 1 \) . Let \( n = {2m} + 1 \) or \( {2m} + 2 \) . Then \( n \geq 3 \) . We notice that\n\n\[ \n{S}_{n - 1} < {a}_{n - 2} + {a}_{n} \n\]\n\nThe inequality writes as \( {2}^{n} + {2}^{\lceil n/2\rceil } + {2}^{\lfloor n/2\rfloor } - 3 < {2}^{n} + {2}... | Yes |
Lemma 2. If \( k \geq 2 \), then \( {2}^{{4k} - 2} \) is representable if and only if \( {2}^{k + 1} \) is representable. | Proof. We have \( {2}^{{4k} - 2} < {a}_{{4k} - 2} \), so in a representation of \( {2}^{{4k} - 2} \) we can have only terms \( {a}_{i} \) with \( i \leq {4k} - 3 \) . Notice that\n\n\[ \n{a}_{0} + \cdots + {a}_{{4k} - 3} = {2}^{{4k} - 2} + {2}^{2k} - 3 < {2}^{{4k} - 2} + {2}^{2k} + {2}^{k} = {2}^{{4k} - 2} + {a}_{2k}.\... | Yes |
Proposition 1.1 (Diamond). Suppose that \( \pi : {D}_{p} \rightarrow {\mathrm{{GL}}}_{2}\left( A\right) \) is a continuous representation where \( A \) is an Artinian local ring with residue field \( k \), a finite field of characteristic \( p \) . Suppose \( \pi \approx \left( \begin{matrix} {\chi }_{1}\varepsilon & *... | Proof (taken from [Dia, Prop. 6.1]). We may replace \( \pi \) by \( \pi \otimes {\chi }_{2}^{-1} \) and we let \( \varphi = {\chi }_{1}{\chi }_{2}^{-1} \) . Then \( \pi \cong \left( \begin{matrix} {\varphi \varepsilon } & t \\ 0 & 1 \end{matrix}\right) \) determines a cocycle \( t : {D}_{p} \rightarrow M\left( 1\right)... | No |
Proposition 1.4. If \( {\rho }_{f,\lambda } \) is associated to a p-divisible group (the ordinary case is allowed) then\n\n(i) \( {\operatorname{pr}}_{n}\left( {{H}_{F}^{1}\left( {{\mathbf{Q}}_{p}, T}\right) }\right) = {H}_{F}^{1}\left( {{\mathbf{Q}}_{p}, T/{\lambda }^{n}}\right) \) and similarly for \( {T}^{ * },{T}^{... | Proof. We first observe that \( {\operatorname{pr}}_{n}\left( {{H}_{F}^{1}\left( {{\mathbf{Q}}_{p}, T}\right) }\right) \subset {H}_{F}^{1}\left( {{\mathbf{Q}}_{p}, T/{\lambda }^{n}}\right) \) . Now from the construction we may identify \( T/{\lambda }^{n} \) with \( {V}_{{\lambda }^{n}} \) . A result of Bloch-Kato ([BK... | Yes |
Proposition 1.6.\n\n\[ \n\\# {H}_{L}^{1}\left( {{\\mathbf{Q}}_{\\sum }/\\mathbf{Q}, X}\\right) /\\# {H}_{{L}^{ * }}^{1}\left( {{\\mathbf{Q}}_{\\sum }/\\mathbf{Q},{X}^{ * }}\\right) = {h}_{\\infty }\\mathop{\\prod }\\limits_{{q \\in \\sum }}{h}_{q} \n\]\n\nwhere\n\[ \n\\left\\{ \\begin{array}{ll} {h}_{q} & = \\# {H}^{0}... | Proof.Adapting the exact sequence proof of Poitou and Tate(cf.[Mi2, Th.4.20]) we get a seven term exact sequence\n\n\[ \n0 \\rightarrow {H}_{L}^{1}\left( {{\\mathbf{Q}}_{\\sum }/\\mathbf{Q}, X}\\right) \\rightarrow {H}^{1}\left( {{\\mathbf{Q}}_{\\sum }/\\mathbf{Q}, X}\\right) \\rightarrow \\mathop{\\prod }\\limits_{{q ... | Yes |
Proposition 1.7. If \( q \notin \sum \), and \( X \) is an arbitrary finite \( \operatorname{Gal}\left( {{\mathbf{Q}}_{\sum }/\mathbf{Q}}\right) \) - module of \( p \) -power order,\n\n\[ \n\# {H}_{{L}^{\prime }}^{1}\left( {{\mathbf{Q}}_{\sum \cup q}/\mathbf{Q}, X}\right) /\# {H}_{L}^{1}\left( {{\mathbf{Q}}_{\sum }/\ma... | Proof. Consider the short exact sequence of inflation-restriction:\n\n\[ \n0 \rightarrow {H}_{L}^{1}\left( {{\mathbf{Q}}_{\sum }/\mathbf{Q}, X}\right) \rightarrow {H}_{{L}^{\prime }}^{1}\left( {{\mathbf{Q}}_{\sum \cup q}/\mathbf{Q}, X}\right) \rightarrow \operatorname{Hom}{\left( \operatorname{Gal}\left( {\mathbf{Q}}_{... | Yes |
Proposition 1.8. If \( q \in \mathcal{M}\left( {q \neq p}\right) \) and \( X = {V}_{{\lambda }^{n}} \) then \( {h}_{q} = 1 \) . | Proof. This is a straightforward calculation. For example if \( q \) is of type (A) then we have\n\n\[ \n{L}_{n, q} = \ker \left\{ {{H}^{1}\left( {{\mathbf{Q}}_{q},{V}_{{\lambda }^{n}}}\right) \rightarrow {H}^{1}\left( {{\mathbf{Q}}_{q},{W}_{{\lambda }^{n}}/{W}_{{\lambda }^{n}}^{0}}\right) \oplus {H}^{1}\left( {{\mathb... | Yes |
Proposition 1.9. (i) If \( X = {V}_{{\lambda }^{n}} \) then\n\n\[ \n{h}_{p}{h}_{\infty } = \# {\left( \mathcal{O}/\lambda \right) }^{3n}\# {H}^{0}\left( {{\mathbf{Q}}_{p},{V}_{{\lambda }^{n}}^{ * }}\right) /\# {H}^{0}\left( {\mathbf{Q},{V}_{{\lambda }^{n}}^{ * }}\right)\n\]\n\nin the unrestricted case. | Proof. Case (i) is trivial. | No |
Lemma 2.2. \( {\dim }_{\mathbf{T}/\mathfrak{m}}{H}^{0}\left( {{X}_{1}{\left( N, p\right) }_{/{\mathbf{F}}_{p}},\Omega }\right) \left\lbrack \mathfrak{m}\right\rbrack = 1 \) . | Proof. First we remark that the action of the Hecke operator \( {U}_{p} \) here is most conveniently defined using an extension from characteristic zero. This is explained below. We will first show that \( {\dim }_{\mathbf{T}/\mathfrak{m}}{H}^{0}\left( {{X}_{1}{\left( N, p\right) }_{/{\mathbf{F}}_{p}},\Omega }\right) \... | Yes |
Proposition 2.7. Suppose that \( \mathfrak{m} \) is a maximal ideal of \( \mathbf{T} = {\mathbf{T}}_{H}\left( N\right) \) associated to an irreducible representation. Suppose that \( q \nmid {Np} \) . Then\n\n\[ \left( {\Delta }_{q}\right) = {\left( q - 1\right) }^{2}\left( {{T}_{q}^{2}-\langle q\rangle {\left( 1 + q\r... | The proof is a trivial generalization of that of Proposition 2.6. | No |
Proposition 4.1. There is an isomorphism\n\n\[ \n{H}_{\text{unr }}^{1}\left( {{\mathbf{Q}}_{\sum }/\mathbf{Q},{Y}^{ * }}\right) \overset{ \sim }{ \rightarrow }\operatorname{Hom}{\left( \operatorname{Gal}\left( {M}_{\infty }/L\left( \nu \right) \right) ,\left( K/\mathcal{O}\right) \left( \nu \right) \right) }^{\operator... | Proof. The sequence is obtained from the inflation-restriction sequence as follows. First we can replace \( {H}^{1}\left( {{\mathbf{Q}}_{\sum }/\mathbf{Q},{Y}^{ * }}\right) \) by \n\n\[ \n{\left\{ {H}^{1}\left( {\mathbf{Q}}_{\sum }/L,\left( K/\mathcal{O}\right) \left( \nu \right) \right) \oplus {H}^{1}\left( {\mathbf{Q... | Yes |
Proposition 4.6.\n\n\[ \left\langle {{f}_{\varphi },{f}_{\varphi }}\right\rangle = \frac{1}{{16}{\pi }^{3}}{N}^{2}\left\{ {\mathop{\prod }\limits_{\substack{{q \mid N} \\ {q \notin {S}_{\varphi }} }}\left( {1 - \frac{1}{q}}\right) }\right\} {L}_{N}\left( {2,{\varphi }^{2}\overline{\widehat{\chi }}}\right) {L}_{N}\left(... | Proof. One begins with a formula of Petterson that for an eigenform of weight 2 on \( {\Gamma }_{1}\left( N\right) \) says\n\n\[ \langle f, f\rangle = {\left( 4\pi \right) }^{-2}\Gamma \left( 2\right) \left( \frac{1}{3}\right) \pi \left\lbrack {{\mathrm{{SL}}}_{2}\left( \mathbf{Z}\right) : {\Gamma }_{1}\left( N\right) ... | Yes |
Proposition 1. Suppose that \( \mathcal{O} \) is a complete discrete valuation ring and that \( \varphi : S \rightarrow T \) is a surjective local \( \mathcal{O} \) -algebra homomorphism between complete local Noetherian \( \mathcal{O} \) -algebras. Suppose further that \( {\mathfrak{p}}_{T} \) is a prime ideal of \( T... | Proof. First we consider the case where \( u = 0 \) . We may assume that the generators \( {x}_{1},\ldots ,{x}_{r} \) lie in \( {\mathfrak{p}}_{T} \) by subtracting their residues in \( T/{\mathfrak{p}}_{T} \rightarrow \mathcal{O} \) . By (ii) we may also write\n\n\[ S \simeq \mathcal{O}\llbracket {x}_{1},\ldots ,{x}_{... | Yes |
Lemma 1. Let the incircle of triangle \( {ABC} \) touch side \( {BC} \) at \( D \), and let \( {DT} \) be a diameter of the circle. If line \( {AT} \) meets \( {BC} \) at \( X \), then \( {BD} = {CX} \) . | Proof. Assume wlog that \( {AB} \geq {AC} \) . Consider the dilation with center \( A \) that carries the incircle to an excircle. The line segment \( {DT} \) is the diameter of the incircle that is perpendicular to \( {BC} \), and therefore its image under the dilation must be the diameter of the excircle that is perp... | Yes |
Lemma 2. Let \( A, B, C, D \) be four distinct point in the plane, such that \( {AC} \) is not parallel to \( {BD} \) . Let lines \( {AC} \) and \( {BD} \) meet at \( X \) . Let the circumcircles of \( {ABX} \) and \( {CDX} \) meet again at \( O \) . Then \( O \) is the center of the unique spiral similarity that carri... | Proof. We use directed angles mod \( \pi \) (i.e., directed angles between lines, as opposed to rays) in order to produce a single proof that works in all configurations. Let \( \angle \left( {{\ell }_{1},{\ell }_{2}}\right) \) denote the angle of rotation that takes line \( {\ell }_{1} \) to \( {\ell }_{2} \) . A usef... | Yes |
Lemma 3. Let \( {ABC} \) be a triangle and \( \Gamma \) its circumcircle. Let the tangent to \( \Gamma \) at \( B \) and \( C \) meet at \( D \) . Then \( {AD} \) coincides with a symmedian of \( \bigtriangleup {ABC} \) . | First proof. Let the reflection of \( {AD} \) across the angle bisector of \( \angle {BAC} \) meet \( {BC} \) at \( {M}^{\prime } \) . Then\n\n\[ \frac{B{M}^{\prime }}{{M}^{\prime }C} = \frac{A{M}^{\prime }\frac{\sin \angle {BA}{M}^{\prime }}{\sin \angle {ABC}}}{A{M}^{\prime }\frac{\sin \angle {CA}{M}^{\prime }}{\sin \... | Yes |
Prove that for every positive integer \( n \), there exists integers \( a \) and \( b \) such that \( 4{a}^{2} + 9{b}^{2} - 1 \) is divisible by \( n \) . | Proof. It suffices to find such an \( a \) and \( b \) modulo any prime power.\n\nFor \( {2}^{k} \), take \( a \equiv 0\left( {\;\operatorname{mod}\;{2}^{k}}\right) \) and \( b \equiv {3}^{-1}\left( {\;\operatorname{mod}\;{2}^{k}}\right) \) .\n\nFor any other \( {p}^{k} \), take \( a \equiv {2}^{-1}\left( {\;\operatorn... | Yes |
Let \( {m}_{1},{m}_{2},\ldots ,{m}_{2013} > 1 \) be 2013 pairwise relatively prime positive integers and \( {A}_{1},{A}_{2},\ldots ,{A}_{2013} \) be 2013 (possibly empty) sets with \( {A}_{i} \subseteq \left\{ {1,2,\ldots ,{m}_{i} - 1}\right\} \) for \( i = 1,2,\ldots ,{2013} \) . Prove that there is a positive integer... | Pick \( {t}_{i} \) for which \( {t}_{i} \equiv 1\left( {\;\operatorname{mod}\;{m}_{i}}\right) \) and \( {t}_{i} \equiv 0\left( {\;\operatorname{mod}\;{m}_{j}}\right) \) for \( i \neq j \) . Look at numbers of the form\n\n\[ \sum {b}_{i}{t}_{i} \]\n\nwhere \( {b}_{i} \in {B}_{i} \), and \( {B}_{i} \) is selected so that... | Yes |
Let \( p \) be an odd prime. Then there exists an integer \( n \) such that \( p \mid {n}^{2} + 1 \) if and only if \( p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \) . | By introducing the notion of order, we will prove that \( p \mid {n}^{2} + 1 \Rightarrow p \equiv 1\left( {\;\operatorname{mod}\;4}\right) \) . By introducing the notion of a primitive root, we will prove the converse direction. Finally, we will write down the generalized version of this \( {n}^{2} + 1 \) lemma using c... | No |
Theorem 2.1 (Fundamental Theorem of Orders)\n\nSuppose \( {a}^{N} \equiv 1\\left( {\\operatorname{mod}p}\\right) \). Then the order of \( a\\left( {\\operatorname{mod}p}\\right) \) divides \( N \). | Proof. Important exercise (mandatory if you haven't seen it before). As a hint, use the division algorithm. | No |
Example 3.3 (Primitive Roots Modulo 11 and 13)\n\nIt turns out that \( g = 2 \) is a primitive root modulo both 11 and 13 . Let’s write this out. | <table><thead><tr><th>\( {2}^{n} \)</th><th>mod 11</th><th>mod13</th></tr></thead><tr><td>\( {2}^{1} \)</td><td>2</td><td>2</td></tr><tr><td>\( {2}^{2} \)</td><td>4</td><td>4</td></tr><tr><td>\( {2}^{3} \)</td><td>8</td><td>8</td></tr><tr><td>\( {2}^{4} \)</td><td>5</td><td>3</td></tr><tr><td>\( {2}^{5} \)</td><td>10</... | Yes |
Because the primitive fourth roots of unity are \( i \) and \( - i \), we have\n\n\[ \n{\Phi }_{4}\left( X\right) = \left( {X - i}\right) \left( {X + i}\right) = {X}^{2} + 1.\n\] | One can actually show \( {\Phi }_{n}\left( X\right) \) always has integer coefficients. (In fact, it’s the polynomial of minimal degree with this property.) | No |
Proposition 4.5 (Cyclotomic Polynomials Divide \( {X}^{n} - 1 \) )\n\nFor any integer \( n \), we have\n\n\[ \n{X}^{n} - 1 = \mathop{\prod }\limits_{{d \mid n}}{\Phi }_{d}\left( X\right) \n\]\n\nIn particular, if \( p \) is a prime then\n\n\[ \n{\Phi }_{p}\left( X\right) = \frac{{X}^{p} - 1}{X - 1} = {X}^{p - 1} + {X}^... | Exercise 4.6. Prove this result. (If you don’t see why, do the case \( n = 4 \) first.) | No |
Theorem 4.8 (Divisors of Cyclotomic Values)\n\nLet \( p \) be a prime, \( n \) a positive integer and \( a \) any integer. Suppose that\n\n\[ \n{\Phi }_{n}\left( a\right) \equiv 0\;\left( {\;\operatorname{mod}\;p}\right)\n\]\n\nThen either\n\n- \( a \) has order \( n \) modulo \( p \), and hence \( p \equiv 1\left( {\;... | Proof. Suppose \( {\Phi }_{n}\left( a\right) \equiv 0\left( {\;\operatorname{mod}\;p}\right) \) . By Proposition 4.5, we deduce that \( {a}^{n} - 1 \equiv 0 \) \( \left( {\;\operatorname{mod}\;p}\right) \) . So the order \( m \) of \( a\left( {\;\operatorname{mod}\;p}\right) \) divides \( n \) . Thus, we have two cases... | Yes |
Find all positive integers \( n \) such that \( n \) divides \( {2}^{n} - 1 \) . | Solution. As you might guess after some experimentation, the only \( n \) which works is \( n = 1 \) . It’s obvious that \( n \) has to be odd (since \( {2}^{n} - 1 \) is always odd). But how can we show this?\n\nLet us first consider any prime \( p \) dividing \( n \) . We get that \( p \mid {2}^{n} - 1 \), or \( {2}^... | No |
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