Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Proposition 8. For every integer \( n \geq 0,\mathrm{\;N}\left( {\mathrm{U}}_{\mathrm{L}}^{\psi \left( n\right) }\right) \subset {\mathrm{U}}_{\mathrm{K}}^{n} \) and \( \mathrm{N}\left( {\mathrm{U}}_{\mathrm{L}}^{\psi \left( n\right) + 1}\right) \subset {\mathrm{U}}_{\mathrm{K}}^{n + 1} \) . | We will prove this result below. | No |
Proposition 9. For \( n = 0 \) (resp. \( n \neq 0 \) ), the homomorphism \( {\mathbf{N}}_{n} \) is induced by a multiplicative (resp. additive) non-constant polynomial \( {\mathbf{P}}_{n} \) such that\n\n\[ d\left( {\mathrm{P}}_{n}\right) = \operatorname{Card}\left( {\mathrm{G}}_{\psi \left( n\right) }\right) \;\text{ ... | Proof of Props. 8 and 9. We argue by induction on the order of G. The case \( G = \{ 1\} \) is trivial. If \( G \neq \{ 1\} \), then since it is a solvable group (Chap. IV,\n\n§2), it has a quotient that is cyclic of prime order. Hence there is a subextension \( {\mathrm{K}}^{\prime }/\mathrm{K} \) of \( \mathrm{L}/\ma... | Yes |
Corollary 1. \( {\mathrm{N}}_{n} \) is injective if and only if \( {\mathrm{G}}_{\psi \left( n\right) } = {\mathrm{G}}_{\psi \left( n\right) + 1} \) . | Obvious. | No |
Corollary 2. \( {\mathrm{N}}_{n} \) is surjective in each of the following three cases:\ni) \( \overline{\mathrm{K}} \) is algebraically closed,\nii) \( \overline{\mathrm{K}} \) is perfect, and \( {\mathrm{G}}_{\psi \left( n\right) } = {\mathrm{G}}_{\psi \left( n\right) + 1} \),\niii) \( {\mathrm{G}}_{\psi \left( n\rig... | Indeed, in case i), \( {\mathrm{P}}_{n} \) is a non-constant polynomial; in case ii), \( {\mathrm{P}}_{n} = c{\mathrm{X}}^{{p}^{r}} \) ; and in case iii), \( {\mathbf{P}}_{n} = c\mathbf{X} \) . In all three cases, these polynomial maps are surjective. | Yes |
Corollary 3. We have \( \mathrm{N}\left( {\mathrm{U}}_{\mathrm{L}}^{\psi \left( n\right) }\right) = {\mathrm{U}}_{\mathrm{K}}^{n} \) if \( {\mathrm{G}}_{\psi \left( n\right) } = \{ 1\} \), and \( \mathrm{N}\left( {\mathrm{U}}_{\mathrm{L}}^{\psi \left( n\right) + 1}\right) = {\mathrm{U}}_{\mathrm{K}}^{n + 1} \) if \( {\... | Same proof as for cor. 3 of §3. | No |
Corollary 4. Let \( v \) be a non-negative real number. Then \( \mathrm{N}\left( {\mathrm{U}}_{\mathrm{L}}^{\psi \left( v\right) }\right) = {\mathrm{U}}_{\mathrm{K}}^{v} \) if either \( {\mathrm{G}}_{\psi \left( v\right) } = \{ 1\} \) or if \( \overline{\mathrm{K}} \) is algebraically closed. | Same proof as for cor. 4 of §3. | No |
Proposition 11. Let \( \mathrm{L}/\mathrm{K} \) be a cyclic, totally ramified extension, with Galois group \( \mathrm{G} \), and let \( \mu \) be the largest integer such that \( {\mathrm{G}}_{\mu } \neq \{ 1\} \) . Then \( {\phi }_{\mathrm{L}/\mathrm{K}}\left( \mu \right) \) is an integer. | It remains to prove prop. 11. Denote by \( w \) the discrete valuation on \( \mathrm{L} \) , and let \( \pi \) be a uniformizer of L. Put\n\n\[ r = \operatorname{Card}\left( \mathrm{G}\right) ,\;{r}^{\prime } = \operatorname{Card}\left( {\mathrm{G}}_{\mu }\right) ,\;k = r/{r}^{\prime }.\]\n\nChoose a generator \( s \) ... | Yes |
Lemma 8. The group \( \mathrm{V}/\mathrm{W} \) is cyclic. | Indeed, the map \( y \mapsto {y}^{s - 1} \) defines by passage to the quotient a homomorphism of \( {\mathrm{L}}^{ * }/{\mathrm{U}}_{\mathrm{L}} = \mathbf{Z} \) onto \( \mathrm{V}/\mathrm{W} \) . | No |
Lemma 9. For \( m \) sufficiently large, \( {\mathrm{V}}_{m} = {\mathrm{W}}_{m} \) . | Let \( t \in \mathrm{L} \) be such that \( \operatorname{Tr}\left( t\right) = 1 \), and let \( {m}_{0} = - w\left( t\right) \) . Let \( x \in {\mathrm{V}}_{m} \), with \( m > {m}_{0} \) ; we will show that \( x \in {\mathrm{W}}_{m} \) . Form the \ | No |
Lemma 10. If \( \varphi \left( m\right) \) is an integer and if \( {\mathrm{G}}_{m} = {\mathrm{G}}_{m + 1} \), then \( {\mathrm{V}}_{m} = {\mathrm{V}}_{m + 1} \) . | Put \( n = \varphi \left( m\right) \), so that \( m = \psi \left( n\right) \) . Let \( x \in {\mathrm{V}}_{m} \), and let \( \bar{x} \) be the image of \( x \) in \( {\mathrm{U}}_{\mathrm{L}}^{m}/{\mathrm{U}}_{\mathrm{L}}^{m + 1} \) . Clearly \( x \) belongs to the kernel of the homomorphism\n\n\[ \n{\mathrm{N}}_{n} : ... | Yes |
Lemma 11. Let \( m \) be a positive integer. If the image of \( {\mathrm{W}}_{m} \) in \( {\mathrm{U}}_{\mathrm{L}}^{m}/{\mathrm{U}}_{\mathrm{L}}^{m + 1} \) is non-trivial, then that image is equal to all of \( {\mathrm{U}}_{\mathrm{L}}^{m}/{\mathrm{U}}_{\mathrm{L}}^{m + 1} \) . | Let \( x \) be an element of \( {\mathrm{W}}_{m} \) not belonging to \( {\mathrm{U}}_{\mathrm{L}}^{m + 1} \) . Then \( x = {y}^{s - 1} \), with \( y \in {\mathrm{U}}_{\mathrm{L}} \) . We may assume \( y \in {\mathrm{U}}_{\mathrm{L}}^{1} \) (otherwise multiply \( y \) by an element of \( {\mathrm{U}}_{\mathrm{K}} \) , w... | Yes |
Lemma 12. Let \( n \) be an integer such that \( {\mathrm{G}}_{\psi \left( {n + 1}\right) } = \{ 1\} \) . Let \( m \) be an integer such that\n\n\[ n < \varphi \left( m\right) < n + 1 \]\n\nThen the images of \( {\mathrm{V}}_{m} \) and \( {\mathrm{W}}_{m} \) are both equal to all of \( {\mathrm{U}}_{\mathrm{L}}^{m}/{\m... | Let \( x \in {\mathrm{U}}_{\mathrm{L}}^{m} \) represent \( \bar{x} \in {\mathrm{U}}_{\mathrm{L}}^{m}/{\mathrm{U}}_{\mathrm{L}}^{m + 1} \) . Then \( \psi \left( n\right) < m < \psi \left( {n + 1}\right) \), whence \( m \geq \psi \left( n\right) + 1 \) . By prop. 8, \( \mathrm{N}x \in {\mathrm{U}}_{\mathrm{K}}^{n + 1} \)... | Yes |
Lemma 13. Let \( m \) be an integer, and let \( n + 1 \) be the smallest integer \( \geq \varphi \left( m\right) \) . If \( {\mathrm{G}}_{\psi \left( {n + 1}\right) } = \{ 1\} \), then \( {\mathrm{V}}_{m} = {\mathrm{W}}_{m} \) . | Let us first show that \( {\mathrm{V}}_{m} = {\mathrm{V}}_{m + 1}{\mathrm{\;W}}_{m} \) . If \( \varphi \left( m\right) \) is integral, \( \varphi \left( m\right) = n + 1 \) , \( \psi \left( {n + 1}\right) = m \), and our assertion results from lemma 10. If \( \varphi \left( m\right) \) is not integral, then \( n < \var... | Yes |
Theorem 1. The function \( {a}_{\mathrm{G}} \) is the character of a linear representation of \( \mathrm{G} \) . | It is clear that \( {a}_{\mathrm{G}} \) is a class function. Thus we can write\n\n\[ \n{a}_{\mathrm{G}} = \sum {c}_{\chi }\chi \n\]\n\nwhere \( \chi \) runs through the set of irreducible characters of \( \mathbf{G} \) . Then\n\n\[ \n{c}_{\chi } = \left( {{a}_{\mathrm{G}},\chi }\right) = \frac{1}{g}\mathop{\sum }\limit... | Yes |
Proposition 1. The function \( {a}_{\mathrm{G}} \) is equal to the function \( {\left( {a}_{{\mathrm{G}}_{0}}\right) }^{ * } \) induced by the corresponding function for the inertia group \( {\mathbf{G}}_{0} \) . | As \( {\mathrm{G}}_{0} \) is normal in \( \mathrm{G} \), we have \( {\left( {a}_{{\mathrm{G}}_{0}}\right) }^{ * }\left( s\right) = 0 = {a}_{\mathrm{G}}\left( s\right) \) if \( s \notin {\mathrm{G}}_{0} \) . If \( s \in {\mathrm{G}}_{0} \) , \( s \neq 1 \), then\n\n\[ \n{\left( {a}_{{\mathrm{G}}_{0}}\right) }^{ * }\left... | Yes |
Proposition 2. Let \( {\mathrm{G}}_{i} \) be the ith ramification group of \( \mathrm{G} \), let \( {u}_{i} \) be the character of the augmentation representation of \( {\mathrm{G}}_{i} \), and let \( {u}_{i}^{ * } \) be the character of \( \mathrm{G} \) induced by \( {u}_{i} \). Then \[ {a}_{\mathrm{G}} = \mathop{\sum... | Put \( {g}_{i} = \operatorname{Card}\left( {\mathrm{G}}_{i}\right) \). We have \( {u}_{i}^{ * }\left( s\right) = 0 \) if \( s \notin {\mathrm{G}}_{i} \), while \( {u}_{i}^{ * }\left( s\right) = - g/{g}_{i} = \) \( - f.{g}_{0}/{g}_{i} \) if \( s \in {G}_{i}, s \neq 1 \), and \( \mathop{\sum }\limits_{{s \in G}}{u}_{i}^{... | Yes |
Corollary 1. If \( \varphi \) is a class function on \( \mathrm{G} \), then\n\n\[ f\left( \varphi \right) = \mathop{\sum }\limits_{{i = 0}}^{\infty }\frac{{g}_{i}}{{g}_{0}}\left( {\varphi \left( 1\right) - \varphi \left( {\mathrm{G}}_{i}\right) }\right) . \] | This follows from prop. 2, taking into account that\n\n\[ \left( {\varphi ,{u}_{i}^{ * }}\right) = \left( {\varphi \mid {\mathrm{G}}_{i},{u}_{i}}\right) = \varphi \left( 1\right) - \varphi \left( {\mathrm{G}}_{i}\right) . \] | Yes |
Corollary 2. If \( \chi \) is a character of \( \mathrm{G} \), then \( f\left( \chi \right) \) is a non-negative rational number. | Indeed, prop. 2 shows that \( {g}_{0}{a}_{\mathrm{G}} \) is the character of a linear representation of \( \mathrm{G} \) ; hence \( {g}_{0}.f\left( \chi \right) \) is an integer \( \geq 0 \) . | No |
Proposition 3. If \( \mathrm{N} \) is a normal subgroup of \( \mathrm{G} \), then\n\n\[ \n{a}_{\mathrm{G}/\mathrm{N}} = \left( {a}_{\mathrm{G}}\right) \natural \text{.} \n\] | That results from prop. 3 of Chap. IV. | No |
Proposition 4. Let \( \mathrm{H} \) be a subgroup of \( \mathrm{G} \) corresponding to the subextension \( {\mathrm{K}}^{\prime }/\mathrm{K} \), and let \( {\mathfrak{d}}_{{\mathrm{K}}^{\prime }/\mathrm{K}} \) be the discriminant of \( {\mathrm{K}}^{\prime }/\mathrm{K} \). Then \[ {a}_{\mathrm{G}} \mid \mathrm{H} = \la... | If \( s \neq 1 \) is an element of \( \mathrm{H} \), then \[ {a}_{\mathrm{G}}\left( s\right) = - {f}_{\mathrm{L}/\mathrm{K}}{i}_{\mathrm{G}}\left( s\right) ,\;{a}_{\mathrm{H}}\left( s\right) = - {f}_{\mathrm{L}/{\mathrm{K}}^{\prime }}{i}_{\mathrm{H}}\left( s\right) ,\;{r}_{\mathrm{H}}\left( s\right) = 0, \] and as \( {... | Yes |
Proposition 5. Let \( \chi \) be a character of degree 1 on \( \mathrm{G} \) . Let \( {c}_{\chi } \) be the largest integer for which the restriction of \( \chi \) to the ramification group \( {\mathrm{G}}_{{c}_{\chi }} \) is not the unit character (if \( \chi = {1}_{\mathrm{G}} \), take \( {c}_{\chi } = - 1 \) ). Then... | If \( i \leq {c}_{\chi } \), then \( \chi \left( {\mathrm{G}}_{i}\right) = 0 \), whence \( \chi \left( 1\right) - \chi \left( {\mathrm{G}}_{i}\right) = 1 \) ; if \( i > {c}_{\chi } \), then \( \chi \left( {\mathrm{G}}_{i}\right) = 1 \) , whence \( \chi \left( 1\right) - \chi \left( {\mathrm{G}}_{i}\right) = 0 \) . Appl... | Yes |
Corollary 1. \( {\mathfrak{d}}_{{\mathrm{K}}^{\prime }/\mathrm{K}} = \mathfrak{f}\left( {{s}_{\mathrm{G}/\mathrm{H}},\mathrm{L}/\mathrm{K}}\right) \) . | By decomposing \( {s}_{\mathrm{G}/\mathrm{H}} \) into a linear combination of irreducible characters, one gets a decomposition of the discriminant \( {\mathfrak{d}}_{\mathbf{K}\prime /\mathbf{K}} \) into a product of conductors. For example, if \( \mathrm{H} = \{ 1\} \), then \( {s}_{\mathrm{G}/\mathrm{H}} = {r}_{\math... | No |
Corollary 2. \( {\mathfrak{d}}_{\mathrm{L}/\mathrm{K}} = \prod \mathfrak{f}{\left( \chi \right) }^{\chi \left( 1\right) } \), the product being taken over all irreducible characters \( \chi \) of \( \mathrm{G} \) . | If \( \mathrm{G} \) is abelian, this formula simplifies to\n\n\[ \n{\mathfrak{d}}_{\mathrm{L}/\mathrm{K}} = \prod \mathfrak{f}\left( \chi \right) \n\] | No |
Proposition 1. If \( \mathrm{A} \) is relatively injective, \( {\mathrm{H}}^{q}\left( {\mathrm{G},\mathrm{A}}\right) = 0 \) for all \( q \geq 1 \) . | As \( A \) is a direct factor of a co-induced module, additivity reduces us to the case where \( \mathrm{A} \) itself is co-induced, i.e., \( \mathrm{A} = {\operatorname{Hom}}_{\mathbf{Z}}\left( {\Lambda ,\mathrm{X}}\right) \) for some abelian group \( \mathrm{X} \) . If \( \mathrm{B} \) is a \( \mathrm{G} \) -module, ... | Yes |
Proposition 2. If \( \mathrm{A} \) is relatively projective, \( {\mathrm{H}}_{q}\left( {\mathrm{G},\mathrm{A}}\right) = 0 \) for all \( q \geq 1 \) . | The proof is analogous to that of prop. 1. In particular, an induced module has trivial homology, which again permits methods of \ | No |
Proposition 3. The automorphisms \( {\sigma }_{t} \) are equal to the identity. | The \( {\sigma }_{t} \) constitute an automorphism of the cohomological functor \( \left\{ {{\mathrm{H}}^{q}\left( {\mathrm{G},\;}\right) ,\delta }\right\} \) ; this automorphism is the identity in dimension zero (clear). Hence, by a general result (cf. [13], Chap. III or [26], no. 2.2) it is the identity in all dimens... | Yes |
Proposition 4. The sequence below is exact:\n\n\[ 0 \rightarrow {\mathrm{H}}^{1}\left( {\mathrm{G}/\mathrm{H},{\mathrm{A}}^{\mathrm{H}}}\right) \xrightarrow[]{\text{ Inf }}{\mathrm{H}}^{1}\left( {\mathrm{G},\mathrm{A}}\right) \xrightarrow[]{\text{ Res }}{\mathrm{H}}^{1}\left( {\mathrm{H},\mathrm{A}}\right) . \] | It is clear that Res \( \circ \operatorname{Inf} = 0 \) (look at the cochains, for example). Thus there are two things to prove:\n\n1. Exactness at \( {\mathrm{H}}^{1}\left( {\mathrm{G}/\mathrm{H},{\mathrm{A}}^{\mathrm{H}}}\right) \) . Let \( f : \mathrm{G}/\mathrm{H} \rightarrow {\mathrm{A}}^{\mathrm{H}} \) be a cocyc... | Yes |
Proposition 5. Given a positive integer \( q \) . Suppose that \( {\mathrm{H}}^{i}\left( {\mathrm{H},\mathrm{A}}\right) = 0 \) for\n\n\[ 1 \leq i \leq q - 1.\]\n\nThen the sequence below is exact:\n\n\[ 0 \rightarrow {\mathrm{H}}^{q}\left( {\mathrm{G}/\mathrm{H},{\mathrm{A}}^{\mathrm{H}}}\right) \xrightarrow[]{\text{ I... | Proof of Prop. 5. Argue by induction on \( q \), the case \( q = 1 \) being prop. 4. Suppose \( q \geq 2 \) . Let \( \mathrm{B} = \operatorname{Hom}\left( {\mathbf{Z}\left\lbrack \mathrm{G}\right\rbrack ,\mathrm{A}}\right) \) be the co-induced module canonically associated to A; an element of B can be identified with a... | Yes |
Proposition 6. If \( n = \operatorname{Card}\left( {\mathrm{G}/\mathrm{H}}\right) \), then \( \operatorname{Cor} \circ \operatorname{Res} = n \) . | For \( q = 0 \), this says that \( {\mathrm{N}}_{\mathrm{G}/\mathrm{H}}\left( a\right) = {na} \) if \( a \in {\mathrm{A}}^{\mathrm{G}} \), which is clear. The general case can be reduced to the case \( q = 0 \) by shifting. | No |
Proposition 7. Let \( \theta : \mathrm{H} \smallsetminus \mathrm{G} \rightarrow \mathrm{G} \) be a system of representations for the homogeneous space \( \mathrm{H} \smallsetminus \mathrm{G} \) of right cosets of \( \mathrm{G}{\;\operatorname{mod}\;.}\mathrm{H} \) . For every \( s \in \mathrm{G} \) and \( t \in \mathrm... | We have recovered the classical definition of the transfer-cf. for example M. Hall [30], p. 202. | No |
Proposition 8. Let \( {x}_{i} \) be a system of representatives of the double cosets \( \mathrm{H}x\mathrm{\;S} \) , and for each \( i \), define \( {f}_{i} = {f}_{{x}_{i}} \) as above. Then | \[ \operatorname{Ver}\left( s\right) = \prod {x}_{i} \cdot {s}^{{f}_{i}} \cdot {x}_{i}^{-1}\;{\;\operatorname{mod}\;.}{\mathrm{H}}^{\prime }.\] | No |
Proposition 1. Let \( 1 \rightarrow \mathrm{A}\overset{i}{ \rightarrow }\mathrm{B}\overset{p}{ \rightarrow }\mathrm{C} \rightarrow 1 \) be an exact sequence of non-abelian G-modules. Then the sequence of pointed sets below is exact: | \[ 1 \rightarrow {\mathrm{H}}^{0}\left( {\mathrm{G},\mathrm{A}}\right) \overset{{i}_{0}}{ \rightarrow }{\mathrm{H}}^{0}\left( {\mathrm{G},\mathrm{B}}\right) \overset{{p}_{0}}{ \rightarrow }{\mathrm{H}}^{0}\left( {\mathrm{G},\mathrm{C}}\right) \overset{\delta }{ \rightarrow }{\mathrm{H}}^{1}\left( {\mathrm{G},\mathrm{A}... | Yes |
Proposition 2. In addition to the hypothesis of prop. 1, assume that A is in the center of \( \mathrm{B} \) . Then the sequence of pointed sets below is exact:\n\n\[ 1 \rightarrow {\mathrm{H}}^{0}\left( {\mathrm{G},\mathrm{A}}\right) \xrightarrow[]{{i}_{0}}{\mathrm{H}}^{0}\left( {\mathrm{G},\mathrm{B}}\right) \xrightar... | The proof consists of a series of checks:\n\n1. Exactness at \( {\mathrm{H}}^{0}\left( {\mathrm{G},\mathrm{A}}\right) \) . Trivial.\n\n2. Exactness at \( {\mathrm{H}}^{0}\left( {\mathrm{G},\mathrm{B}}\right) \) . We have \( {p}_{0} \circ {i}_{0} = 1 \) by functoriality (where \ | No |
Proposition 1. If \( \mathrm{A} \) is relatively projective, \( {\widehat{\mathrm{H}}}_{0}\left( {\mathrm{G},\mathrm{A}}\right) = 0 = {\widehat{\mathrm{H}}}^{0}\left( {\mathrm{G},\mathrm{A}}\right) \) . | It suffices to prove it for \( \mathrm{A} \) induced. We give the proof for \( {\widehat{\mathrm{H}}}^{0} \) . By definition of \ | No |
Proposition 2. The restriction maps form a morphism of cohomological functors. | Let \( 0 \rightarrow \mathrm{A} \rightarrow \mathrm{B} \rightarrow \mathrm{C} \rightarrow 0 \) be an exact sequence of \( \mathrm{G} \) -modules. We must check the commutativity of the diagram\n\n\[\n\begin{array}{l} {\widehat{\mathrm{H}}}^{q}\left( {\mathrm{G},\mathrm{C}}\right) \overset{\delta }{ \rightarrow }{\wideh... | Yes |
Proposition 3. The corestriction maps form a morphism of cohomological functors. | More precisely, Cor is the unique morphism of cohomological functors that coincides in degree -1 with the map \( {}_{{\mathrm{N}}_{\mathrm{H}}}\mathrm{A}/{\mathrm{I}}_{\mathrm{H}}\mathrm{A} \rightarrow {}_{{\mathrm{N}}_{\mathrm{G}}}\mathrm{A}/{\mathrm{I}}_{\mathrm{G}}\mathrm{A} \) induced by the \( {\text{ inclusion }}... | Yes |
Proposition 4. If \( n = \operatorname{Card}\left( {\mathrm{G}/\mathrm{H}}\right) \), then \( \operatorname{Cor} \circ \operatorname{Res} = n \) . | DIRECT PROOF. If \( {f}_{n} \) denotes multiplication by \( n \), then the morphism of cohomological functors Cor \( \circ \) Res \( - {f}_{n} \) is zero in dimension 0 (trivial to check), hence is zero in all dimensions. | No |
Corollary 1. If \( g \) is the order of \( \mathrm{G} \), then all the groups \( {\widehat{\mathrm{H}}}^{q}\left( {\mathrm{G},\mathrm{A}}\right) \) are annihilated by \( g \) . | Apply prop. 4 with \( \mathrm{H} = \{ 1\} \), remarking that the \( {\widehat{\mathrm{H}}}^{q}\left( {\mathrm{H},\mathrm{A}}\right) \) are all zero. | No |
Corollary 2. If \( \mathrm{A} \) is a finitely generated abelian group (and is a G-module), then the \( {\widehat{\mathrm{H}}}^{q}\left( {\mathrm{G},\mathrm{A}}\right) \) are finite groups. | Indeed, the definition of these groups in terms of chains and cochains shows that they are finitely generated groups; by corollary 1 , they are torsion groups, hence finite. | Yes |
Proposition 5. If \( \mathrm{G} \) is a finite group, there exists one and only one family of homomorphisms (called cup product)\n\n\[ \n{\widehat{\mathrm{H}}}^{p}\left( {\mathrm{G},\mathrm{A}}\right) \otimes {\widehat{\mathrm{H}}}^{q}\left( {\mathrm{G},\mathrm{B}}\right) \rightarrow {\widehat{\mathrm{H}}}^{p + q}\left... | Properties iii) and iv) allow the use of \ | No |
Proposition 6. The cohomological functor \( \\left\\{ {{\\mathrm{H}}^{q}\\left( {\\mathrm{\\;K}\\left( \\;\\right) }\\right) ,\\delta }\\right\\} \) is isomorphic to the functor \( \\left\\{ {{\\mathrm{H}}^{q}\\left( {\\mathrm{G},\\;} \\right) ,\\delta }\\right\\} \) . | First of all, it is clear that \( {\\widehat{\\mathrm{H}}}^{0}\\left( {\\mathrm{G},\\mathrm{A}}\\right) = {\\mathrm{H}}^{0}\\left( {\\mathrm{K}\\left( \\mathrm{A}\\right) }\\right) ,{\\widehat{\\mathrm{H}}}^{-1}\\left( {\\mathrm{G},\\mathrm{A}}\\right) = {\\mathrm{H}}^{-1}\\left( {\\mathrm{K}\\left( \\mathrm{A}\\right)... | Yes |
Proposition 7. Given an exact sequence \( 0 \rightarrow \mathrm{A} \rightarrow \mathrm{B} \rightarrow \mathrm{C} \rightarrow 0 \) of \( \mathrm{G} \) -modules for which at least two of the three Herbrand quotients \( h\left( \mathrm{\;A}\right), h\left( \mathrm{\;B}\right), h\left( \mathrm{C}\right) \) are defined. The... | This follows from the exact hexagon written above. (More generally, when \( {2n} \) finite groups form an exact \( {2n} \) -gon, the alternating product of their orders is equal to 1.) | No |
Proposition 8. If \( \mathrm{A} \) is a finite \( \mathrm{G} \) -module, then \( h\left( \mathrm{\;A}\right) = 1 \) . | The exact sequence\n\n\[\n0 \rightarrow {\mathrm{A}}^{\mathrm{G}} \rightarrow \mathrm{A}\overset{\mathrm{D}}{ \rightarrow }\mathrm{A} \rightarrow {\mathrm{A}}_{\mathrm{G}} \rightarrow 0\n\]\n\nshows first of all that \( {\mathrm{A}}^{\mathrm{G}} \) and \( {\mathrm{A}}_{\mathrm{G}} \) have the same number of elements. T... | Yes |
Lemma 1. Let \( 0 \rightarrow {\mathrm{A}}^{\prime } \rightarrow \mathrm{A} \rightarrow {\mathrm{A}}^{\prime \prime } \rightarrow 0 \) be an exact sequence of \( \mathrm{G} \) -modules such that \( \varphi \left( {\mathrm{A}}^{\prime }\right) \) and \( \varphi \left( {\mathrm{A}}^{\prime \prime }\right) \) are defined.... | Clearly \( \varphi \left( \mathrm{A}\right) \) and \( h\left( \mathrm{\;A}\right) \) are defined and satisfy \( \varphi \left( \mathrm{A}\right) = \varphi \left( {\mathrm{A}}^{\prime }\right) .\varphi \left( {\mathrm{A}}^{\prime \prime }\right) \) and \( h\left( \mathrm{\;A}\right) = h\left( {\mathrm{\;A}}^{\prime }\ri... | Yes |
Lemma 2. Let \( \mathrm{A} \) be a \( \mathrm{G} \) -module for which \( \varphi \left( \mathrm{A}\right) \) is defined. Then there is an exact sequence of \( \mathrm{G} \) -modules\n\n\[ 0 \rightarrow {\mathrm{A}}^{\prime } \rightarrow \mathrm{A} \rightarrow {\mathrm{A}}^{\prime \prime } \rightarrow 0 \]\n\nsuch that ... | By hypothesis, A/pA is finite. Hence there exists a subgroup of A that is finitely generated and maps onto \( \mathrm{A}/p\mathrm{\;A} \) ; we may assume this subgroup to be stable under \( \mathrm{G} \) (otherwise replace it with the sum of its transforms by the elements of \( \mathrm{G} \) ). Call it \( {\mathrm{A}}^... | Yes |
Lemma 1. Suppose \( \mathrm{G} \) is a p-group acting on a finite set \( \mathrm{E} \), and let \( {\mathrm{E}}^{\mathrm{G}} \) be the subset of elements fixed by \( \mathrm{G} \). Then\n\n\[ \operatorname{Card}\left( {\mathrm{E}}^{\mathrm{G}}\right) \equiv \operatorname{Card}\left( \mathrm{E}\right) \;{\;\operatorname... | Indeed, \( \mathrm{E} - {\mathrm{E}}^{\mathrm{G}} \) is the disjoint union of orbits \( \mathrm{G}x \) not reduced to a single point, each having cardinality equal to the index of its stabilizer in \( G \), which is divisible by \( p \). | Yes |
Lemma 2. If a p-group acts on a p-group of order \( > 1 \), then the fixed points form a subgroup of order \( > 1 \). | Indeed, the number of fixed points is divisible by \( p \) (lemma 1). | No |
Theorem 1. The center of a p-group of order \( > 1 \) has order \( > 1 \) . | Apply the preceding lemma, letting the group act on itself by inner automorphisms. | No |
Theorem 2. Every linear representation \( \neq 0 \) of a p-group over a field of characteristic \( p \) contains the unit representation. | Let \( \mathrm{E} \) be the representation space. Let \( x \) be a non-zero element of \( \mathrm{E} \) , \( \mathrm{H} \) the subgroup of \( \mathrm{E} \) generated by the \( s.x, s \in \mathrm{G};\mathrm{H} \) is a finite dimensional vector space over the prime field \( {\mathbf{F}}_{p} \) . Applying lemma 2 to \( \m... | Yes |
Theorem 3 (Sylow). Let \( \mathrm{G} \) be a group of order \( n = {p}^{m}q \), with \( p \) prime and \( \left( {p, q}\right) = 1 \) . Then there exist subgroups of \( \mathrm{G} \) having order \( {p}^{m} \) (called Sylow \( p \) - subgroups); they are all conjugate to one another, and every p-group contained in \( \... | Proof (after G. A. Miller and H. Wielandt). Let \( \mathrm{E} \) be the family of all subsets \( \mathrm{X} \) of \( \mathrm{G} \) having \( {p}^{m} \) elements. The group \( \mathrm{G} \) operates on \( \mathrm{E} \) by translations, and\n\n\[ \operatorname{Card}\left( \mathrm{E}\right) = \left( \begin{matrix} n \\ {p... | No |
Lemma 3. If \( n = {p}^{m}q \), with \( \left( {p, q}\right) = 1 \), then\n\n\[ \left( \begin{matrix} n \\ {p}^{m} \end{matrix}\right) \equiv q\;{\;\operatorname{mod}\;.}p \] | Indeed, let \( \mathrm{X} \) and \( \mathrm{Y} \) be indeterminates over a field of characteristic \( p \) . Then\n\n\( {\left( \mathrm{X} + \mathrm{Y}\right) }^{n} = {\left( \mathrm{X} + \mathrm{Y}\right) }^{{p}^{m}q} = {\left( {\mathrm{X}}^{{p}^{m}} + {\mathrm{Y}}^{{p}^{m}}\right) }^{q} = {\mathrm{X}}^{{p}^{m}q} + q{... | Yes |
Theorem 4. Let \( \\mathrm{G} \) be a finite group, \( p \) a prime number, and \( {\\mathrm{G}}_{p} \) a Sylow \( p \) - subgroup of \( \\mathrm{G} \). Then for every \( \\mathrm{G} \) -module \( \\mathrm{A} \) and every \( n \\in \\mathbf{Z} \), the restriction homomorphism \[ \\operatorname{Res} : {\\widehat{\\mathr... | Given \( x \) in the kernel of Res. If \( q = \\operatorname{Card}\\left( {\\mathrm{G}/{\\mathrm{G}}_{p}}\\right) \), then \[ q.x = \\operatorname{Cor} \\circ \\operatorname{Res}\\left( x\\right) = 0\\;\\text{ (Chap. VIII, prop. 4). } \] But if \( x \) belongs to the \( p \) -primary component of \( {\\widehat{\\mathrm... | Yes |
Proposition 1. If \( \mathrm{A} \) is induced (resp. relatively projective), then \( \mathrm{T}\left( {\mathrm{A},\mathrm{B}}\right) \) is induced (resp. relatively projective), hence cohomologically trivial. | We may assume that \( \mathrm{A} \) is induced (passing to a direct factor otherwise); then \( \mathrm{A} \) is the direct sum of the \( s.{\mathrm{A}}^{\prime } \) for some subgroup \( {\mathrm{A}}^{\prime } \) . The group \( \mathrm{T}\left( {\mathrm{A},\mathrm{B}}\right) \) is then the direct sum of the \( \mathrm{T... | Yes |
Lemma 4. Let \( \mathrm{G} \) be a p-group and \( \mathrm{A} \) a \( \mathrm{G} \) -module such that \( p\mathrm{\;A} = 0 \) . Then the three following conditions are equivalent:\ni) \( \mathrm{A} = 0 \) .\nii) \( {\mathrm{H}}^{0}\left( {\mathrm{G},\mathrm{A}}\right) = 0 \) .\niii) \( {\mathrm{H}}_{0}\left( {\mathrm{G}... | The implications i) \( \Rightarrow \) ii) and i) \( \Rightarrow \) iii) are trivial. The implication ii) \( \Rightarrow \) i) has been proved in theorem 2. Let us show that iii) \( \Rightarrow \) i). Let \( {\mathrm{A}}^{\prime } = \operatorname{Hom}\left( {\mathrm{A},{\mathbf{F}}_{p}}\right) \) be the dual of \( \math... | Yes |
Lemma 5. With the hypotheses of lemma 4, suppose that \( {\mathrm{H}}_{1}\left( {\mathrm{G},\mathrm{A}}\right) = 0 \) . Then \( \mathrm{A} \) is a free module over the algebra \( \Lambda = {\mathbf{F}}_{p}\left\lbrack \mathrm{G}\right\rbrack \) . | Let \( r \) be the augmentation ideal of \( \Lambda \) . Then \( A/{rA} = {H}_{0}\left( {G, A}\right) \), and this is a vector space over \( {\mathbf{F}}_{p} \) . Let \( {h}_{\lambda } \) be a basis of this vector space, and lift it to a family \( {a}_{\lambda } \in \mathrm{A} \) . Since the \( {h}_{\lambda } \) genera... | Yes |
Theorem 5. Let \( \mathrm{G} \) be a p-group and \( \mathrm{A} \) a \( \mathrm{G} \) -module annihilated by \( p \) . The following conditions are equivalent:\n\ni) There exists an integer \( q \) such that \( {\widehat{\mathrm{H}}}^{q}\left( {\mathrm{G},\mathrm{A}}\right) = 0 \),\n\nii) A is cohomologically trivial,\n... | Obviously it suffices to prove i) \( \Rightarrow \) iv). The shifting procedure already used several times enables us to construct a G-module B, annihilated by \( p \) ,\n\nsuch that \( {\widehat{\mathrm{H}}}^{n}\left( {\mathrm{G},\mathrm{A}}\right) = {\widehat{\mathrm{H}}}^{n - q - 2}\left( {\mathrm{G},\mathrm{B}}\rig... | No |
Theorem 6. Let \( \mathrm{G} \) be a p-group and let \( \mathrm{A} \) be a \( \mathrm{G} \) -module without p-torsion. The following conditions are equivalent:\n\ni) \( {\widehat{\mathrm{H}}}^{q}\left( {\mathrm{G},\mathrm{A}}\right) = 0 \) for two consecutive values of \( q \) ,\n\nii) A is cohomologically trivial,\n\n... | Since A has no \( p \) -torsion, there is an exact sequence\n\n\[ 0 \rightarrow \mathrm{A}\overset{p}{ \rightarrow }\mathrm{\;A} \rightarrow \mathrm{A}/p\mathrm{\;A} \rightarrow 0. \]\n\nPassing to the cohomology gives the exact sequence\n\n\[ {\widehat{\mathrm{H}}}^{q}\left( {\mathrm{G},\mathrm{A}}\right) \overset{p}{... | Yes |
Theorem 7. Let \( \mathrm{G} \) be a finite group, \( \mathrm{A} \) a \( \mathbf{Z} \) -free \( \mathrm{G} \) -module, and \( {\mathrm{G}}_{p} \) a Sylow p-subgroup of \( \mathbf{G} \), for each \( p \) . The following conditions are equivalent:\n\ni) For every prime number \( p \), the \( {\mathrm{G}}_{p} \) -module \... | We must show that i) implies ii). Write A as a quotient of a free \( \mathbf{Z}\left\lbrack \mathrm{G}\right\rbrack \) - module \( \mathrm{L} \):\n\n\[ 0 \rightarrow \mathrm{N} \rightarrow \mathrm{L} \rightarrow \mathrm{A} \rightarrow 0. \]\n\nThe \( \mathbf{Z} \) -module \( \mathrm{A} \) being free yields the exact se... | Yes |
Lemma 6. Let \( 0 \rightarrow {\mathrm{X}}_{1} \rightarrow {\mathrm{X}}_{2} \rightarrow \cdots \rightarrow {\mathrm{X}}_{n} \rightarrow 0 \) be an exact sequence of \( \mathrm{G} \) - modules. If all but one of the \( {\mathrm{X}}_{i} \) are cohomologically trivial, then that one also is. | Put \( {\mathrm{N}}_{i} = \operatorname{Ker}\left( {{\mathrm{X}}_{i} \rightarrow {\mathrm{X}}_{i + 1}}\right) ,{\mathrm{N}}_{0} = {\mathrm{N}}_{n + 1} = 0 \) . Then there are \( n + 1 \) exact sequences \( \left( {\mathrm{E}}_{i}\right) \) \[ 0 \rightarrow {\mathrm{N}}_{i} \rightarrow {\mathrm{X}}_{i} \rightarrow {\mat... | Yes |
Theorem 8. Let \( \mathrm{A} \) be any \( \mathrm{G} \) -module. The following are equivalent:\ni) For every prime \( p,{\widehat{\mathrm{H}}}^{q}\left( {{\mathrm{G}}_{p},\mathrm{\;A}}\right) = 0 \) for two consecutive values of \( q \) (that may depend on \( p \) ),\nii) A is cohomologically trivial,\niii) There exist... | We have iv) \( \Rightarrow \) iii) trivially, iii) \( \Rightarrow \) ii) by lemma 6, and ii) \( \Rightarrow \) i) trivially. Let us show i) \( \Rightarrow \) iv): Let \( 0 \rightarrow \mathrm{R} \rightarrow \mathrm{L} \rightarrow \mathrm{A} \rightarrow 0 \) be an exact sequence of \( \mathrm{G} \) -modules, with \( \ma... | Yes |
Theorem 9. Let A, B be G-modules, with A cohomologically trivial. In order that \( \mathrm{A} \otimes \mathrm{B} \) (resp. \( \operatorname{Hom}\left( {\mathrm{A},\mathrm{B}}\right) \), resp. \( \operatorname{Hom}\left( {\mathrm{B},\mathrm{A}}\right) \) ) be cohomologically trivial, it is necessary and sufficient that ... | By th. 8, iv), A has a resolution by projective modules\n\n\[ 0 \rightarrow {\mathrm{P}}_{1} \rightarrow {\mathrm{P}}_{0} \rightarrow \mathrm{A} \rightarrow 0. \]\n\nHence there is an exact sequence\n\n\[ 0 \rightarrow \operatorname{Tor}\left( {\mathrm{A},\mathrm{B}}\right) \rightarrow {\mathrm{P}}_{1} \otimes \mathrm{... | Yes |
Lemma 7. Let \( \mathrm{G} \) be a finite group, and let \( \mathrm{A} \) be an injective \( \mathbf{Z}\left\lbrack \mathrm{G}\right\rbrack \) -module. Then \( \mathrm{A} \) is \( \mathbf{Z} \) -injective, i.e., divisible. | We must show that the functor \( {\operatorname{Hom}}_{\mathbf{Z}}\left( {\mathrm{C},\mathrm{A}}\right) \) is exact in \( \mathrm{C} \) . If \( \Lambda = \mathbf{Z}\left\lbrack \mathrm{G}\right\rbrack \) , there is a functorial isomorphism\n\n\[ \n{\operatorname{Hom}}_{\mathbf{Z}}\left( {\mathrm{C},\mathrm{A}}\right) =... | Yes |
Theorem 10 (Dual to Theorem 7). Let \( \mathrm{G} \) be a finite group, and let \( \mathrm{A} \) be a \( \mathbf{Z} \) - injective G-module. In order that A be cohomologically trivial, it is necessary and sufficient that \( \mathrm{A} \) be \( \mathbf{Z}\left\lbrack \mathrm{G}\right\rbrack \) -injective. | Sufficiency is trivial. To see the necessity, embed A in a Z[G]-injective module I, obtaining an exact sequence\n\n\[ 0 \rightarrow \mathrm{A} \rightarrow \mathrm{I} \rightarrow \mathrm{R} \rightarrow 0 \]\n\nSince \( \mathrm{A} \) is \( \mathbf{Z} \) -injective, this yields the exact sequence\n\n\[ 0 \rightarrow {\ope... | Yes |
Theorem 11 (Dual to Theorem 8). In order that a G-module A be cohomologically trivial, it is necessary and sufficient that there be an exact sequence \( 0 \rightarrow \mathrm{A} \rightarrow {\mathrm{I}}_{0} \rightarrow {\mathrm{I}}_{1} \rightarrow 0 \), where the \( {\mathrm{I}}_{i} \) are injective \( \mathbf{Z}\left\... | As before, there is an exact sequence\n\n\[ 0 \rightarrow \mathrm{A} \rightarrow {\mathrm{I}}_{0} \rightarrow \mathrm{R} \rightarrow 0 \]\n\nwith \( {\mathrm{I}}_{0}\mathbf{Z}\left\lbrack \mathrm{G}\right\rbrack \) -injective. Since \( \mathrm{A} \) is cohomologically trivial, so is \( \mathrm{R} \) ; on the other hand... | No |
Theorem 13. Let \( \mathrm{G} \) be a finite group, \( \mathrm{A},\mathrm{B},\mathrm{C} \) three \( \mathrm{G} \) -modules, and let \( \varphi : \mathrm{A} \times \mathrm{B} \rightarrow \mathrm{C} \) be a G-invariant bilinear map. Let \( q \in \mathbf{Z}, a \in {\widehat{H}}^{q}\left( {\mathrm{G},\mathrm{A}}\right) \) ... | We first treat the case \( q = 0 \) . The class \( a \in {\widehat{\mathrm{H}}}^{0}\left( {\mathrm{G},\mathrm{A}}\right) \) can be represented by an element \( a \in {\mathrm{A}}^{G} \) . Putting \( f\left( b\right) = \varphi \left( {a, b}\right) \), we obtain a homomorphism of G-modules\n\n\[ f : \mathbf{B} \rightarro... | Yes |
Theorem 14. Let \( \mathrm{G} \) be a finite group, \( \mathrm{A} \) a \( \mathrm{G} \) -module, and \( a \in {\mathrm{H}}^{2}\left( {\mathrm{G},\mathrm{A}}\right) \) . Let \( {\mathrm{G}}_{p} \) be a Sylow p-subgroup of \( \mathrm{G} \), for each prime \( p \), and suppose that\n\n1) \( {\mathrm{H}}^{1}\left( {{\mathr... | Apply th. 13 with \( \mathrm{B} = \mathbf{Z},\mathrm{C} = \mathrm{A}, q = 2,\varphi : \mathrm{A} \times \mathbf{Z} \rightarrow \mathrm{A} \) being the obvious map. Take \( {n}_{p} = - 1 \) . For \( n = - 1 \), hypothesis 1) shows that the cup product is surjective; for \( n = 0 \), hypothesis 2) shows that it is biject... | Yes |
Proposition 1. For every integer \( n,{\widehat{\mathrm{H}}}^{n}\left( {\mathrm{G},\mathrm{K}}\right) = 0 \) . | Indeed, the normal basis theorem (Bourbaki, Alg., Chap. V, §10) shows that \( \mathrm{K} \) is an induced module, and we know that the cohomology of such a module is trivial.\n\n[If one wishes to avoid the normal basis theorem, one can simply remark that \( \mathrm{K} \) contains an element of trace 1, which implies th... | Yes |
Proposition 2. \( {\mathrm{H}}^{1}\left( {\mathrm{G},{\mathrm{K}}^{ * }}\right) = 0 \) . | Let \( s \mapsto {a}_{s} \) be a 1-cocycle. If \( c \in \mathrm{K} \), form the \ | No |
Proposition 3. \( {\mathrm{H}}^{1}\left( {\mathrm{G},\mathrm{{GL}}\left( {n,\mathrm{\;K}}\right) }\right) = \{ 1\} \) . | The proof is analogous to that of prop. 2. Let \( {a}_{s} \) be a 1-cocycle, \( c \in {\mathbf{M}}_{n}\left( \mathbf{K}\right) \) any matrix. Again form the Poincaré series\n\n\[ b = \mathop{\sum }\limits_{{s \in \mathrm{G}}}{a}_{s} \cdot s\left( c\right) \]\n\nand check that \( s\left( b\right) = {a}_{s}^{-1}.b \) ; t... | Yes |
Proposition 4. The map \( \theta \) just defined is bijective. | We first show \( \theta \) is injective. Let \( \left( {{\mathrm{V}}_{1}^{\prime },{x}_{1}^{\prime }}\right) \) and \( \left( {{\mathrm{V}}_{2}^{\prime },{x}_{2}^{\prime }}\right) \) correspond to the same cocycle \( {p}_{s} \), and let \( {f}_{1},{f}_{2} \) be the corresponding K-isomorphisms. Then \( {f}_{1}^{-1} \ci... | Yes |
Corollary 1. The set \( {\mathrm{H}}^{1}\left( {\mathrm{G},{\mathbf{O}}_{\mathbf{K}}\left( \Phi \right) }\right) \) is in bijective correspondence with the set of classes of quadratic \( k \) -forms that are \( \mathrm{K} \) -isomorphic to \( \Phi \) . | This interpretation of \( {\mathrm{H}}^{1}\left( {\mathrm{G},{\mathbf{O}}_{\mathrm{K}}\left( \Phi \right) }\right) \) allows the construction of examples where this set is non-trivial (e.g., \( k = \mathbf{R},\mathrm{K} = \mathbf{C} \) ). | No |
Proposition 5. The homomorphisms \( {f}_{q} \) are independent of the choice of \( f \) . | Indeed, two such choices differ by an element of \( G \), and prop. 3 of Chap. VII. \( §5 \) can be applied to this element. | No |
Proposition 6. Let \( \mathrm{L}/k \) be a Galois extension containing the Galois extension \( \mathrm{K}/k \) . Then there is an exact sequence\n\n\[\n0 \rightarrow {\mathrm{H}}^{2}\left( {\mathrm{\;K}/k}\right) \rightarrow {\mathrm{H}}^{2}\left( {\mathrm{\;L}/k}\right) \rightarrow {\mathrm{H}}^{2}\left( {\mathrm{\;L}... | Let \( \mathrm{G} = \mathrm{G}\left( {\mathrm{L}/k}\right) \), and let \( \mathrm{H} = \mathrm{G}\left( {\mathrm{L}/\mathrm{K}}\right) \) . Since \( {\mathrm{H}}^{1}\left( {\mathrm{H},{\mathrm{L}}^{ * }}\right) = 0 \), we can apply prop. 5 of Chap. VII, \( §6 \), with \( q = 2 \) ; we get the exact sequence\n\n\[\n0 \r... | Yes |
Proposition 7. Let \( k \) be a field and \( \mathrm{A} \) a finite dimensional \( k \) -algebra. The following conditions are equivalent:\n\na) A has no non-trivial two-sided ideal, and its center is \( k \) .\n\nb) If \( \mathrm{K} \) is the algebraic closure of \( k \), then the extended algebra is isomorphic to a m... | For the proof, see Bourbaki, Alg., Chap. VIII, §§5, 10. | No |
Proposition 8. There is a canonical bijection\n\n\\[ \n\\theta : \\mathrm{A}\\left( {n,\\mathrm{\\;K}/k}\\right) \\rightarrow {\\mathrm{H}}^{1}\\left( {\\mathrm{G},\\mathbf{{PGL}}\\left( {n,\\mathrm{\\;K}}\\right) }\\right) .\n\\] | On the other hand, the exact sequence (*) defines (cf. Chap. VII, appendix) a “coboundary” operator\n\n\\[ \n{\\Delta }_{n} : {\\mathrm{H}}^{1}\\left( {\\mathrm{G},\\mathbf{{PGL}}\\left( {n,\\mathrm{\\;K}}\\right) }\\right) \\rightarrow {\\mathrm{H}}^{2}\\left( {\\mathrm{G},{\\mathrm{K}}^{ * }}\\right) .\n\\]\n\nCompos... | Yes |
Proposition 9. The homomorphism \( \delta : \mathrm{A}\left( {\mathrm{K}/k}\right) \rightarrow {\mathrm{H}}^{2}\left( {\mathrm{\;K}/k}\right) \) is bijective. | We have already seen that it is injective. The next lemma shows that it is surjective. | No |
Lemma 1. If \( n = \left\lbrack {\mathrm{\;K} : k}\right\rbrack \), then the map \( {\delta }_{n} : \mathrm{A}\left( {n,\mathrm{\;K}/k}\right) \rightarrow {\mathrm{H}}^{2}\left( {\mathrm{\;K}/k}\right) \) is surjective. | (Compare with Bourbaki, Alg., Chap. VIII, §10, prop. 7.)\n\nBy prop. 8 and the definition of \( {\delta }_{n} \), it suffices to show that \( {\Delta }_{n} \) is surjective, i.e., that every 2-cocycle \( {a}_{s, t} \) with values in \( {\mathrm{K}}^{ * } \) can be written\n\n\[ {a}_{s, t} = {p}_{s}s\left( {p}_{t}\right... | Yes |
Proposition 10. If the field \( k \) has property \( {\mathrm{C}}_{1} \), then the Brauer group of every finite extension \( \mathrm{K} \) of \( k \) is zero. | Indeed, let \( \mathrm{D} \) be a division algebra with center \( \mathrm{K} \) and of degree \( {r}^{2} \) over \( \mathrm{K} \) . Let \( s = \left\lbrack {\mathrm{\;K} : k}\right\rbrack \) . If \( x \in \mathrm{D} \), let \( \operatorname{Nrd}\left( x\right) \in \mathrm{K} \) be its reduced norm (Bourbaki, Alg., Chap... | Yes |
For a given field \( k \), the following are equivalent:\n\n1) The Brauer group of every finite separable extension of \( k \) is zero.\n\n2) If \( \mathrm{K} \) is finite separable over \( k \), and \( \mathrm{L}/\mathrm{K} \) is a finite Galois extension, then the \( \mathrm{G}\left( {\mathrm{L}/\mathrm{K}}\right) \)... | Suppose 2) holds, then \( {\widehat{\mathrm{H}}}^{0}\left( {\mathrm{G}\left( {\mathrm{L}/\mathrm{K}}\right) ,{\mathrm{L}}^{ * }}\right) = 0 \), whence 3) holds, and\n\n\[{\mathrm{H}}^{2}\left( {\mathrm{G}\left( {\mathrm{L}/\mathrm{K}}\right) ,{\mathrm{L}}^{ * }}\right) = 0\]\n\nwhence 1) holds by passage to the limit o... | Yes |
For every Galois extension \( \mathrm{F}/\mathrm{E} \) and every integer \( n \), the homomorphism\n\n\[ \n{\theta }^{n}\left( {\mathrm{\;F}/\mathrm{E}}\right) : {\widehat{\mathrm{H}}}^{n}\left( {{\mathrm{G}}_{\mathrm{F}/\mathrm{E}},\mathbf{Z}}\right) \rightarrow {\widehat{\mathrm{H}}}^{n + 2}\left( {{\mathrm{G}}_{\mat... | If \( \mathrm{F} \supset {\mathrm{E}}^{\prime } \supset \mathrm{E} \), with \( \mathrm{F}/\mathrm{E} \) Galois, the homomorphisms \( {\theta }^{n} \) commute with restriction and corestriction: that follows from the above formulas for the \( {u}_{\mathrm{F}/\mathrm{E}} \) and known formulas for the cup product. For exa... | Yes |
Proposition 2. Let \( \chi \in \operatorname{Hom}\left( {{\mathrm{G}}_{\mathrm{F}/\mathrm{E}},\mathbf{Q}/\mathbf{Z}}\right) \) be a character of degree 1 of the group \( {\mathrm{G}}_{\mathrm{F}/\mathrm{E}} \) (or, what amounts to the same thing, of the group \( {\mathrm{G}}_{\mathrm{F}/\mathrm{E}}^{a} \) ). For any \(... | To simplify the notation, we set \( {s}_{a} = \left( {a,\mathrm{\;F}/\mathrm{E}}\right) \) ; if \( s \in {\mathrm{G}}_{\mathrm{F}/\mathrm{E}} \), we denote by \( \bar{s} \) the canonical image of \( s \) in \( {\widehat{\mathrm{H}}}^{-2}\left( {{\mathrm{G}}_{\mathrm{F}/\mathrm{E}},\mathbf{Z}}\right) = {\mathrm{G}}_{\ma... | Yes |
Proposition 3. Let \( {\mathrm{E}}^{\prime }/\mathrm{E} \) be any extension, and let \( {\mathrm{E}}^{\prime \prime } \) be the largest abelian extension of \( \mathrm{E} \) contained in \( {\mathrm{E}}^{\prime } \) . Then \[ {\mathrm{N}}_{{\mathrm{E}}^{\prime }/\mathrm{E}}\left( {\mathrm{A}}_{{\mathrm{E}}^{\prime }}\r... | Let \( \mathrm{F} \) be a Galois extension of \( \mathrm{E} \) containing \( {\mathrm{E}}^{\prime } \), and let \( \mathrm{G} = {\mathrm{G}}_{\mathrm{F}/\mathrm{E}} \) be its Galois group. Let \( \mathrm{H} = {\mathrm{G}}_{\mathrm{F}/{\mathrm{E}}^{\prime }} \) . Then the Galois group of \( \mathrm{F}/{\mathrm{E}}^{\pri... | Yes |
Proposition 4. The map \( \mathrm{F} \rightarrow {\mathrm{I}}_{\mathrm{F}} \) is a bijection of the set of abelian extensions of \( \mathrm{E} \) onto the set of norm groups of \( {\mathrm{A}}_{\mathrm{E}} \) ; this correspondence reverses inclusion and satisfies\n\n\[{\mathrm{I}}_{\mathrm{F}.{\mathrm{F}}^{\prime }} = ... | If \( \mathrm{F} \) and \( {\mathrm{F}}^{\prime } \) are abelian extensions, so is \( \mathrm{F}.{\mathrm{F}}^{\prime } \), and \( {\mathrm{I}}_{\mathrm{F}.{\mathrm{F}}^{\prime }} \subset {\mathrm{I}}_{\mathrm{F}} \cap {\mathrm{I}}_{{\mathrm{F}}^{\prime }} \) ; conversely, if \( a \in {\mathrm{I}}_{\mathrm{F}} \cap {\m... | Yes |
Proposition 5. For every extension \( \mathrm{F}/\mathrm{E} \), we have \( {\mathrm{N}}_{\mathrm{F}/\mathrm{E}}{\mathrm{D}}_{\mathrm{F}} = {\mathrm{D}}_{\mathrm{E}} \) . | The inclusion \( {\mathrm{N}}_{\mathrm{F}/\mathrm{E}}{\mathrm{D}}_{\mathrm{F}} \subset {\mathrm{D}}_{\mathrm{E}} \) follows from the transitivity of the norms. Conversely, let \( a \in {\mathbf{D}}_{\mathrm{E}} \), and let \( {\mathrm{F}}^{\prime } \) be an extension of \( \mathrm{F} \) ; denote by \( \mathrm{K}\left( ... | Yes |
Proposition 6. For every field \( \mathrm{E} \), the group \( {\mathrm{D}}_{\mathrm{E}} \) is divisible and equal to \( \bigcap n.{\mathrm{A}}_{\mathrm{E}} \) . | First we show that for every \( \mathrm{E} \) and every prime \( p,{\mathrm{D}}_{\mathrm{E}} = p.{\mathrm{D}}_{\mathrm{E}} \) . Let \( a \in {\mathrm{D}}_{\mathrm{E}} \) and let \( \mathrm{F} \) be an extension of \( \mathrm{E} \) containing \( {\mathrm{E}}_{p} \) (i.e.,\ | No |
Theorem 2. Suppose axioms III-1, III-2, III-3 hold. For a subgroup of \( {\mathrm{A}}_{\mathrm{E}} \) to be a norm group, it is necessary and sufficient that it be closed and of finite index. | We know that these conditions are necessary. Let I be a subgroup satisfying them; if \( \left( {{\mathrm{A}}_{\mathrm{E}} : \mathrm{I}}\right) = n \), then \( n.{\mathrm{A}}_{\mathrm{E}} \subset \mathrm{I} \), whence \( {\mathrm{D}}_{\mathrm{E}} \subset \mathrm{I} \) by prop. 6 . If \( \mathrm{N} \) runs through the fa... | Yes |
Given G-modules A, B and \( a \in {\mathrm{A}}^{\mathrm{G}} \), let \( {f}_{a} : \mathbf{Z} \rightarrow \mathrm{A} \) be the G-homomorphism such that \( {f}_{a}\left( 1\right) = a \) . Let \( x \in {\widehat{\mathrm{H}}}^{n}\left( {\mathrm{G},\mathrm{B}}\right) \) . Then the cup product \[ {\bar{a}}^{0}.x \in {\widehat... | We will give the proof for \( n \geq 0 \), for example. When \( n = 0 \), the very definition of cup product gives \( {\bar{a}}^{0}.x = \left( {{f}_{a} \otimes 1}\right) \left( x\right) \) . Use induction on \( n \) : we know that there is an exact sequence of G-modules \[ 0 \rightarrow \mathrm{B} \rightarrow {\mathrm{... | Yes |
Lemma 2. Given \( a \in \mathrm{A} \) with \( \mathrm{N}a = 0 \), and let \( f \) be a 1-cocycle of \( \mathrm{G} \) with values in \( \mathrm{B},\bar{f} \in {\mathrm{H}}^{1}\left( {\mathrm{G},\mathrm{B}}\right) \) its cohomology class. Then\n\n\[ \n{\bar{a}}_{0} \cdot \bar{f} = {\bar{c}}^{0}\;\text{ in }{\widehat{\mat... | Consider an exact sequence\n\n\[ \n0 \rightarrow \mathrm{B} \rightarrow {\mathrm{B}}^{\prime } \rightarrow {\mathrm{B}}^{\prime \prime } \rightarrow 0\n\]\n\nas above. Since \( {\mathrm{H}}^{1}\left( {\mathrm{G},{\mathrm{B}}^{\prime }}\right) = 0 \), there exists \( {b}^{\prime } \in {\mathrm{B}}^{\prime } \) such that... | Yes |
Lemma 3. Let \( \\mathrm{B} \) be a \( \\mathrm{G} \) -module and \( f : \\mathrm{G} \\rightarrow \\mathrm{B} \) a 1-cocycle, \( \\bar{f} \\in {\\mathrm{H}}^{1}\\left( {\\mathrm{G},\\mathrm{B}}\\right) \) its cohomology class. Then for every \( s \\in \\mathbf{G} \) ,\n\n\\[ \n\\bar{s} \\cdot \\bar{f} = {\\overline{f\\... | The homomorphism \( d : {\\widehat{\\mathbf{H}}}^{-1}\\left( {\\mathbf{G},\\mathbf{B}}\\right) \\rightarrow {\\widehat{\\mathbf{H}}}^{0}\\left( {\\mathbf{G},\\mathbf{I} \\otimes \\mathbf{B}}\\right) \) is an isomorphism. Thus it will suffice to show that the images under \( d \) of \( \\bar{s}.\\bar{f} \) and \( \\bar{... | Yes |
Lemma 4. Let \( \mathrm{B} \) be a \( \mathrm{G} \) -module, \( u : \mathrm{G} \times \mathrm{G} \rightarrow \mathrm{B} \) a 2-cocycle, \( \bar{u} \in {\mathrm{H}}^{2}\left( {\mathrm{G},\mathrm{B}}\right) \) its cohomology class. Then for all \( s \in \mathbf{G} \) , \[ \bar{s} \cdot \bar{u} = {\bar{a}}^{0},\;\text{ wi... | Once again introduce an exact sequence \( 0 \rightarrow \mathbf{B} \rightarrow {\mathbf{B}}^{\prime } \rightarrow {\mathbf{B}}^{\prime \prime } \rightarrow 0 \) of \( \mathrm{G} \) - modules, with \( {\mathbf{B}}^{\prime } \) induced. Since \( {\mathbf{H}}^{2}\left( {\mathbf{G},{\mathbf{B}}^{\prime }}\right) = 0 \), th... | Yes |
Theorem 1. Suppose that the residue field \( \overline{\mathrm{K}} \) is perfect. Then each element of the Brauer group \( {\mathbf{B}}_{\mathbf{K}} \) of \( \mathbf{K} \) is split by a finite unramified extension of \( \mathbf{K} \) . | Let \( {\mathrm{K}}_{nr} \) be the maximal unramified extension of \( \mathrm{K} \) . We know that its Brauer group is zero (Chap. X, \( §7 \), example b)). If \( a \in {\mathbf{B}}_{\mathbf{K}} \), the image of \( a \) in \( {\mathrm{B}}_{{\mathrm{K}}_{nr}} \) is zero; as \( {\mathrm{K}}_{nr} \) is the directed union ... | Yes |
Lemma 1. Suppose \( \overline{\mathrm{K}} \) is perfect and \( n \geq 2 \) . Then there is a commutative subfield \( \mathrm{L} \) of \( \mathrm{D} \), containing \( \mathrm{K} \), that is unramified over \( \mathrm{K} \) and distinct from \( \mathrm{K} \) . | Suppose that such a field did not exist. Then for every commutative extension \( \mathrm{L} \) of \( \mathrm{K} \) within \( \mathrm{D} \), the residue field \( \mathrm{L} \) would be equal to \( \overline{\mathrm{K}} \) (otherwise, by cor. 3 to th. 3, Chap. III, §5, L would contain an unramified extension distinct fro... | No |
Proposition 2. Suppose \( \overline{\mathrm{K}} \) is perfect. Then there is a maximal subfield of \( \mathrm{D} \) that is unramified over \( \mathrm{K} \) . | Argue by induction on \( n \), the case \( n = 1 \) being trivial. For \( n \geq 2 \), lemma 1 furnishes an unramified extension \( {\mathrm{K}}^{\prime } \) of \( \mathrm{K} \) within \( \mathrm{D} \) . Let \( {\mathrm{D}}^{\prime } \) be the centralizer of \( {\mathrm{K}}^{\prime } \) in D. Then \( {\mathrm{D}}^{\pri... | Yes |
Proposition 3. The group \( {\mathrm{B}}_{\mathrm{K}} \) is the union of the subgroups \( {\mathrm{H}}^{2}\left( {\mathrm{\;L}/\mathrm{K}}\right) \), as \( \mathrm{L} \) runs through the family of finite unramified Galois extensions of \( \mathbf{K} \) . | Indeed, these extensions correspond to the subextensions \( \mathrm{L} \) of \( {\overline{\mathrm{K}}}_{nr} \) that are finite and Galois over \( \overline{\mathrm{K}} \), and the union of these \( \overline{\mathrm{L}} \) is \( {\overline{\mathrm{K}}}_{nr} \) ; the union of the \( \mathrm{L} \) is therefore \( {\math... | No |
Lemma 2. For all \( q \geq 1,{\mathrm{H}}^{q}\left( {\mathrm{\;g},{\mathrm{U}}_{\mathrm{L}}^{1}}\right) = 0 \) . | Filter \( {\mathrm{U}}_{\mathrm{L}}^{1} \) by the \( {\mathrm{U}}_{\mathrm{L}}^{n} \) ; the quotients \( {\mathrm{U}}_{\mathrm{L}}^{n}/{\mathrm{U}}_{\mathrm{L}}^{n + 1} \) are isomorphic as g-modules to the additive group \( \mathrm{L} \) ; therefore they have trivial cohomology (Chap. X, \( §1 \), prop. 1), and lemma ... | Yes |
Lemma 3. Let \( \mathfrak{g} \) be a finite group, \( \mathrm{M} \) a \( \mathfrak{g} \) -module filtered by a decreasing sequence \( {\mathrm{M}}_{n}, n \geq 1 \), of submodules, with \( {\mathrm{M}}_{1} = \mathrm{M} \) . Suppose that \( \mathrm{M} \) is complete Hausdorff in the topology defined by the \( {\mathbf{M}... | Let \( \varphi \left( {{g}_{1},\ldots ,{g}_{q}}\right) \) be a \( q \) -cocycle of \( g \) with values in \( \mathbf{M} \) . Since\n\n\[{\mathrm{H}}^{q}\left( {\mathrm{\;g},{\mathrm{M}}_{1}/{\mathrm{M}}_{2}}\right) = 0\]\n\nthere is a \( \left( {q - 1}\right) \) -cochain \( {\psi }_{1} \) of \( g \) with values in \( {... | Yes |
Proposition 1. Let \( {\mathrm{A}}^{\prime } \) be the subgroup of \( \mathrm{A} \) consisting of those \( a \in \mathrm{A} \) for which there is a positive integer \( n \) satisfying \( \left( {1 + \mathrm{F} + \cdots + {\mathrm{F}}^{n - 1}}\right) a = 0 \) . Then \[ {\mathrm{H}}^{1}\left( {\mathrm{\;g},\mathrm{\;A}}\... | (The isomorphism is obtained by assigning to each 1-cocyle \( \varphi : \mathrm{g} \rightarrow \mathrm{A} \) the coset of \( \varphi \left( 1\right) \) in \( {\mathrm{A}}^{\prime }/\left( {\mathrm{F} - 1}\right) \mathrm{A} \) .) The proposition follows from passage to the limit, using formula (*) and the determination ... | No |
Proposition 2. If \( \mathrm{A} \) is either a divisible group or a torsion group, then \( {\mathrm{H}}^{2}\left( {\mathrm{\;g},\mathrm{\;A}}\right) = 0 \) . | Suppose first \( \mathrm{A} \) is finite. Then \( {\mathrm{H}}^{2}\left( {\mathrm{\;g}/{\mathrm{g}}_{n},{\mathrm{\;A}}^{\mathrm{g}n}}\right) = {\mathrm{A}}^{\mathrm{g}}/{\mathrm{N}}_{n}{\mathrm{A}}^{\mathrm{g}n} \), with\n\n\[{\mathrm{N}}_{n} = 1 + \mathrm{F} + \cdots + {\mathrm{F}}^{n - 1}.\n\]\n\nLet \( m \) be a pos... | Yes |
Proposition 4. Let \( k \) be a quasi-finite field, \( \mathrm{F} \) the free generator of its Galois group \( \mathrm{g} = \mathrm{G}\left( {{k}_{s}/k}\right) \). a) If \( w \in {K}_{s}^{ * } \) is a root of unity, there exists \( y \in {k}_{s}^{ * } \) such that \( w = {y}^{\mathrm{F} - 1} \) (i.e., \( w = \mathrm{F}... | We know \( {\mathrm{H}}^{1}\left( {\mathrm{g},{k}_{s}^{ * }}\right) = 0 \) (Chap. X, prop. 2); if we set \( \mathrm{A} = {k}_{s}^{ * } \), then \( {\mathrm{A}}^{\prime } = \) \( \left( {\mathrm{F} - 1}\right) \mathrm{A} \) by prop. 1; since \( w \) belongs to the torsion subgroup of \( \mathrm{A} \), we have \( w \in {... | Yes |
Proposition 5. The Brauer group of a quasi-finite field is zero. | Again put \( \mathfrak{g} = \mathrm{G}\left( {{k}_{s}/k}\right) \) . The \( \mathfrak{g} \) -module \( {k}_{s}^{ * } \) is divisible (since \( {k}_{s} \) is algebraically closed). By prop. 2, \( {\mathrm{H}}^{2}\left( {\mathrm{\;g},{k}_{s}^{ * }}\right) = 0 \) . | No |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.