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For \( f, g \in {\mathcal{F}}_{\varpi }, u \in {\mathcal{O}}_{K}^{ \times } \), then the map \( {\Lambda }_{f} \rightarrow {\Lambda }_{g}, a \mapsto {\left\lbrack u\right\rbrack }_{g, f}\left( a\right) \) is an isomorphism of \( {\mathcal{O}}_{K} \) -modules. | We have \( {\left\lbrack u\right\rbrack }_{g, f}\left( {a{ + }_{{F}_{f}}b}\right) = {\left\lbrack u\right\rbrack }_{g, f}\left( a\right) { + }_{{F}_{g}}{\left\lbrack u\right\rbrack }_{g, f}\left( b\right) \), and \( {\left\lbrack u\right\rbrack }_{g, f}\left( {{\left\lbrack \alpha \right\rbrack }_{f}\left( a\right) }\r... | Yes |
Lemma 1.4.3. The extension \( {K}_{\varpi, n} \) is independent of the choice of \( f \in {\mathcal{F}}_{\varpi } \), in particular, \( {K}_{\varpi, n} \) is finite over \( K \) . | Proof. Let \( f, g \in {\mathcal{F}}_{\varpi } \), and \( u \in {\mathcal{O}}_{K}^{ \times } \) . We have seen that \( {\left\lbrack u\right\rbrack }_{g, f} : {\Lambda }_{f}\overset{ \sim }{ \rightarrow }{\Lambda }_{g} \) as \( {\mathcal{O}}_{K} \) -modules. Thus \( {\left\lbrack u\right\rbrack }_{g, f} \) induces an i... | Yes |
Example 1.4.5. Suppose \( K = {\mathbb{Q}}_{p} \), and \( f\left( T\right) = {\left( 1 + T\right) }^{p} - 1 \) . We have | \[ {\Lambda }_{n} = \left\{ {x \in \overline{{\mathbb{Q}}_{p}} \mid {\left( 1 + x\right) }^{{p}^{n}} - 1 = 0}\right\} = {\left\{ {\zeta }_{{p}^{n}}^{i} - 1\right\} }_{1 \leq i \leq {p}^{n} - 1}, \] and hence \( {\mathbb{Q}}_{p}\left( {\Lambda }_{n}\right) = {\mathbb{Q}}_{p}\left( {\zeta }_{{p}^{n}}\right) \) . | Yes |
Lemma 1.4.6. \( {K}_{\varpi ,1} \) is a totally ramified extension of \( K \) of degree \( \left( {q - 1}\right) \) . | Proof. By definition, \( f\left( T\right) /T = {T}^{q - 1} + \cdots + \varpi \) is an Eisenstein polynomial, the lemma follows. | No |
Proposition 1.4.7. We have \( {\Lambda }_{n} \cong {\mathcal{O}}_{K}/{\varpi }^{n} \) as \( {\mathcal{O}}_{K} \) -module. | Proof. It is clear that \( {\Lambda }_{n} \) is annihilated by \( {\varpi }^{n} \) and \( \left| {\Lambda }_{n}\right| \leq {q}^{n} \) . Together with the above lemma, we see \( \left| {\Lambda }_{1}\right| = q \) and the morphism \( {\mathcal{O}}_{K}/\varpi \rightarrow {\Lambda }_{1}, x \mapsto x{\alpha }_{1} \) for a... | Yes |
Proposition 1.4.8. \( {\alpha }_{n} \) is a uniformizer of \( {K}_{\varpi, n} \), and \( {K}_{\varpi, n} \) is totally ramified over \( K \) of degree \( {q}^{n - 1}\left( {q - 1}\right) \) . | Proof. We use induction on \( n \) . The polynomial \( f\left( T\right) /T = {T}^{q - 1} + \cdots + \varpi \) is an Eisenstein polynomial over \( K \), we deduce hence \( {K}_{\varpi ,1} = K\left( {\alpha }_{1}\right) \) is totally ramified over \( K \) of degree \( q - 1 \), and \( {\alpha }_{1} \) is a uniformizer of... | Yes |
Proposition 1.4.10. (1) The map (1.4) is independent of the choice of \( {\alpha }_{n} \) . | Proof. Let \( {\alpha }_{n}^{\prime } \) be another generator of \( {\Lambda }_{n} \), and \( \mu \in {\left( {\mathcal{O}}_{K}/{\varpi }^{n}\right) }^{ \times } \) such that \( {\alpha }_{n}^{\prime } = {\left\lbrack \mu \right\rbrack }_{f}\left( {\alpha }_{n}\right) \) . Let \( {\iota }_{n} \) (resp. \( {\iota }_{n}^... | Yes |
Proposition 1.4.11. For \( n \geq 1 \), there exists \( \alpha \in {K}_{\varpi, n} \) such that \( {N}_{{K}_{\varpi, n}/K}\left( \alpha \right) = \varpi \) . | Proof. Let \( {\alpha }_{n} \in {\Lambda }_{n} \) be a generator. The minimal polynomial \( r\left( x\right) \) of \( {\alpha }_{n} \) has the form \( {x}^{\left( {q - 1}\right) {q}^{n - 1}} + \cdots + \varpi \) . Thus \( {N}_{{K}_{\varpi, n}/K}\left( \alpha \right) = {\left( -1\right) }^{\left( {q - 1}\right) {q}^{n -... | Yes |
Corollary 1.5.4. The morphism \( {\rho }_{K} \) is continuous and injective. | Proof. By Theorem 1.5.1 (b), \( {\rho }_{K}^{-1}\left( {\operatorname{Gal}\left( {{K}^{\mathrm{{ab}}}/L}\right) }\right) = {N}_{L/K}\left( {L}^{ \times }\right) \) that is open by Theorem 1.5.3. Hence \( {\rho }_{K} \) is continuous. By Theorem 1.5.3, \( \operatorname{Ker}{\rho }_{K} \) is contained in any open subgrou... | Yes |
Proposition 1.5.5. A norm group is an open subgroup of \( {K}^{ \times } \) of finite index. | Proof. Let \( L \) be a finite extension of \( K \) of degree \( d \), then \( {N}_{L/K}\left( {L}^{ \times }\right) \supset {\left( {K}^{ \times }\right) }^{d} \) . We show \( {\left( {K}^{ \times }\right) }^{d} \) (hence \( {N}_{L/K}\left( {L}^{ \times }\right) \) ) is an open subgroup of finite index in \( {K}^{ \ti... | No |
Theorem 1.5.6. \( {K}^{\mathrm{{unr}}}{K}_{\varpi } \) and \( {\rho }_{\varpi } \) are both independent of the choice of \( \varpi \) . | In the rest of the section, we prove the theorem. Let \( {\varpi }_{1},{\varpi }_{2} \) be two uniformizers of \( K \) . Let \( f \in {\mathcal{F}}_{{\varpi }_{1}} \) and \( g \in {\mathcal{F}}_{{\varpi }_{2}} \) . We want to show \( {K}_{{\varpi }_{1}}{K}^{\mathrm{{ur}}} = {K}_{{\varpi }_{2}}{K}^{\mathrm{{ur}}} \) and... | No |
Lemma 1.5.7. There exists \( \theta \left( T\right) \in T{\mathcal{O}}_{\breve{K}}\left\lbrack \left\lbrack T\right\rbrack \right\rbrack \smallsetminus {T}^{2}{\mathcal{O}}_{\breve{T}}\left\lbrack \left\lbrack T\right\rbrack \right\rbrack \) such that\n\n\[ \sigma \left( \theta \right) = \theta \circ {\left\lbrack u\ri... | Proof of Theorem 1.5.6. We explain how to deduce Theorem 1.5.6 from the above lemma.\n\nBy (1.6), we have\n\n\[ {\left\lbrack {u}^{n}\right\rbrack }_{f} \circ \underset{n}{\underbrace{f \circ \cdots \circ f}} \circ {\theta }^{-1} = {\theta }^{-1} \circ \underset{n}{\underbrace{g \circ \cdots \circ g}}. \]\n\nSo if \( {... | No |
Let \( G = \{ 1\} \), then \( \mathbb{Q}/\mathbb{Z} \) is an injective \( G \) -module (that is an injective object in the category of abelian groups). | In fact, let \( M \hookrightarrow N \), and \( f : M \rightarrow \mathbb{Q}/\mathbb{Z} \) . Let \( S \) be the set consisting of \( \left( {{M}^{\prime },{f}_{{M}^{\prime }}}\right) \) where \( {M}^{\prime } \supset M \) is a submodule of \( N \), and \( {f}_{{M}^{\prime }} : M \rightarrow \mathbb{Q}/\mathbb{Z} \) is a... | Yes |
Suppose \( G = \{ 1\} \), for any abelian group \( M \), let \( F \) be a free abelian group such that \( F \rightarrow M \), and let \( N \) be the kernel of the projection. Then we have | We deduce \( M \hookrightarrow \left( {F{ \otimes }_{\mathbb{Z}}\mathbb{Q}}\right) /N \), where the latter is injective by Lemma 2.1.3. So \( {\mathcal{{Mod}}}_{\{ 1\} } = \) \( \mathcal{A}\mathrm{b} \) has enough injective objects. | No |
Lemma 2.1.6. The functor is left exact, i.e. given an exact sequence \( 0 \rightarrow {M}_{1} \rightarrow {M}_{2} \rightarrow \) \( {M}_{3} \rightarrow 0 \) in \( {\mathcal{{Mod}}}_{G} \), then \( 0 \rightarrow {M}_{1}^{G} \rightarrow {M}_{2}^{G} \rightarrow {M}_{3}^{G} \) is exact. | Proof. It is clear that \( {M}_{1}^{G} \hookrightarrow {M}_{2}^{G} \) . Let \( x \in {M}_{2}^{G} \), and suppose \( x \) is sent to 0 in \( {M}_{3} \), by the given exact sequence we see \( x \in {M}_{1} \) . However, we have \( {M}_{2}^{G} \cap {M}_{1} = {M}_{1}^{G} \) . The lemma follows. | No |
Lemma 2.1.7. Let \( f : M \rightarrow N \) be a morphism of \( G \) -modules, \( 0 \rightarrow M \rightarrow {I}^{ \bullet } \) be an exact sequence of \( G \) -modules \( {I}^{ \bullet } \), and \( {J}^{ \bullet } \) be an injective resolution of \( N \) . Then there exists a commutative diagram (of morphisms in \( {\... | Proof. We inductively construct \( {f}^{i} \) . The existence of \( {f}^{0} \) follows from the injectivity of \( {J}^{0} \) . Now suppose the maps \( {\left\{ {f}^{j}\right\} }_{j \leq i} \) have been constructed. In particular, we have\n\n \) are independent of the choice of \( {f}_{i} \) . | Proof. It suffices to show if \( f = 0 \), then \( {H}^{i}\left( f\right) = 0 \) for any choice of \( {f}_{i} \) . We claim there exists \( {g}^{i} : {I}^{i + 1} \rightarrow {J}^{i} \) such that \( {f}^{i} = {d}_{J}^{i - 1} \circ {g}^{i - 1} + {g}^{i} \circ {d}_{I}^{i}\left( {{g}^{-1} = 0}\right) \) :\n\n![00eeb6ce-d10... | Yes |
Corollary 2.1.9. (1) For \( M \in {\mathcal{{Mod}}}_{G},{H}^{i}\left( {G, M}\right) \) is independent of the choice of the injective resolution \( {I}^{ \bullet } \) of \( M \) . | Proof. (1) Let \( {I}^{ \bullet },{J}^{ \bullet } \) be two resolutions of \( M \), let \( {H}_{I}^{i} \) and \( {H}_{J}^{i} \) be the cohomology group of \( {\left( {I}^{ \bullet }\right) }^{G} \) and \( {\left( {J}^{ \bullet }\right) }^{G} \) respectively. The identity map on \( M \) induces \( \alpha : {H}_{I}^{i} \... | Yes |
Proposition 2.1.12. Let \( {I}^{ \bullet } \) be an acyclic resolution of \( M \), i.e. there is an exact sequence \( 0 \rightarrow M \rightarrow {I}^{0} \rightarrow {I}^{1}\cdots \) with \( {I}^{i} \) all acyclic. Then \( {H}^{i}\left( {\left( {I}^{ \bullet }\right) }^{G}\right) \cong {H}^{i}\left( {G, M}\right) \) fo... | Proof. Let \( {M}_{0} \mathrel{\text{:=}} M \), and we inductively construct \( {M}_{i} \mathrel{\text{:=}} {I}^{i - 1}/{M}_{i - 1} \hookrightarrow {I}^{i} \) . We have thus an exact sequence \( 0 \rightarrow {M}_{i} \rightarrow {I}^{i} \rightarrow {M}_{i + 1} \rightarrow 0 \), that induces \( 0 \rightarrow {H}^{0}\lef... | Yes |
Lemma 2.2.1. \( \mathbb{Z}\left\lbrack G\right\rbrack \) is a free \( \mathbb{Z}\left\lbrack H\right\rbrack \) -module, and consequently, the functor \( {\mathcal{{Mod}}}_{H} \rightarrow \) \( {\mathcal{{Mod}}}_{G}, M \mapsto \mathbb{Z}\left\lbrack G\right\rbrack { \otimes }_{\mathbb{Z}\left\lbrack H\right\rbrack }M \)... | Proof. Let \( R \) be a set of representatives of the right cosets \( H \) in \( G \), then \( {\mathbb{Z}}_{G} \cong { \oplus }_{g \in R}\mathbb{Z}\left\lbrack H\right\rbrack {e}_{g} \). | No |
Lemma 2.2.2. (1) There is a natural isomorphism \( {M}^{H}\overset{ \sim }{ \rightarrow }{\left( {\operatorname{Ind}}_{H}^{G}M\right) }^{H} \) . | Proof. (1) We have a natural map \( {M}^{H} \hookrightarrow {\operatorname{Ind}}_{H}^{G}M, m \mapsto \left\lbrack {g \mapsto m}\right\rbrack \) . It is clear the image is contained in \( {\left( {\operatorname{Ind}}_{H}^{G}M\right) }^{G} \) . Let \( f \in {\left( {\operatorname{Ind}}_{H}^{G}M\right) }^{G} \), we deduce... | Yes |
Lemma 2.2.3. We have \( \mathbb{Z}\left\lbrack G\right\rbrack { \otimes }_{\mathbb{Z}\left\lbrack H\right\rbrack }M \cong {\operatorname{Ind}}_{H}^{G}M \), in particular, we have\n\n\[{\operatorname{Hom}}_{H}\left( {N, M}\right) \cong {\operatorname{Hom}}_{G}\left( {{\operatorname{Ind}}_{H}^{G}N, M}\right)\]\n\n\[{\ope... | Proof. For \( {e}_{g} \otimes m \in \mathbb{Z}\left\lbrack G\right\rbrack { \otimes }_{\mathbb{Z}}M \), consider the induced map\n\n\[G \rightarrow M,{g}^{\prime } \mapsto \left\{ {\begin{array}{ll} {g}^{\prime }{gm} & {g}^{\prime }g \in H \\ 0 & \text{ otherwise } \end{array}.}\right.\]\n\nOne can check the map lies i... | Yes |
Proposition 2.2.5. If \( M \in {\mathcal{{Mod}}}_{H} \) is injective, then \( {\operatorname{Ind}}_{H}^{G}M \) (hence \( \mathbb{Z}\left\lbrack G\right\rbrack { \otimes }_{\mathbb{Z}\left\lbrack H\right\rbrack }M \) ) is injective in \( {\mathcal{{Mod}}}_{G} \) . | Proof. Let \( {M}_{1} \hookrightarrow {M}_{2} \) be an injection in \( {\mathcal{{Mod}}}_{G} \), and \( {I}_{H} \) be an injective \( H \) -module. Then we have\n\n\n\nhence the top map is surjective. The proposition f... | No |
Corollary 2.2.6. The category \( {\mathcal{{Mod}}}_{G} \) has enough injective objects. | Proof. Let \( M \in {\mathcal{{Mod}}}_{G} \) . Forgetting the \( G \) -action, we view \( M \) as an object in \( \mathcal{A}\mathrm{b} = {\mathcal{{Mod}}}_{\{ 1\} } \) . Let \( I \in \mathcal{A}\mathrm{b} \) be an injective object such that \( f : M \hookrightarrow I \) (in \( \mathcal{A}\mathrm{b} \) ). By Frobenius ... | Yes |
Corollary 2.2.7 (Shapiro’s lemma). Let \( H \subset G \) and \( N \in {\mathcal{{Mod}}}_{H} \) . There is a canonical isomorphism \[ {H}^{i}\left( {G,{\operatorname{Ind}}_{H}^{G}N}\right) \overset{ \sim }{ \rightarrow }{H}^{i}\left( {H, N}\right) \] | Proof. Let \( 0 \rightarrow N \rightarrow {I}^{ \bullet } \) be an injective resolution of \( N \) in \( {\mathcal{{Mod}}}_{H} \) . Then by Proposition 2.2.5 and the fact \( {\operatorname{Ind}}_{H}^{G} - \) is exact, we see \( 0 \rightarrow {\operatorname{Ind}}_{H}^{G}N \rightarrow \left( {{\operatorname{Ind}}_{N}^{G}... | Yes |
Corollary 2.2.9. Let \( M \in {\mathcal{{Mod}}}_{G} \) . If \( M \) is a finitely generated abelian group, then \( {H}^{i}\left( {G, M}\right) \) is a finitely generated abelian group. | Proof. We have an injection \( M \hookrightarrow {\operatorname{Ind}}_{\{ 1\} }^{G}M \) . As \( M \) is finitely generated, we see \( {\operatorname{Ind}}_{\{ 1\} }^{G}M \) is also finitely generated. We then deduce that \( M \) admits an acyclic resolution \( {I}^{ \bullet } \) consisting of \( G \) -modules that are ... | Yes |
Corollary 2.2.10. Let \( L/K \) be a finite Galois extension. Then \( {H}^{i}\left( {\operatorname{Gal}\left( {L/K}\right), L}\right) = 0 \), for all \( i > 0 \) . | Proof. By the normal basis theorem, there exists \( \alpha \in L \) such that \( \{ g\left( \alpha \right) {\} }_{g \in \operatorname{Gal}\left( {L/K}\right) } \) form a basis of \( L \) over \( K \) . We see as \( \operatorname{Gal}\left( {L/K}\right) \) -module, \( \mathbb{Z}\left\lbrack {\operatorname{Gal}\left( {L/... | Yes |
Corollary 2.2.11. Let \( H \subset G, M \in {\mathcal{{Mod}}}_{G} \) . There are natural morphisms Res : \( {H}^{i}\left( {G, M}\right) \rightarrow {H}^{i}\left( {H, M}\right) \) (called restrictions) and Cor : \( {H}^{i}\left( {H, M}\right) \rightarrow {H}^{i}\left( {G, M}\right) \) (called corestrictions). Moreover, ... | Proof. By Frobeinus reciprocty, we have a natural \( G \) -equivariant morphism \( \iota : M \rightarrow \) \( {\operatorname{Ind}}_{H}^{G}M, m \mapsto \left\lbrack {g \mapsto {gm}}\right\rbrack \), that induces Res \( : {H}^{i}\left( {G, M}\right) \rightarrow {H}^{i}\left( {G,{\operatorname{Ind}}_{H}^{G}M}\right) \con... | Yes |
Corollary 2.2.12. Let \( M \) be a finite \( G \) -module, if \( \left( {\left| M\right| ,\left| G\right| }\right) = 1 \), then \( {H}^{i}\left( {G, M}\right) = 0 \) for all \( i > 0 \) . | Proof. As \( \left( {\left| M\right| ,\left| G\right| }\right) = 1 \), multiplying \( \left| G\right| \) is an isomorphism on \( M \) hence is an isomorphism on \( {H}^{i}\left( {G, M}\right) \) for all \( i \) . Applying the above corollary to \( H = \{ 1\} \), Cor \( \circ \operatorname{Res} = \left| G\right| \) : \(... | Yes |
Proposition 2.2.13. Let \( {M}_{1} \in {\mathcal{{Mod}}}_{{G}_{1}} \), and \( {M}_{2} \in {\mathcal{{Mod}}}_{{G}_{2}} \) . Let \( f : {M}_{2} \rightarrow {M}_{1} \) be a morphism in \( {\mathcal{{Mod}}}_{{G}_{1}} \) . Then \( f \) induces natural morphisms \[ {H}^{i}\left( {{G}_{2},{M}_{2}}\right) \rightarrow {H}^{i}\l... | Proof. First for any \( N \in {\mathcal{{Mod}}}_{{G}_{2}} \), as the \( {G}_{1} \) -action on \( N \) factors through \( {G}_{2} \), there is a natural injection \[ {N}^{{G}_{2}} \hookrightarrow {N}^{{G}_{1}} \] Let \( {I}_{{G}_{2}}^{ \bullet } \) be an injective resolution of \( {M}_{2} \) in \( {\mathcal{{Mod}}}_{{G}... | Yes |
Proposition 2.2.15 (Inflation-Restriction). The following sequence is exact\n\n\[ \n0 \rightarrow {H}^{1}\left( {G/H,{M}^{H}}\right) \overset{\text{ inf }}{ \rightarrow }{H}^{1}\left( {G, M}\right) \overset{\text{ Res }}{ \rightarrow }{H}^{1}\left( {H, M}\right) .\n\] | Proof. One can use cochains to directly prove (2.5) (that we will leave as an exercise for the next section). Assume now (2.5) holds. Recall we have a natural \( G \) -equivariant injection \( M \rightarrow {\operatorname{Ind}}_{\{ 1\} }^{G}M, m \mapsto \left\lbrack {g \mapsto {gm}}\right\rbrack \), and let \( N \mathr... | No |
Lemma 2.3.1. We have \( {N}^{i} \cong {\operatorname{Ind}}_{\{ 1\} }^{G}{N}_{0}^{i} \) as \( G \) -module. | Proof. Recall \( {\operatorname{Ind}}_{\{ 1\} }^{G}{N}_{0}^{i} \) is isomorphic to \( \left\{ {f : G \rightarrow {N}_{0}^{i}}\right\} \) with the \( G \) -action given by \( \left( {gf}\right) \left( {g}^{\prime }\right) \mathrel{\text{:=}} f\left( {{g}^{-1}{g}^{\prime }}\right) \) . Consider the map\n\n\[ \left\{ {f :... | Yes |
Lemma 2.3.2. We have an exact sequence of \( G \) -modules\n\n\[ 0 \rightarrow M\overset{\iota }{ \rightarrow }{N}^{0}\overset{{d}_{1}^{0}}{ \rightarrow }{N}^{1} \rightarrow \cdots {N}^{i}\overset{{d}_{1}^{i}}{ \rightarrow }{N}^{i + 1} \rightarrow \cdots \] | Together with Lemma 2.3.1, we see (2.9) gives an acyclic resolution of \( M \) . Now we apply the functor \( {\left( -\right) }^{G} \) to the resolution. We have \( {C}^{i}\left( {G, M}\right) \mathrel{\text{:=}} \left\{ {{G}^{i} \rightarrow M}\right\} \underset{ \sim }{\overset{{j}_{i}}{ \rightarrow }}{\left( {N}^{i}\... | Yes |
Proposition 2.3.4 (Hilbert’s theorem 90). Let \( L/K \) be a finite Galois extension, then \( {H}^{1}\left( {\operatorname{Gal}\left( {L/K}\right) ,{L}^{ \times }}\right) = \{ 1\} \) . | Proof. Let \( c : \operatorname{Gal}\left( {L/K}\right) \rightarrow {L}^{ \times } \) be a cocycle, i.e. \( c\left( {{g}_{1}{g}_{2}}\right) = {g}_{1}\left( {c\left( {g}_{2}\right) }\right) c\left( {g}_{1}\right) \) . For \( x \in L \) , consider \( {a}_{x} \mathrel{\text{:=}} \mathop{\sum }\limits_{{g \in \operatorname... | Yes |
Lemma 2.4.1. \( {I}_{G} \cong { \oplus }_{g \in G\smallsetminus \{ 1\} }\mathbb{Z}\left( {{e}_{g} - 1}\right) \) (note \( 1 = {e}_{1} \in \mathbb{Z}\left\lbrack G\right\rbrack \) ). | Proof. Let \( \alpha = \mathop{\sum }\limits_{{g \in G}}{a}_{g}{e}_{g} \in {I}_{G} \) with \( {a}_{g} \in \mathbb{Z} \), by definition we have \( \mathop{\sum }\limits_{{g \in G}}{a}_{g} = 0 \) . Hence \( \alpha = \mathop{\sum }\limits_{{g \in G}}{a}_{g}\left( {{e}_{g} - 1}\right) \) . Together with the fact \( \left\{... | Yes |
Lemma 2.4.2. The category \( {\mathcal{{Mod}}}_{G} \) has enough projective objects, i.e. for any \( M \in {\mathcal{{Mod}}}_{G} \) , there exists a projective object \( P \in {\mathcal{{Mod}}}_{G} \) such that \( P \rightarrow M \) . | Proof. Any free \( \mathbb{Z}\left\lbrack G\right\rbrack \) -module is projective. For any \( M \in {\mathcal{{Mod}}}_{G} \), we have a surjective morphism \( { \oplus }_{m \in M}\mathbb{Z}{\left\lbrack G\right\rbrack }_{m} \rightarrow M \), where \( \mathbb{Z}{\left\lbrack G\right\rbrack }_{m} \cong \mathbb{Z}\left\lb... | Yes |
Proposition 2.4.5. Cor \( \circ \) Res \( = \left\lbrack {G : H}\right\rbrack \) . | Suppose \( H \) is a normal subgroup of \( G \) . For \( M \in {\mathcal{{Mod}}}_{G},{M}_{H} \cong M{ \otimes }_{\mathbb{Z}\left\lbrack H\right\rbrack }\mathbb{Z} \) inherits from \( M \) a natural \( G \) -action that factors through \( G/H \) . By Proposition 2.4.4(4), the \( G \) -equivariant morphism \( M \rightarr... | No |
Proposition 2.4.6. The following sequence is exact | \[ \n{H}_{1}\left( {H, M}\right) \overset{\text{ Cor }}{ \rightarrow }{H}_{1}\left( {G, M}\right) \overset{\text{ Coinf }}{ \rightarrow }{H}_{1}\left( {G/H,{M}_{H}}\right) \rightarrow 0. \n\] | No |
Lemma 2.4.7. The map \( \kappa : G \rightarrow {I}_{G}/{I}_{G}^{2}, g \mapsto {e}_{g} - 1 \) is a group homomorphism, and induces an isomorphism \( {G}^{\mathrm{{ab}}}\overset{ \sim }{ \rightarrow }{I}_{G}/{I}_{G}^{2} \) . | Proof. We have \( {e}_{gh} - 1 - \left( {{e}_{g} - 1 + {e}_{h} - 1}\right) = {e}_{gh} - {e}_{g} - \left( {{e}_{h} - 1}\right) = \left( {{e}_{g} - 1}\right) \left( {{e}_{h} - 1}\right) \in {I}_{G}^{2} \) , for \( g, h \in G \) . Hence \( \kappa \) is a group homomoprhism. It is also clear that \( \kappa \) is surjective... | Yes |
Lemma 2.4.9. Let \( H \subset G \) be a subgroup, then \( \operatorname{Cor} : {H}_{1}\left( {H,\mathbb{Z}}\right) \rightarrow {H}_{1}\left( {G,\mathbb{Z}}\right) \) coincides with the natural map \( {H}^{\mathrm{{ab}}} \rightarrow {G}^{ab} \) and \( \operatorname{Res} : {H}_{1}\left( {G,\mathbb{Z}}\right) \rightarrow ... | Proof. By the \( G \) -equivariant exact sequence \( 0 \rightarrow {I}_{G} \rightarrow \mathbb{Z}\left\lbrack G\right\rbrack \rightarrow \mathbb{Z} \rightarrow 0 \), we have a commutative diagram\n\n\[\n\begin{array}{l} {H}_{1}\left( {H,\mathbb{Z}}\right) \rightarrow {H}_{0}\left( {H,{I}_{G}}\right) \rightarrow \mathbb... | No |
Lemma 2.5.1. The map \( {\mathcal{N}}_{G} \) is a morphism of \( G \) -modules, and \( \operatorname{Im}\left( {\mathcal{N}}_{G}\right) \subset {M}^{G},{I}_{G}M \subset \) \( \operatorname{Ker}\left( {\mathcal{N}}_{G}\right) \) . | Proof. For \( m \in M, h \in G,{\mathcal{N}}_{G}\left( {hm}\right) = \mathop{\sum }\limits_{{g \in G}}{ghm} = h\mathop{\sum }\limits_{{g \in G}}\left( {{h}^{-1}{gh}}\right) m = h\mathop{\sum }\limits_{{g \in G}}{gm} = \) \( h{\mathcal{N}}_{G}\left( m\right) \) . For \( \alpha = \mathop{\sum }\limits_{{g \in M}}{gm} \in... | Yes |
Proposition 2.5.3. Let \( 0 \rightarrow {M}_{1} \rightarrow {M}_{2} \rightarrow {M}_{3} \rightarrow 0 \) be an exact sequence in \( {\mathcal{{Mod}}}_{G} \) . Then there is a natural long exact sequence\n\n\[ \cdots \rightarrow {H}_{T}^{-2}\left( {G,{M}_{1}}\right) \rightarrow {H}_{T}^{-2}\left( {G,{M}_{2}}\right) \rig... | Proof. We have a commutative diagram (for example, to see \( {H}_{1}\left( {G,{M}_{3}}\right) \rightarrow {H}_{0}\left( {G,{M}_{1}}\right) \overset{{\mathcal{N}}_{G}}{ \rightarrow } \) \( {H}^{0}\left( {G,{M}_{1}}\right) \) is zero, one uses the fact \( {H}^{0}\left( {G,{M}_{1}}\right) \rightarrow {H}^{0}\left( {G,{M}_... | Yes |
Proposition 2.5.5. Let \( H \) be a subgroup of \( G \) and \( M \in {\mathcal{{Mod}}}_{H} \), then \( {H}_{T}^{i}\left( {G,{\operatorname{Ind}}_{H}^{G}M}\right) \cong \) \( {H}_{T}^{i}\left( {H, M}\right) \) . In particular, \( {H}_{T}^{i}\left( {G,{\operatorname{Ind}}_{\{ 1\} }^{G}M}\right) = 0 \) . | Proof. We only need to show the isomorphism for \( i = - 1,0 \) . Recall the isomorphism \( {\iota }^{0} : {H}^{0}\left( {H, M}\right) \rightarrow {H}^{0}\left( {G,{\operatorname{Ind}}_{H}^{G}M}\right) \) is induced by the \( H \) -equivariant morphism \( {\operatorname{Ind}}_{H}^{G}M \rightarrow M \) , \( f \mapsto f\... | No |
Theorem 2.5.7. Let \( G \) be a finite cyclic group, \( M \in {\mathcal{{Mod}}}_{G} \) . Then there is a canonical (up to the choice of a generator of \( G \) ) functorial isomoprhism \( {H}_{T}^{i}\left( {G, M}\right) \overset{ \sim }{ \rightarrow }{H}_{T}^{i + 2}\left( {G, M}\right) \) . | Proof. Let \( h \) be a generator of \( G \), we have an exact sequence (of \( G \) -modules)\n\n\[ 0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Z}\left\lbrack G\right\rbrack \rightarrow \mathbb{Z}\left\lbrack G\right\rbrack \rightarrow \mathbb{Z} \rightarrow 0 \]\n\nwhere \( \mathbb{Z} \rightarrow \mathbb{Z}\left\lbra... | Yes |
Lemma 2.5.10. Suppose \( G \) is finite cyclic, and \( M \in {\mathcal{{Mod}}}_{G} \) has finite cardinality. Then \( h\left( M\right) = 1 \) . | Proof. Let \( h \in G \) be a generator, we have an exact sequence (of finite abelian groups)\n\n\[ 0 \rightarrow {M}^{G} \rightarrow M\xrightarrow[]{m \mapsto {hm} - m}M \rightarrow {M}_{G} \rightarrow 0. \]\n\nWe deduce hence \( \left| {M}^{G}\right| = \left| {M}_{G}\right| \) . On the other hand, we have by definiti... | Yes |
Lemma 2.6.1. \( {d}_{M{ \otimes }_{\mathbb{Z}}N}^{i + j}\left( {f \cup {f}^{\prime }}\right) = {d}_{M}^{i}\left( f\right) \cup {f}^{\prime } + {\left( -1\right) }^{i}f \cup {d}_{N}^{j}{f}^{\prime } \) . | Proof. We have\n\n\[ \n{d}_{M{ \otimes }_{\mathbb{Z}}N}^{i + j}\left( {f \cup {f}^{\prime }}\right) \left( {{g}_{0},\cdots ,{g}_{i + j}}\right) \n\] \n\n\[ \n= {g}_{0}\left( {\left( {f \cup {f}^{\prime }}\right) \left( {{g}_{1},\cdots ,{g}_{i + j}}\right) }\right) + \mathop{\sum }\limits_{{k = 1}}^{{i + j}}\left( {f \c... | Yes |
Theorem 2.6.3. The collection of maps\n\n\\[ \n{\\left\\{ {H}^{i}\\left( G, M\\right) { \\otimes }_{\\mathbb{Z}}{H}^{j}\\left( G, N\\right) \\overset{ \\cup }{ \\rightarrow }{H}^{i + j}\\left( G, M{ \\otimes }_{\\mathbb{Z}}N\\right) \\right\\} }_{\\begin{matrix} {i, j \\in {\\mathbb{Z}}_{ \\geq 0}} \\ {M, N \\in {\\mat... | Sketch of proof. One can directly check these conditions, using the following description of the \\( \\delta \\) -maps in term of cochains. If we have an exact sequence \\( 0 \\rightarrow {M}_{1} \\rightarrow M \\rightarrow {M}_{2} \\rightarrow 0 \\) in \\( {\\mathcal{{Mod}}}_{G} \\), then we naturally get exact sequen... | Yes |
Proposition 2.6.4. We have a commutative diagram\n\n\[ \n{H}^{i}\left( {G, M}\right) \otimes {H}^{j}\left( {G, N}\right) \overset{ \cup }{ \rightarrow }{H}^{i + j}\left( {G, M{ \otimes }_{\mathbb{Z}}N}\right) \]\n\n\[ \ns \downarrow \;s \downarrow \]\n\n\[ \n{H}^{j}\left( {G, N}\right) \otimes {H}^{i}\left( {G, M}\righ... | Proof. We use induction on \( i, j \) . The case \( i = j = 0 \) is clear. Suppose it holds for \( i, j \) . Using the exact sequence in (2.16), we deduce a surjective \( \delta \) -map \( {H}^{i}\left( {G,{M}^{\prime }}\right) \rightarrow {H}^{i + 1}\left( {G, M}\right) \) . We have commutative diagrams ![00eeb6ce-d10... | Yes |
Proposition 2.6.5. Let \( {M}_{1},{M}_{2},{M}_{3} \in {\operatorname{Mod}}_{G},\alpha \in {H}^{i}\left( {G,{M}_{1}}\right) ,\beta \in {H}^{j}\left( {G,{M}_{2}}\right) ,\gamma \in \) \( {H}^{k}\left( {G,{M}_{3}}\right) \), then\n\n\[ \left( {\alpha \cup \beta }\right) \cup \gamma = \alpha \cup \left( {\beta \cup \gamma ... | Proof. The proposition follows from the explicit formula (and the uniqueness in Theorem 2.6.3). | No |
Proposition 2.6.7. Let \( M, N \in {\mathcal{{Mod}}}_{G} \). (1) Let \( H \) be a subgroup of \( G,\alpha \in {H}^{i}\left( {G, M}\right) ,\beta \in {H}^{j}\left( {G, N}\right) \), then \[ \operatorname{Res}\left( {\alpha \cup \beta }\right) = \operatorname{Res}\left( \alpha \right) \cup \operatorname{Res}\left( \beta ... | Proof. (1) (2) follow by explicit formulas (or induction on degrees as for (3)). (3) follows by induction on degrees: one first checks it holds for \( i = j = 0 \), then uses (2.16) and induction on \( i \) to prove it holds for \( i \geq 0 \) and \( j = 0 \) (as in the proof of Theorem 2.6.3); finally one uses (2.16) ... | No |
Theorem 2.6.8. There is a unique collection of maps \[ {\left\{ {H}_{T}^{i}\left( G, M\right) { \otimes }_{\mathbb{Z}}{H}_{T}^{j}\left( G, N\right) \overset{ \cup }{ \rightarrow }{H}_{T}^{i + j}\left( G, M{ \otimes }_{\mathbb{Z}}N\right) \right\} }_{\begin{matrix} {i, j \in \mathbb{Z}} \\ {M, N \in \mathcal{M}{\operato... | Proof. The existence and uniqueness can follow by a dimension shifting argument, similarly as in the proof of Theorem 2.6.3. | No |
The map \( {H}_{T}^{0}\left( {G, M}\right) \otimes {H}_{T}^{-1}\left( {G, N}\right) \overset{ \cup }{ \rightarrow }{H}_{T}^{-1}\left( {G, M{ \otimes }_{\mathbb{Z}}N}\right) \) is given by \( \left( {a, b}\right) \mapsto \) \( a \otimes b \). | Indeed, as \( a \in {M}^{G} \), and \( {\mathcal{N}}_{G}\left( b\right) = 0 \), we see \( {\mathcal{N}}_{G}\left( {a \otimes b}\right) = 0 \) (so that the map is well-defined). | No |
Proposition 2.6.10. We have a commutative diagram\n\n\[ \n{H}_{T}^{i}\left( {G, M}\right) \otimes {H}_{T}^{j}\left( {G, N}\right) \overset{ \cup }{ \rightarrow }{H}_{T}^{i + j}\left( {G, M{ \otimes }_{\mathbb{Z}}N}\right) \n\]\n\n\[ \ns \downarrow \;s \downarrow \;. \n\]\n\n(2.18)\n\n\[ \n{H}_{T}^{j}\left( {G, N}\right... | We have as in Proposition 2.6.7 (1) (3): | No |
Lemma 3.1.2. Keep the assumption of the theorem, then the restriction map \( {H}^{2}\left( {G, M}\right) \rightarrow \) \( {H}^{2}\left( {H, M}\right) \) is surjective for \( H \leq G \) . | Proof. The lemma follows easily from the fact \( \operatorname{Cor} \circ \operatorname{Res} = \left\lbrack {G : H}\right\rbrack \), and that \( {H}^{2}\left( {G, M}\right) \) (resp. \( {H}^{2}\left( {H, M}\right) \) ) is cyclic of order \( \left| G\right| \) (resp. \( \left| H\right| \) ). | Yes |
Lemma 3.1.4. Let \( G \) be a finite group, \( N \in {\mathcal{{Mod}}}_{G} \). Suppose \[ {H}^{1}\left( {H, N}\right) = {H}^{2}\left( {H, N}\right) = 0\text{ for all }H \leq G, \] then \( {H}_{T}^{i}\left( {G, N}\right) = 0 \) for all \( i \). | Proof. (1) Suppose first \( G \) is cyclic, then \( {H}_{T}^{i}\left( {G, N}\right) \cong {H}_{T}^{i + 2}\left( {G, N}\right) \). By (3.4), we deduce \( {H}_{T}^{i}\left( {G, N}\right) = 0 \) for all \( i \). (2) Suppose \( G \) is solvable, and we use induction on the order of \( G \). Induction hypothesis: if \( \lef... | Yes |
Lemma 3.2.1. \( {H}_{T}^{0}\left( {\operatorname{Gal}\left( {{k}_{L}/k}\right) ,{k}_{L}^{ \times }}\right) = \{ 1\} \) | Proof. One can directly prove \( {N}_{{k}_{L}/k}\left( {k}_{L}^{ \times }\right) = {k}^{ \times } \), or use the fact \( h\left( {k}_{L}^{ \times }\right) = 1\left( {k}_{L}^{ \times }\right. \) is finite) and \( {H}_{T}^{1}\left( {\operatorname{Gal}\left( {{k}_{L}/k}\right) ,{k}_{L}^{ \times }}\right) = \{ 1\} . | No |
Lemma 3.2.2. We have \( {N}_{L/K}\left( {\mathcal{O}}_{L}^{ \times }\right) = {\mathcal{O}}_{K}^{ \times } \) . | Proof. By the above lemma, for \( x \in {\mathcal{O}}_{K}^{ \times } \), there exists \( {y}_{1} \in {\mathcal{O}}_{L}^{ \times } \) such that \( {N}_{L/K}\left( {y}_{1}\right) \equiv x \) \( \left( {\;\operatorname{mod}\;{\varpi }_{K}}\right) \), or equivalently, \( {N}_{L/K}\left( {y}_{1}\right) /x \equiv 1\left( {\;... | Yes |
Lemma 3.2.3. We have \( {H}_{T}^{1}\left( {\operatorname{Gal}\left( {L/K}\right) ,{\mathcal{O}}_{L}^{ \times }}\right) = \{ 1\} \) . | Proof. As \( {H}_{T}^{1}\left( {\operatorname{Gal}\left( {L/K}\right) ,{L}^{ \times }}\right) = \{ 1\} \), for any \( f \in {Z}^{1}\left( {\operatorname{Gal}\left( {L/K}\right) ,{\mathcal{O}}_{L}^{ \times }}\right) \subset {Z}^{1}\left( {\operatorname{Gal}\left( {L/K}\right) ,{L}^{ \times }}\right) \) there exists \( \... | Yes |
Lemma 3.2.5. There exists a finite free \( {\mathcal{O}}_{K} \) -submodule \( V \) of \( {\mathcal{O}}_{L} \) of rank \( \left\lbrack {L : K}\right\rbrack \), stable by \( \operatorname{Gal}\left( {L/K}\right) \), such that \( {H}^{i}\left( {\operatorname{Gal}\left( {L/K}\right), V}\right) = 0 \) for all \( i > 0 \) . | Proof. Recall there exists \( \alpha \in L \) such that \( L = { \oplus }_{\sigma \in \operatorname{Gal}\left( {L/K}\right) }{K\sigma }\left( \alpha \right) \) . Multiplying \( \alpha \) be a certain power of \( {\varpi }_{K} \), we can and do assume \( \alpha \in {\mathcal{O}}_{L} \) . Put \( V \mathrel{\text{:=}} { \... | Yes |
Lemma 3.2.6. There exists an open subgroup \( W \subset {\mathcal{O}}_{L}^{ \times } \), stable by \( \operatorname{Gal}\left( {L/K}\right) \) such that \( {H}^{i}\left( {\operatorname{Gal}\left( {L/K}\right), W}\right) = 1 \) for all \( i > 0 \) . | Proof. We use the notation in the proof of the above lemma. Multiplying \( \alpha \) by a certain power of \( {\varpi }_{K} \), we can and do assume that for all \( x \in V \), the power series \( \exp \left( x\right) \mathrel{\text{:=}} \mathop{\sum }\limits_{{i = 0}}^{\infty }\frac{{x}^{i}}{n!} \) converges. Let \( W... | Yes |
Corollary 3.2.7. Suppose \( L/K \) is cyclic. We have \( h\left( {L}^{ \times }\right) = \left\lbrack {L : K}\right\rbrack \), consequently, the group \( {H}^{2}\left( {\operatorname{Gal}\left( {L/K}\right) ,{L}^{ \times }}\right) \) is finite of order \( \left\lbrack {L : K}\right\rbrack \) . | Proof. By the above lemma, we deduce \( h\left( {\mathcal{O}}_{K}^{ \times }\right) = h\left( W\right) h\left( {{\mathcal{O}}_{K}^{ \times }/W}\right) = 1 \) . We also have \( h\left( \mathbb{Z}\right) = \) \( \left| {{H}_{T}^{0}\left( {\operatorname{Gal}\left( {L/K}\right) ,\mathbb{Z}}\right) }\right| = \left\lbrack {... | Yes |
Corollary 3.2.8. Let \( L \) be a finite Galois extension of \( K \), we have \( \left| {{H}^{2}\left( {\operatorname{Gal}\left( {L/K}\right) ,{L}^{ \times }}\right) }\right| \leq \) \( \left\lbrack {L : K}\right\rbrack \) . | Proof. Recall \( \operatorname{Gal}\left( {L/K}\right) \) is solvable (by ramification theory). By Hilbert’s theorem 90, \( {H}^{1}\left( {\operatorname{Gal}\left( {L/{K}^{\prime }}\right) ,{L}^{ \times }}\right) = 0 \) for any subextension \( {K}^{\prime }/K \) . The corollary then follows from the above lemma by usin... | No |
Corollary 3.2.10. Let \( L/K \) be a finite Galois extension, taking cup-product with a generator of \( {H}^{2}\left( {\operatorname{Gal}\left( {L/K}\right) ,{L}^{ \times }}\right) \) induces a canonical isomorphism | \[ \operatorname{Gal}{\left( L/K\right) }^{\mathrm{{ab}}}\overset{ \sim }{ \rightarrow }{K}^{ \times }/{N}_{L/K}\left( {L}^{ \times }\right) \] | Yes |
Lemma 3.2.12. Let \( G \) be finite group, there are canonical isomorphisms \( {H}^{i}\left( {G,\mathbb{Q}/\mathbb{Z}}\right) \rightarrow \) \( {H}^{i + 1}\left( {G,\mathbb{Z}}\right) \) for \( i \geq 1 \) . | Proof. Consider the exact sequence of (trivial) \( G \) -modules:\n\n\[ 0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Q} \rightarrow \mathbb{Q}/\mathbb{Z} \rightarrow 0 \]\n\n(3.9)\n\nBy the same argument as in the proof of Corollary 2.2.12 (using restriction and corestric-tion), we see \( {H}^{i}\left( {G,\mathbb{Q}}\r... | No |
We may try to understand \( {\rho }_{L/K} \) in the case \( L/K \) is unramified. In this case, \( {K}^{ \times }/{N}_{L/K}\left( {L}^{ \times }\right) \) is generated by an arbitrary uniformizer \( {\varpi }_{K} \) of \( K \), and we want to describe the element \( {\rho }_{L/K}\left( {\varpi }_{K}\right) \). | The composition of the second column is equal to \( {\operatorname{inv}}_{L/K} \), and the composition of the third column sends \( {\varpi }_{K} \) to \( 1/\left\lbrack {L : K}\right\rbrack \) . Let \( \chi \in {H}_{T}^{1}\left( {\operatorname{Gal}\left( {L/K}\right) ,\mathbb{Q}/Z}\right) \) be the element correspondi... | Yes |
Corollary 3.3.5. Let \( M \supset L \supset K \) be finite abelian extensions, then \( {\left. {\rho }_{M/K}\left( a\right) \right| }_{L} = {\rho }_{L/K}\left( a\right) \) for all \( a \in {K}^{ \times } \) . | Proof. By Pontryagain duality, it suffices to show for any \( \chi \in {H}^{1}\left( {\operatorname{Gal}\left( {L/K}\right) ,\mathbb{Q}/\mathbb{Z}}\right) \), we have \( \chi \left( {{\rho }_{L/K}\left( a\right) }\right) = \chi \left( {\left. {\rho }_{M/K}\left( a\right) \right| }_{L}\right) \) . By the previous propos... | Yes |
Lemma 3.3.7. For \( a \in {K}^{ \times },{\left. {\rho }_{K}\left( a\right) \right| }_{{K}_{\varpi {K}^{\mathrm{{ur}}}}} = {\rho }_{\mathrm{{LT}}}\left( a\right) \) . | Proof. It suffices to show the equality for all uniformizers \( {\varpi }^{\prime } \) of \( K \) (since they generate \( {K}^{ \times } \) ). Recall we have \( {\rho }_{\mathrm{{LT}}}\left( {\varpi }^{\prime }\right) = \left\{ \begin{array}{ll} \mathrm{{id}} & \text{ on }{K}_{{\varpi }^{\prime }} \\ {\sigma }_{K} & \t... | Yes |
Theorem 3.3.8. (1) We have \( {K}_{\varpi }{K}^{\mathrm{{ur}}} = {K}^{\mathrm{{ab}}} \), and \( {\rho }_{K} \) is unique satisfying the given properties in Corollary 3.3.6. | Proof. We put \( {K}_{n, m} \mathrel{\text{:=}} {K}_{\varpi, n}{K}_{m}^{\mathrm{{ur}}} \), where \( {K}_{m}^{\mathrm{{ur}}} \) denotes the unramified extension of \( K \) of degree \( m \) . We see \( \left( {1 + {\varpi }^{n}{\mathcal{O}}_{K}}\right) \left\langle {\varpi }^{m}\right\rangle \subset {K}^{ \times } \) fi... | Yes |
Proposition 3.3.9. Let \( L \) be a finite extension of \( K \). Then we have the following commutative diagrams (where \( {\phi }_{L} \) sends uniformizers to \( {\sigma }_{L} = {\sigma }_{K}^{\left\lbrack {k}_{L} : k\right\rbrack } \in \operatorname{Gal}\left( {{L}^{\mathrm{{unr}}}/L}\right) \) ):\n\n\[ \n\begin{matr... | Proof. We prove the first commutative diagram leaving the other two as exercises. Let \( M \supset L \) be a finite Galois extension of \( K \), it suffices to show the following diagram commutes\n\n\[ \n{L}^{ \times }/{N}_{M/L}\left( {M}^{ \times }\right) \overset{{\rho }_{M/L}}{ \rightarrow }\operatorname{Gal}{\left(... | No |
If \( K \) is a finite extension of \( {\mathbb{Q}}_{p} \), we see \( \left( {{K}^{ \times },{\operatorname{inv}}_{K}}\right) \) is a class formation. | Let \( \left( {A,\text{inv}}\right) \) be a class formation. Let \( L/K \) be a finite Galois extension, \( {\alpha }_{L/K} \mathrel{\text{:=}} \) \( {\operatorname{inv}}_{L/K}^{-1}\left( {1/\left\lbrack {L : K}\right\rbrack }\right) \in {H}^{2}\left( {\operatorname{Gal}\left( {L/K}\right) ,{A}_{L}}\right) \) . By Tate... | No |
Lemma 4.2.3. Let \( L/K \) be a finite separable extension of \( K \), and let \( E \) be the maximal abelian extension of \( K \) in \( L \), then \( {\mathcal{N}}_{L} = {\mathcal{N}}_{E} \) . | Proof. It suffices to show \( {\mathcal{N}}_{E} \subset {\mathcal{N}}_{L} \) . Let \( a \in {\mathcal{N}}_{E} \) . Let \( M \) be a finite Galois extension of \( K \) containing \( L, G \mathrel{\text{:=}} \operatorname{Gal}\left( {M/K}\right), H \mathrel{\text{:=}} \operatorname{Gal}\left( {M/L}\right) \), and let \( ... | Yes |
Proposition 4.2.6. For any finite abelian extensions \( L, M \) of \( K \), the followings hold.\n\n(1) \( {\mathcal{N}}_{L} \cap {\mathcal{N}}_{M} = {\mathcal{N}}_{LM} \) . | Proof. (1) \( \supset \) is clear. We have\n\n\n\nhence \( {\mathcal{N}}_{L} \cap {\mathcal{N}}_{M} = {\mathcal{N}}_{LM} \) . | No |
Let \( K \) be a finite extension of \( {\mathbb{Q}}_{p} \), we show \( \left( {{\bar{K}}^{ \times },\text{inv }}\right) \) is a topological class formation where \( {\bar{K}}^{ \times } \) is equipped with the p-adic topology. | Condition (1) is clear. For (3), one can take \( {U}_{L} \mathrel{\text{:=}} {\mathcal{O}}_{L}^{ \times } \), then (3) follows by considering unramified extensions of \( L \) . For (2), the map \( {L}^{ \times } \rightarrow {L}^{ \times }, x \mapsto {x}^{p} \) has compact kernel. As we knew \( {D}_{L} = 1 \), the secon... | No |
Lemma 4.2.14. Let \( L \) be a finite separable extension of \( K \), then \( {N}_{L/K}{D}_{L} = {D}_{K} \) . | Proof. It is clear that \( {N}_{L/K}{D}_{L} \subset {D}_{K} \) . Let \( a \in {D}_{K} \), and consider \( {N}_{L/K}^{-1}\left( a\right) \), that is a compact subset of \( {A}_{L} \) by Property 1. For any finite separable extension \( M/L \), as \( a \in \) \( {N}_{M/K}\left( {A}_{M}\right) \), we deduce \( {N}_{M/L}\l... | Yes |
Proposition 4.2.15. The group \( {D}_{K} \) is divisible, and \( {D}_{K} = { \cap }_{n}n{A}_{K} \) . | Proof. To show \( {D}_{K} \) is divisible, it suffices to show \( {\phi }_{p} : {D}_{K} \rightarrow {D}_{K}, x \mapsto {px} \) is surjective for any prime number \( p \) . For any \( x \in {D}_{K},{\phi }_{p}^{-1}\left( x\right) \) is closed hence compact. For any finite separable extension \( L/K \), there exists \( y... | Yes |
Lemma 5.1.2. Let \( g \in \operatorname{Gal}\left( {L/K}\right) \), then \( g \) sends \( {L}_{w} \hookrightarrow {K}_{v}{ \otimes }_{K}L \cong \mathop{\prod }\limits_{{w \mid v}}{L}_{w} \) to \( {L}_{g\left( w\right) } \) (where \( {L}_{w} \) is viewed as a \( {K}_{v} \) vector subspace of \( \mathop{\prod }\limits_{{... | Proof. Let \( x = \mathop{\sum }\limits_{i}{a}_{i} \otimes {\alpha }_{i} \in {K}_{v}{ \otimes }_{K}L \), and suppose \( x \) is sent to \( {L}_{w} \hookrightarrow \mathop{\prod }\limits_{{w \mid v}}{L}_{w} \) . Thus \( \mathop{\sum }\limits_{i}{a}_{i}{\iota }_{{w}^{\prime }}\left( {\alpha }_{i}\right) = 0 \) for all \(... | Yes |
Lemma 5.1.3. We have \( {\mathbb{A}}_{L}^{\mathrm{{Gal}}\left( {L/K}\right) } = {\mathbb{A}}_{K} \) and \( {I}_{L}^{\mathrm{{Gal}}\left( {L/K}\right) } = {I}_{K} \) . | Proof. As \( {I}_{L} \cap {\mathbb{A}}_{K} = {I}_{K} \), it suffices to prove the statement for \( {\mathbb{A}}_{L} \) . We first show \( \left( {{K}_{v}{ \otimes }_{K}}\right. \) \( L{)}^{\operatorname{Gal}\left( {L/K}\right) } = {K}_{v} \) for any place \( v \) of \( K \) . The direction \ | No |
Corollary 5.1.4. \( {H}^{0}\left( {\operatorname{Gal}\left( {L/K}\right) ,{\mathbb{C}}_{L}}\right) \cong {\mathbb{C}}_{K} \) . | Proof. Consider the exact sequence \( 1 \rightarrow {L}^{ \times } \rightarrow {I}_{L} \rightarrow {\mathbb{C}}_{L} \rightarrow 1 \) . Taking \( \operatorname{Gal}\left( {L/K}\right) \) - cohomology (and using the above lemma), we obtain\n\n\[ 1 \rightarrow {K}^{ \times } \rightarrow {I}_{K} \rightarrow {\mathbb{C}}_{L... | No |
Lemma 5.2.1. The composition \( {\jmath }_{w} : \operatorname{Gal}\left( {\mathcal{L}/{K}_{v}}\right) \overset{ \sim }{ \rightarrow }\operatorname{Gal}\left( {{L}_{w}/{K}_{v}}\right) \rightarrow \operatorname{Gal}\left( {L/K}\right) \) is independent of the choice of \( w \) . | Proof. Let \( {w}^{\prime } \) be another place of \( L \) dividing \( v,\sigma \in \operatorname{Gal}\left( {L/K}\right) \) such that \( \sigma \left( w\right) = {w}^{\prime } \) . For \( g \in \operatorname{Gal}\left( {\mathcal{L}/{K}_{v}}\right) ,{\jmath }_{w}\left( g\right) : L \rightarrow L \) is the map satisfyin... | Yes |
Lemma 5.2.2. Let \( L/K \) be finite abelian, \( \alpha = \left( {\alpha }_{v}\right) \in {I}_{K} \), then \( {\rho }_{{L}_{w}/{K}_{v}}\left( {\alpha }_{v}\right) = 1 \) for all but finitely many places \( v \) of \( K \) . | Proof. We only need to consider the non-archemedean places. However, for all but finitely many non-archimedean places \( v \), we have \( {\alpha }_{v} \in {\mathcal{O}}_{{K}_{v}}^{ \times } \) and \( {L}_{w}/{K}_{v} \) is unramified. Hence \( {\alpha }_{v} \in {N}_{{L}_{w}/{K}_{v}}\left( {L}_{w}^{ \times }\right) \) a... | Yes |
Lemma 5.2.3. Let \( L \subset M \) be finite abelian extensions of \( K,\alpha \in {I}_{K} \), then \( {\Phi }_{L/K}\left( \alpha \right) = \) \( {\left. {\Phi }_{M/K}\left( \alpha \right) \right| }_{L} \) . | Proof. For any place \( v \) of \( K \), let \( w \mid v \) be a place of \( L \) and \( \widetilde{w} \mid w \) a place of \( M \) . By the local artin reciprocity law, we have \( {\left. {\rho }_{{M}_{\widetilde{w}}/{K}_{v}}\left( {\alpha }_{v}\right) \right| }_{{L}_{w}} = {\rho }_{{L}_{w}/{K}_{v}}\left( {\alpha }_{v... | Yes |
Proposition 5.2.6. Let \( L/K \) be a finite extension. Then the following diagram commutes\n\n\[ \n\begin{matrix} {I}_{L}\overset{{\Phi }_{L}}{ \rightarrow }\operatorname{Gal}{\left( \bar{K}/L\right) }^{\mathrm{{ab}}} & {I}_{K}\overset{{\Phi }_{K}}{ \rightarrow }\operatorname{Gal}{\left( \bar{K}/K\right) }^{\mathrm{{a... | Proof. Exercise. | No |
Proposition 5.2.8. Let \( L/K \) be a finite extension, then \( {N}_{L/K}{I}_{L} \) is an open subgroup of \( {I}_{K} \) . | Proof. Let \( S \) be a finite set of places of \( K \) containing all the archimedean places and those that ramify in \( L \), and \( {S}_{L} \mathrel{\text{:=}} \{ w \mid v, v \in S\} \) . By Lemma 4.2.3, \( {N}_{L/K}{I}_{L,{S}_{L}} = \mathop{\prod }\limits_{{v \in S}}{U}_{v} \times \) \( \mathop{\prod }\limits_{{v \... | No |
Lemma 5.3.1. Suppose \( S \) contains a finite set \( {S}_{0} \) of finite places \( w \) of \( L \) such that \( {\left\{ {\mathfrak{p}}_{w}\right\} }_{w \in {S}_{0}} \) can generate the ideal class group of \( L \), then \( {I}_{L} = {L}^{ \times }{I}_{L, S} \) . | Proof. Recall we have \( {I}_{L}/\left( {{L}^{ \times }\left( {\mathop{\prod }\limits_{{w \mid \infty }}{L}_{w}^{ \times } \times \mathop{\prod }\limits_{{w \nmid \infty }}{\mathcal{O}}_{{L}_{w}}^{ \times }}\right) }\right) \overset{ \sim }{ \rightarrow }{\mathrm{{Cl}}}_{L} \) . By the assumption on \( S \) , the induc... | Yes |
Lemma 5.3.2. We have \( {\left( L{ \otimes }_{K}{K}_{v}\right) }^{ \times } \cong {\operatorname{Ind}}_{{D}_{w}}^{\operatorname{Gal}\left( {L/K}\right) }{L}_{w}^{ \times } \) and \( \mathop{\prod }\limits_{{{w}^{\prime } \mid v}}{\mathcal{O}}_{{w}^{\prime }}^{ \times } \cong {\operatorname{Ind}}_{{D}_{w}}^{\operatornam... | Proof. The natural \( {D}_{w} \) -equivariant injection \( {L}_{w}^{ \times } \hookrightarrow \mathop{\prod }\limits_{{{w}^{\prime } \mid v}}{L}_{{w}^{\prime }}^{ \times } \) (resp. \( {\mathcal{O}}_{w}^{ \times } \hookrightarrow \mathop{\prod }\limits_{{{w}^{\prime } \mid v}}{\mathcal{O}}_{{w}^{\prime }}^{ \times } \)... | Yes |
Proposition 5.3.4. Let \( S \) be a finite set of places of \( K \) containing all the archimedean places and the places that ramify in \( L/K \) . Then we have for all \( i \in \mathbb{Z} \) :\n\n\[ \n{H}_{T}^{i}\left( {\operatorname{Gal}\left( {L/K}\right) ,{I}_{L, S}}\right) \cong { \oplus }_{v \in S}{H}_{T}^{i}\lef... | Proof. First group cohomology commutes with product (one can see this using cochains, or using the fact that a direct product of injective objects is still injective): \( {H}^{i}\left( {G,\mathop{\prod }\limits_{i}{M}_{i}}\right) \cong \) \( \mathop{\prod }\limits_{i}{H}^{i}\left( {G,{M}_{i}}\right) \) . We also have \... | Yes |
Proposition 5.3.6. Whave \( {H}_{T}^{i}\left( {\operatorname{Gal}\left( {L/K}\right) ,{I}_{L}}\right) \cong { \oplus }_{v}{H}_{T}^{i}\left( {{D}_{w},{L}_{w}^{ \times }}\right) \) . | Proof. We have \( {I}_{L} = \mathop{\lim }\limits_{{ \rightarrow S}}{I}_{L, S} \) where \( S \) runs through finite set of places of \( K \) satisfying the condition in Proposition 5.3.4. The proposition then follows from Proposition 5.3.4 and the fact that Tate cohomology commutes with direct limit: taking (finite) gr... | No |
Corollary 5.3.11. Let \( L/K \) be a finite abelian extension. Then there exists infinitely many places of \( K \) that do not split completely in \( L \) . | Proof. It is easy to reduce to the cyclic case. Assume hence \( L/K \) cyclic. Suppose the statement does not hold. Let \( S \) be the finite set of places of \( K \) containing \( {S}_{\infty } \), the places that ramify in \( L \) and the places that do not split. For any \( v \notin S, w \mid v \), we have \( {L}_{w... | Yes |
Corollary 5.3.12. Let \( L/K \) be a finite abelian extension, \( S \) be a finite set of places of \( K \) containing \( {S}_{\infty } \) and those that ramify in \( L \) . Then \( {\left\{ {\operatorname{Frob}}_{v}\right\} }_{v \notin S} \) generated \( \operatorname{Gal}\left( {L/K}\right) \) . | Proof. Let \( H \) be the subgroup generated by \( {\left\{ {\operatorname{Frob}}_{v}\right\} }_{v \notin S} \), and \( M \mathrel{\text{:=}} {L}^{H} \) . By assumption, for all \( v \notin S \), and \( w \mid v \) a place in \( M \), we have \( {M}_{w} = {K}_{v} \) . Thus any \( v \notin S \) splits in \( M \) . By th... | Yes |
Corollary 5.3.13. Let \( {L}_{1},\cdots ,{L}_{t} \) be cyclic extensions of \( K \) of prime degree \( p \), such that each \( {L}_{i} \) is disjoint from the composition of \( {L}_{j} \) for \( j \neq i \) . Then there are infinitely many places of \( K \) that are inert in \( {L}_{1} \) and splits completely in \( {L... | Proof. Let \( L \mathrel{\text{:=}} {L}_{1}\cdots {L}_{t} \) and \( {L}^{\prime } \mathrel{\text{:=}} {L}_{2}\cdots {L}_{t} \), then \( \operatorname{Gal}\left( {L/{L}^{\prime }}\right) \cong \mathbb{Z}/p\mathbb{Z} \) . By the above corollary, there exists infinitely many finite places \( w \) of \( {L}^{\prime } \) su... | Yes |
Lemma 5.4.4. Let \( L/K \) be cyclic of order \( p \), suppose the statements in Theorem holds for the extension \( L\left( {\zeta }_{p}\right) /K\left( {\zeta }_{p}\right) \), then they hold for \( L/K \) . | Proof. As \( p \nmid \left\lbrack {K\left( {\zeta }_{p}\right) : K}\right\rbrack = : d,\left\lbrack {L\left( {\zeta }_{p}\right) : K\left( {\zeta }_{p}\right) }\right\rbrack = p \) and \( \operatorname{Gal}\left( {L\left( {\zeta }_{p}\right) /K\left( {\zeta }_{p}\right) }\right) \overset{ \sim }{ \rightarrow }\operator... | Yes |
Lemma 5.4.5. Let \( \alpha \in {K}^{ \times } \) and \( v \) be a non-archimedean place of \( K, v \nmid p \) . Then \( {K}_{v}\left( {\alpha }^{\frac{1}{p}}\right) \) is unramified over \( {K}_{v} \) if and only if there exist \( {x}_{v} \in {\mathcal{O}}_{v}^{ \times },{y}_{v} \in {\left( {K}_{v}^{ \times }\right) }^... | Proof. The \ | No |
Lemma 5.4.6. Let \( \Delta \mathrel{\text{:=}} {\left( {L}^{ \times }\right) }^{p} \cap {O}_{K, S}^{ \times } \) . Then \( L = K\left( {\Delta }^{\frac{1}{p}}\right) \) . | Proof. Let \( x \in {\left( {L}^{ \times }\right) }^{p} \cap {K}^{ \times } \) such that \( L = K\left( {x}^{\frac{1}{p}}\right) \) . For \( v \notin S,{K}_{v}\left( {x}^{\frac{1}{p}}\right) \) is unramified over \( {K}_{v} \) . So \( {\operatorname{val}}_{{K}_{v}}\left( x\right) = {\operatorname{val}}_{{K}_{v}\left( {... | Yes |
Lemma 5.4.9. We have \( \left\lbrack {{I}_{K} : {K}^{ \times }{J}_{K, S, T}}\right\rbrack = p \) . | Proof. Note \( {I}_{K} = {K}^{ \times }{I}_{K, S \cup T} \) . We have an exact sequence\n\n\[ 1 \rightarrow \frac{{I}_{K, S \cup T} \cap \left( {{K}^{ \times }{J}_{K, S, T}}\right) }{{J}_{K, S, T}} \rightarrow \frac{{I}_{K, S \cup T}}{{J}_{K, S, T}} \rightarrow \frac{{K}^{ \times }{I}_{K, S \cup T}}{{K}^{ \times }{J}_{... | Yes |
Lemma 5.4.10. We have \( {J}_{K, S, T} \cap {K}^{ \times } = {\left( {O}_{K, S \cup T}^{ \times }\right) }^{p} \) . | Proof. \ | No |
For \( \alpha \in {I}_{K},\chi \in \operatorname{Hom}\left( {\operatorname{Gal}\left( {L/K}\right) ,\mathbb{Q}/\mathbb{Z}}\right) \overset{\delta }{ \rightarrow }{H}^{2}\left( {\operatorname{Gal}\left( {L/K}\right) ,\mathbb{Z}}\right) \), we have \[ {\operatorname{inv}}_{L/K}\left( {\bar{\alpha } \cup \delta \left( \ch... | For a place \( v \) of \( K \) and \( w \mid v \), we have commutative diagrams \[ {H}_{T}^{0}\left( {\operatorname{Gal}\left( {{L}_{w}/{K}_{v}}\right) ,\mathop{\prod }\limits_{{{w}^{\prime } \mid v}}{L}_{{w}^{\prime }}^{ \times }}\right) \times {H}_{T}^{2}\left( {\operatorname{Gal}\left( {{L}_{w}/{K}_{v}}\right) ,\mat... | Yes |
For any \( n \in {\mathbb{Z}}_{ \geq 1}, a \in {\mathbb{Q}}^{ \times } \), we have \( {\Phi }_{\mathbb{Q}}\left( a\right) \left( {\zeta }_{n}\right) = {\zeta }_{n} \). | It sufficient to prove the case where \( a \) is a prime number \( q \) or -1, and \( n = {p}^{k} \) . First suppose \( q \neq p \):\n\n- for a prime number \( \ell \) different from \( p \) and \( q \), the extension \( {\mathbb{Q}}_{\ell }\left( {\zeta }_{{p}^{k}}\right) \) is unramified, and \( q \in {\mathbb{Z}}_{\... | Yes |
Lemma 5.5.3. Suppose \( L \subset K\left( {\zeta }_{n}\right) \), then \( {\Phi }_{L/K}\left( a\right) = 1 \) for all \( a \in {K}^{ \times } \) . | Proof. By Proposition 5.2.6, we have \( {\left. {\Phi }_{K\left( {\zeta }_{n}\right) /K}\left( a\right) \right| }_{\mathbb{Q}\left( {\zeta }_{n}\right) } = {\Phi }_{\mathbb{Q}\left( {\zeta }_{n}\right) /\mathbb{Q}}\left( {{N}_{K/\mathbb{Q}}\left( a\right) }\right) \) . | No |
Lemma 5.5.4. Let \( G \) be a finite cyclic group of order \( n,\chi : G \rightarrow \mathbb{Q}/Z \), and \( \delta \) be the connecting map for \( 0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Q} \rightarrow \mathbb{Q}/\mathbb{Z} \rightarrow 0 \) . Let \( g \) be a generator of \( G \), viewed as an element, denote by ... | Proof. The lemma follows from Lemma 3.3.2 and the following commutative diagram:\n\n\[\n\begin{array}{l} {H}_{T}^{-2}\left( {G,\mathbb{Z}}\right) \times {H}_{T}^{2}\left( {G,\mathbb{Z}}\right) \overset{ \cup }{ \rightarrow }{H}_{T}^{0}\left( {G,\mathbb{Z}}\right) \overset{ \sim }{ \rightarrow }{H}_{T}/n\mathbb{Z} \\ \p... | Yes |
Corollary 5.5.5. Keep the situation as in the above lemma, and suppose \( \chi \) is injective. Let \( A \) be a \( G \) -module, then the map \[ {H}_{T}^{i}\left( {G, A}\right) \rightarrow {H}_{T}^{i + 2}\left( {G, A}\right), c \mapsto \delta \left( \chi \right) \cup c \] is an isomorphism. | Proof. The composition \( {H}_{T}^{i + 2}\left( {G, A}\right) \overset{\cup {u}_{g}}{ \rightarrow }{H}_{T}^{i}\left( {G, A}\right) \overset{\cup \delta \left( \chi \right) }{ \rightarrow }{H}_{T}^{i + 2}\left( {G, A}\right) \) is given by \( c \mapsto \) \( c \cup \left( {{u}_{g} \cup \delta \left( \chi \right) }\right... | Yes |
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