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Let $k$ be a number field. Let $M$ be a nonzero finite $\mathrm{Gal}({\bar k}/k)$-module, i.e., a nonzero discrete finite abelian group with a continuous action of $\mathrm{Gal}({\bar k}/k)$.

Theorem. The group $H^1(k,M)$ is infinite.

In this proof I use Chebotarev's theorem (of course!) and a finiteness/surjectivity result from Sansuc's paper.

Let $M'=\mathrm{Hom}(M,\mathbb{G}_{m,k})$ denote the Cartier dual of $M$. The action of $\mathrm{Gal}({\bar k}/k)$ on $M'$ defines a continuous homomorphism $\varphi\colon \mathrm{Gal}({\bar k}/k)\to\mathrm{Aut}\, M'$. We denote by $\Gamma$ the image of $\varphi$ and by $K\subset {\bar k}$ the subfield corresponding to $\ker\varphi$, then $\Gamma=\mathrm{Gal}(K/k)$.

For a finite set $S$ of places of $k$ we consider the localization homomorphism $$ \lambda_S\colon H^1(k,M)\to\prod_{v\in S} H^1(k_v,M), $$ where $k_v$ denotes the completion of $k$ at $v$. We set $$Ч_S=\mathrm{coker}\,\lambda_S$$ ( Ч is to be read "Cheh" as in "Chebotarev").

Let $S_0$ denote the set of places $v$ of $k$ for which a decomposition group $\Gamma_v\subset \Gamma$ (defined up to conjugation) is not cyclic. All $v\in S_0$ are ramified in $K$.

Lemma. For any finite set $S$ of places of $k$, the canonical surjective map $$ Ч_S\to Ч_{S\cap S_0} $$ is bijective.

See Sansuc's paper, the formula after Lemma 1.5. It follows from the lemma that if $S\cap S_0=\emptyset$, then $Ч_S=0$ and hence, the homomorphism $\lambda_S$ is surjective in this case (see also Theorem 9.2.3(vii) from the book by Neukirch, Schmidt, and Wingberg "Cohomology of number fields").

Proof of the theorem. By the Chebotarev density theorem, for any natural $n$ there exists a finite set $S$ of cardinality $n$ consisting of finite places $v$ of $k$ with $\Gamma_v=1$. Clearly $S\cap S_0=\emptyset$, hence the localization homomorphism $\lambda_S$ is surjective. For $v\in S$, the local Galois group $\mathrm{Gal}({\bar k}_v/k_v)$ acts on $M$ trivially. Since $\mathrm{Gal}({\bar k}_v/k_v)$ has $\widehat{\mathbb{Z}}$ as a quotient group, we have $$\# H^1(k_v,M)=\#\mathrm{Hom}(\mathrm{Gal}({\bar k}_v/k_v),M)\ge \#\mathrm{Hom}(\widehat{\mathbb{Z}},M)=\# M.$$ Since $\lambda_S$ is surjective, $$\mathrm{Card}\, H^1(k,M)\ge \# \prod_{v\in S} H^1(k_v,M)\ge(\#M)^n.$$ Since this is true for any natural $n$, and $\# M>1$, we conclude that $H^1(k,M)$ is infinite.

Let $k$ be a number field. Let $M$ be a nonzero finite $\mathrm{Gal}({\bar k}/k)$-module, i.e., a nonzero discrete finite abelian group with a continuous action of $\mathrm{Gal}({\bar k}/k)$.

Theorem. The group $H^1(k,M)$ is infinite.

In this proof I use Chebotarev's theorem (of course!) and a finiteness/surjectivity result from Sansuc's paper.

Let $M'=\mathrm{Hom}(M,\mathbb{G}_{m,k})$ denote the Cartier dual of $M$. The action of $\mathrm{Gal}({\bar k}/k)$ on $M'$ defines a continuous homomorphism $\varphi\colon \mathrm{Gal}({\bar k}/k)\to\mathrm{Aut}\, M'$. We denote by $\Gamma$ the image of $\varphi$ and by $K\subset {\bar k}$ the subfield corresponding to $\ker\varphi$, then $\Gamma=\mathrm{Gal}(K/k)$.

For a finite set $S$ of places of $k$ we consider the localization homomorphism $$ \lambda_S\colon H^1(k,M)\to\prod_{v\in S} H^1(k_v,M), $$ where $k_v$ denotes the completion of $k$ at $v$. We set $$Ч_S=\mathrm{coker}\,\lambda_S$$ ( Ч is to be read "Cheh" as in "Chebotarev").

Let $S_0$ denote the set of places $v$ of $k$ for which a decomposition group $\Gamma_v\subset \Gamma$ (defined up to conjugation) is not cyclic. All $v\in S_0$ are ramified in $K$.

Lemma. For any finite set $S$ of places of $k$, the canonical surjective map $$ Ч_S\to Ч_{S\cap S_0} $$ is bijective.

See Sansuc, J.-J. Groupe de Brauer et arithmétique des groupes algébriques linéaires sur un corps de nombres. J. Reine Angew. Math. 327 (1981), 12–80, the formula after Lemma 1.5. It follows from the lemma that if $S\cap S_0=\emptyset$, then $Ч_S=0$ and hence, the homomorphism $\lambda_S$ is surjective in this case (see also Theorem 9.2.3(vii) from the book by Neukirch, Schmidt, and Wingberg "Cohomology of number fields").

Proof of the theorem. By the Chebotarev density theorem, for any natural $n$ there exists a finite set $S$ of cardinality $n$ consisting of finite places $v$ of $k$ with $\Gamma_v=1$. Clearly $S\cap S_0=\emptyset$, hence the localization homomorphism $\lambda_S$ is surjective. For $v\in S$, the local Galois group $\mathrm{Gal}({\bar k}_v/k_v)$ acts on $M$ trivially. Since $\mathrm{Gal}({\bar k}_v/k_v)$ has $\widehat{\mathbb{Z}}$ as a quotient group, we have $$\# H^1(k_v,M)=\#\mathrm{Hom}(\mathrm{Gal}({\bar k}_v/k_v),M)\ge \#\mathrm{Hom}(\widehat{\mathbb{Z}},M)=\# M.$$ Since $\lambda_S$ is surjective, $$\mathrm{Card}\, H^1(k,M)\ge \# \prod_{v\in S} H^1(k_v,M)\ge(\#M)^n.$$ Since this is true for any natural $n$, and $\# M>1$, we conclude that $H^1(k,M)$ is infinite.

Let $k$ be a number field. Let $M$ be a nonzero finite $\mathrm{Gal}({\bar k}/k)$-module, i.e., a nonzero discrete finite abelian group with a continuous action of $\mathrm{Gal}({\bar k}/k)$.

Theorem. The group $H^1(k,M)$ is infinite.

In this proof I use Chebotarev's theorem (of course!) and a finiteness/surjectivity result from Sansuc's paper.

The action of $\mathrm{Gal}({\bar k}/k)$ on $M$ defines a continuous homomorphism $\varphi\colon \mathrm{Gal}({\bar k}/k)\to\mathrm{Aut}\, M$. We denote by $\Gamma$ the image of $\varphi$ and by $K\subset {\bar k}$ the subfield corresponding to $\ker\varphi$, then $\Gamma=\mathrm{Gal}(K/k)$.

For a finite set $S$ of places of $k$ we consider the localization homomorphism $$ \lambda_S\colon H^1(k,M)\to\prod_{v\in S} H^1(k_v,M), $$ where $k_v$ denotes the completion of $k$ at $v$. We set $$Ч_S=\mathrm{coker}\,\lambda_S$$ ( Ч is to be read "Cheh" as in "Chebotarev").

Let $S_0$ denote the set of places $v$ of $k$ for which a decomposition group $\Gamma_v\subset \Gamma$ (defined up to conjugation) is not cyclic. All $v\in S_0$ are ramified in $K$.

Lemma. For any finite set $S$ of places of $k$, the canonical surjective map $$ Ч_S\to Ч_{S\cap S_0} $$ is bijective.

See Sansuc's paper, the formula after Lemma 1.5. It follows from the lemma that if $S\cap S_0=\emptyset$, then $Ч_S=0$ and hence, the homomorphism $\lambda_S$ is surjective in this case.

Proof of the theorem. By the Chebotarev density theorem, for any natural $n$ there exists a finite set $S$ of cardinality $n$ consisting of finite places $v$ of $k$ with $\Gamma_v=1$. Clearly $S\cap S_0=\emptyset$, hence the localization homomorphism $\lambda_S$ is surjective. For $v\in S$, the local Galois group $\mathrm{Gal}({\bar k}_v/k_v)$ acts on $M$ trivially. Since $\mathrm{Gal}({\bar k}_v/k_v)$ has $\widehat{\mathbb{Z}}$ as a quotient group, we have $$\# H^1(k_v,M)=\#\mathrm{Hom}(\mathrm{Gal}({\bar k}_v/k_v),M)\ge \#\mathrm{Hom}(\widehat{\mathbb{Z}},M)=\# M.$$ Since $\lambda_S$ is surjective, $$\mathrm{Card}\, H^1(k,M)\ge \# \prod_{v\in S} H^1(k_v,M)\ge(\#M)^n.$$ Since this is true for any natural $n$, and $\# M>1$, we conclude that $H^1(k,M)$ is infinite.

Let $k$ be a number field. Let $M$ be a nonzero finite $\mathrm{Gal}({\bar k}/k)$-module, i.e., a nonzero discrete finite abelian group with a continuous action of $\mathrm{Gal}({\bar k}/k)$.

Theorem. The group $H^1(k,M)$ is infinite.

In this proof I use Chebotarev's theorem (of course!) and a finiteness/surjectivity result from Sansuc's paper.

Let $M'=\mathrm{Hom}(M,\mathbb{G}_{m,k})$ denote the Cartier dual of $M$. The action of $\mathrm{Gal}({\bar k}/k)$ on $M'$ defines a continuous homomorphism $\varphi\colon \mathrm{Gal}({\bar k}/k)\to\mathrm{Aut}\, M'$. We denote by $\Gamma$ the image of $\varphi$ and by $K\subset {\bar k}$ the subfield corresponding to $\ker\varphi$, then $\Gamma=\mathrm{Gal}(K/k)$.

For a finite set $S$ of places of $k$ we consider the localization homomorphism $$ \lambda_S\colon H^1(k,M)\to\prod_{v\in S} H^1(k_v,M), $$ where $k_v$ denotes the completion of $k$ at $v$. We set $$Ч_S=\mathrm{coker}\,\lambda_S$$ ( Ч is to be read "Cheh" as in "Chebotarev").

Let $S_0$ denote the set of places $v$ of $k$ for which a decomposition group $\Gamma_v\subset \Gamma$ (defined up to conjugation) is not cyclic. All $v\in S_0$ are ramified in $K$.

Lemma. For any finite set $S$ of places of $k$, the canonical surjective map $$ Ч_S\to Ч_{S\cap S_0} $$ is bijective.

See Sansuc's paper, the formula after Lemma 1.5. It follows from the lemma that if $S\cap S_0=\emptyset$, then $Ч_S=0$ and hence, the homomorphism $\lambda_S$ is surjective in this case (see also Theorem 9.2.3(vii) from the book by Neukirch, Schmidt, and Wingberg "Cohomology of number fields").

Proof of the theorem. By the Chebotarev density theorem, for any natural $n$ there exists a finite set $S$ of cardinality $n$ consisting of finite places $v$ of $k$ with $\Gamma_v=1$. Clearly $S\cap S_0=\emptyset$, hence the localization homomorphism $\lambda_S$ is surjective. For $v\in S$, the local Galois group $\mathrm{Gal}({\bar k}_v/k_v)$ acts on $M$ trivially. Since $\mathrm{Gal}({\bar k}_v/k_v)$ has $\widehat{\mathbb{Z}}$ as a quotient group, we have $$\# H^1(k_v,M)=\#\mathrm{Hom}(\mathrm{Gal}({\bar k}_v/k_v),M)\ge \#\mathrm{Hom}(\widehat{\mathbb{Z}},M)=\# M.$$ Since $\lambda_S$ is surjective, $$\mathrm{Card}\, H^1(k,M)\ge \# \prod_{v\in S} H^1(k_v,M)\ge(\#M)^n.$$ Since this is true for any natural $n$, and $\# M>1$, we conclude that $H^1(k,M)$ is infinite.

Let $k$ be a number field. Let $M$ be a nonzero finite $\mathrm{Gal}({\bar k}/k)$-module, i.e., a nonzero discrete finite abelian group with a continuous action of $\mathrm{Gal}({\bar k}/k)$.

Theorem. The group $H^1(k,M)$ is infinite.

The action of $\mathrm{Gal}({\bar k}/k)$ on $M$ defines a continuous homomorphism $\varphi\colon \mathrm{Gal}({\bar k}/k)\to\mathrm{Aut}\, M$. We denote by $\Gamma$ the image of $\varphi$ and by $K\subset {\bar k}$ the subfield corresponding to $\ker\varphi$, then $\Gamma=\mathrm{Gal}(K/k)$.

For a finite set $S$ of places of $k$ we consider the localization homomorphism $$ \lambda_S\colon H^1(k,M)\to\prod_{v\in S} H^1(k_v,M), $$ where $k_v$ denotes the completion of $k$ at $v$. We set $$Ч_S=\mathrm{coker}\,\lambda_S$$ ( Ч is to be read "Cheh" as in "Chebotarev").

Let $S_0$ denote the set of places $v$ of $k$ for which a decomposition group $\Gamma_v\subset \Gamma$ (defined up to conjugation) is not cyclic. All $v\in S_0$ are ramified in $K$.

Lemma. For any finite set $S$ of places of $k$, the canonical surjective map $$ Ч_S\to Ч_{S\cap S_0} $$ is bijective.

See Sansuc's paper, the formula after Lemma 1.5. It follows from the lemma that if $S\cap S_0=\emptyset$, then $Ч_S=0$ and hence, the homomorphism $\lambda_S$ is surjective in this case.

Proof of the theorem. By the Chebotarev density theorem, for any natural $n$ there exists a finite set $S$ of cardinality $n$ consisting of finite places $v$ of $k$ with $\Gamma_v=1$. Clearly $S\cap S_0=\emptyset$, hence the localization homomorphism $\lambda_S$ is surjective. For $v\in S$, the local Galois group $\mathrm{Gal}({\bar k}_v/k_v)$ acts on $M$ trivially. Since $\mathrm{Gal}({\bar k}_v/k_v)$ has $\widehat{\mathbb{Z}}$ as a quotient group, we have $$\# H^1(k_v,M)=\#\mathrm{Hom}(\mathrm{Gal}({\bar k}_v/k_v),M)\ge \#\mathrm{Hom}(\widehat{\mathbb{Z}},M)=\# M.$$ Since $\lambda_S$ is surjective, $$\mathrm{Card}\, H^1(k,M)\ge \# \prod_{v\in S} H^1(k_v,M)\ge(\#M)^n.$$ Since this is true for any natural $n$, and $\# M>1$, we conclude that $H^1(k,M)$ is infinite.

Let $k$ be a number field. Let $M$ be a nonzero finite $\mathrm{Gal}({\bar k}/k)$-module, i.e., a nonzero discrete finite abelian group with a continuous action of $\mathrm{Gal}({\bar k}/k)$.

Theorem. The group $H^1(k,M)$ is infinite.

In this proof I use Chebotarev's theorem (of course!) and a finiteness/surjectivity result from Sansuc's paper.

The action of $\mathrm{Gal}({\bar k}/k)$ on $M$ defines a continuous homomorphism $\varphi\colon \mathrm{Gal}({\bar k}/k)\to\mathrm{Aut}\, M$. We denote by $\Gamma$ the image of $\varphi$ and by $K\subset {\bar k}$ the subfield corresponding to $\ker\varphi$, then $\Gamma=\mathrm{Gal}(K/k)$.

For a finite set $S$ of places of $k$ we consider the localization homomorphism $$ \lambda_S\colon H^1(k,M)\to\prod_{v\in S} H^1(k_v,M), $$ where $k_v$ denotes the completion of $k$ at $v$. We set $$Ч_S=\mathrm{coker}\,\lambda_S$$ ( Ч is to be read "Cheh" as in "Chebotarev").

Let $S_0$ denote the set of places $v$ of $k$ for which a decomposition group $\Gamma_v\subset \Gamma$ (defined up to conjugation) is not cyclic. All $v\in S_0$ are ramified in $K$.

Lemma. For any finite set $S$ of places of $k$, the canonical surjective map $$ Ч_S\to Ч_{S\cap S_0} $$ is bijective.

See Sansuc's paper, the formula after Lemma 1.5. It follows from the lemma that if $S\cap S_0=\emptyset$, then $Ч_S=0$ and hence, the homomorphism $\lambda_S$ is surjective in this case.

Proof of the theorem. By the Chebotarev density theorem, for any natural $n$ there exists a finite set $S$ of cardinality $n$ consisting of finite places $v$ of $k$ with $\Gamma_v=1$. Clearly $S\cap S_0=\emptyset$, hence the localization homomorphism $\lambda_S$ is surjective. For $v\in S$, the local Galois group $\mathrm{Gal}({\bar k}_v/k_v)$ acts on $M$ trivially. Since $\mathrm{Gal}({\bar k}_v/k_v)$ has $\widehat{\mathbb{Z}}$ as a quotient group, we have $$\# H^1(k_v,M)=\#\mathrm{Hom}(\mathrm{Gal}({\bar k}_v/k_v),M)\ge \#\mathrm{Hom}(\widehat{\mathbb{Z}},M)=\# M.$$ Since $\lambda_S$ is surjective, $$\mathrm{Card}\, H^1(k,M)\ge \# \prod_{v\in S} H^1(k_v,M)\ge(\#M)^n.$$ Since this is true for any natural $n$, and $\# M>1$, we conclude that $H^1(k,M)$ is infinite.