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](https://mathoverflow.net/a/239461/4149). There is a comment of nfdc to that answer, but I think that this question deserves an answer rather than just a comment! 



Infiniteness of the Galois cohomology over a number field with coefficients in a finite Galois module
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. 


