Let $k$ be a number field and $M$ be a nonzero finite discrete $\mathrm{Gal}(\bar k/k)$-module. Is it true that $H^1(k,M)$ is infinite?
This would complete the answer of Daniel Loughran. There is a comment of nfdc to that answer, but I think that this question deserves an answer rather than just a comment!
In fact, $H^1(K,M)$ is infinite not only to the family of number fields, but also to the more general family of Hilbertian fields (those fields which satisfies the Hilbert irreducibility theorem). Below is a proof based on embedding problems:
Let $K$ be Hilbertian with absolute Galois group $G_K$ and $M$ a finite nontrivial $G_K$module.
A finite embedding problem for a field $K$ is a pair of two epimorphisms (all morphisms in this answer are assumed to continuous) $$ \mathcal{E}=(\alpha\colon G_K\to G ; \beta \colon \Gamma\to G) $$ with $\Gamma$ finite. A solution of $\mathcal{E}$, in this answer, is an epimorphism $\psi\colon G_K \to \Gamma$ with $\alpha = \beta\circ \psi$.
If $M$ is a finite $G_K$-module, then from continuity of the action, there exists an open subgroup $U$ of $G_K$ that acts trivially on $M$, and so the action factors through $G:=G_K/U$; let $\alpha\colon G_K\to G$ be the quotient map. Then, we have the semi-direct product $\Gamma = M\rtimes G$ and the projection map $\beta\colon \Gamma\to G$, which form the embedding problem $$ (1)\qquad \mathcal{E}=(\alpha\colon G_K \to G,\beta\colon \Gamma\to G), \qquad \Gamma =M\rtimes G, $$ We note that $\mathcal{E}$ is split in the sense that $\beta$ has a group theoretical section and $\mathcal{E}$ has an abelian kernel in the sense that $M = \ker\beta$ is abelian.
In a similar fashion, for any $n>0$, we may construct the embedding problem $$ \mathcal{E}_n = (\alpha\colon G_K ; \beta_n \colon \Gamma_n \to G), $$ where $\Gamma_n = M^n\rtimes G$ and $\beta_n$ is the quotient map. To be precise, here $G$ acts on $M^n$ coordinate-wise; i.e., $g.(m_1,\ldots, m_n) := (g.m_1,\ldots, g.m_n)$.
Theorem (Ikeda, [Fried-Jarden, Field Arithmetic 2nd edtn, Prop. 16.4.5]) If $K$ is Hilbertian, then any split embedding problem with abelian kernel has a solution; in particular, $\mathcal{E}_n$ has a solution $\Psi \colon G_K\to \Gamma_n$.
We note that $\Psi$ gives rise to $n$ distinct solutions of $\mathcal{E}$:
Proposition Under the notation above, let $\pi_i\colon \Gamma_n\to \Gamma$ be the projection map given by $\pi_i((m_1,\ldots,m_n)g) = (m_i,g)$. Then,
a) $\psi_i = \pi_i\circ \Psi$ is a solution of $\mathcal{E}$.
b) For all $i\neq j$ one has $\ker \psi_i \ker \psi_j = \ker \alpha$, and, in particular, if $M\neq 0$, then $\ker \psi_i\neq \ker \psi_j$.
Proof. a) is immediate.
For b), put $M'_i = \ker \pi$, so $M'_i = \{(m_1,\ldots, m_n)\in M^n : m_i=0\}$. Then if $i\neq j$, then \begin{equation}\label{eq:EQ} (2) \qquad M_i'M_j'=M^n. \end{equation} Note that we have $$ (3) \Psi^{-1}(M_i') =\ker \psi_i \quad \mbox{and} \quad \Psi^{-1}(M^n) = \Psi^{-1}(\ker\beta_n) = \ker (\beta_n\circ \Psi) = \ker \alpha, $$
Since $\Psi\colon G_K\to \Gamma_n$ is surjective, it induces a bijection between subgroups of $G_K$ containing $\ker \Psi$ and subgroups of $\Gamma_n$; namely, $H\leq \Gamma_n$ corresponds to $\Psi^{-1}(H)$. So (2) and (3) give that $$(4)\qquad \ker \psi_i \ker \psi_j = \ker \alpha. $$ If $M\neq 0$, then $M_i'\neq M^n$, and so $\ker \psi_i\lneq \ker\alpha$. Thus, (4) implies that $\ker\psi_i\neq \ker\psi_j$, and in particular $\psi_i\neq \psi_j$. QED
Corollary If $M\neq 0$, then the embedding problem $\mathcal{E}$ given in (1) has infinitely many solutions.
Each solution $\psi$ of $\mathcal{E}$ given in (1) induces a $1$-cocyle in $c\in H^1(K,M)$. Indeed, $c$ is defined by
$$
\psi(g) = (c(g),\alpha(g))\in \Gamma
$$
(recall that $G=G_K/U$ and $\alpha\colon G_K\to G$ is the quotient map) or equivalently by
$$
c(g) = \psi(g) \alpha(g)^{-1}.
$$
Obviously, these formulas imply that the map from solutions to cocycles is injective, and so the Corollary gives infinitely many cocycles, and so $H^1(K,M)$ is also infinite.
An easy proof:
Let $L/K$ be a Galois extension with Galois group $G$ such that $M$ becomes trivial over $L$. Thus we can also think of $M$ as a $G$-module.
The kernel of $H^1(K,M[p]) \rightarrow H^1(K,M)$ is a quotient of $H^0(K,M/M[p])$ which is finite, so it suffices to assume that $M = M[p]$ for some prime $p$.
By inflation restriction, one is reduced to showing that $H^1(L,M)^{G}$ is infinite; the boundary terms $H^1(G,M^{G})$ and $H^2(G,M^G)$ are finite.
As a $G_{L}$-module, $M = M[p] = (\mathbb{Z}/p\mathbb{Z})^n$ for some $n$. Thus $$H^1(G_L,M) = \mathrm{Hom}(G_L,M) = \mathrm{Hom}(G_L,\mathbb{Z}/p\mathbb{Z})^n.$$ What we want to show is that this contains many copies for which the action of $G$ corresponds to the action of $G$ on $M$.
It suffices to show that $\mathrm{Hom}(G_L,\mathbb{Z}/p\mathbb{Z})$ contains infinitely many copies of the regular representation $\mathbb{Z}/p\mathbb{Z}[G]$ of $G$.
Let $S$ be a collection of primes $q$ which have the property that they are $1$ modulo $p$ and that they all split completely in $M$. I will show that $\mathrm{Hom}(G_{L,S},\mathbb{Z}/p \mathbb{Z})$, for $S$ sufficiently large, contains many copies of the regular representation $\mathbb{Z}/p \mathbb{Z}[G]$.
The assumption on $q$ implies that $(\mathcal{O}_L/q)^{\times} \otimes \mathbb{Z}/p \mathbb{Z}$ is isomorphic to $\mathbb{Z}/p \mathbb{Z}[G]$.
By class field theory, there is an exact sequence:
$$\mathcal{O}^{\times}_L \rightarrow \prod_S (\mathcal{O}_L/q)^{\times} \rightarrow \Gamma_S \rightarrow \mathrm{Cl}(\mathcal{O}_L) \rightarrow 0,$$
where $\Gamma_S$ is the ray class group of $L$ of conductor $S$. The first and last terms are finitely generated, so it's easy enough to see that $\Gamma_S/p \Gamma_S$ (which is dual to $\mathrm{Hom}(G_{L,S},\mathbb{Z}/p\mathbb{Z})$) contains at least $|S| - C$ copies of the regular representation $\mathbb{Z}/p\mathbb{Z}[G]$ of $G$ for some $C$ depending only on $p$ and $L$.
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.
