Galois groups over p-adic fields Let $k$ be a finite Galois extension of $\mathbb{Q}_p$.
Do we know what groups $G$ show up as $G=Gal(k/\mathbb{Q}_p)$, as $p$ varies?
 A: The inverse Galois problem is true for $K=\mathbb{C}(t)$, i.e., any finite group $G$ occurs as a Galois group over $K$. If $K$ is a $p$-adic field, then this is not true, since any polynomial over $K$ is solvable and hence only solvable groups can occur. It is not known which solvable groups exactly do occur. Several conditions are given in the answer of this MO-question, which is linked to the one Franz Lemmermeyer has given. For example, one restriction is, e.g. that the groups occuring as a Galois extension of $K/\mathbf{Q}_p$ have to be generated by $\leq n(K)$ elements for some $n(K) \in \mathbf{N}$ depending on $K$.
A: As is well known, ramification theory implies that the Galois group of a finite normal extension of local fields must be solvable. In the global case, Shafarevitch’s famous theorem (proved by intricate techniques of embedding problems) asserts that any solvable group can be realized as the Galois group of an extension of $\mathbf Q$. It is all the more disappointing that the local analogue over $\mathbf Q_{p}$ is not true, because $\mathbf Q_{p}$ has not enough « degrees of liberty ». Explanation : for clarity, it’s better to start from a p-adic field K, of degree N over $\mathbf Q_{p}$ ; the first classical family of solvable groups being the q-groups,  where q is a prime, equal to p or not, let us look for (normal) q-extensions L/K. To catch all these extensions, put them in a big net : consider the maximal pro-q-extension of K, with Galois group G(K). We have a good hold on the description of G(K) by (topological) generators and relations. For simplicity, suppose p and q odd. Then the minimal number of generators (resp. relations) is the $\mathbf F_{q}$-dimension of $H^1(G(K), \mathbf F_{q})$ (resp. of $H^2$). The $H^1$ is not hard to calculate, its dimension is equal to 1 + $\delta$ if q $\neq p$ , N + 1 + $\delta$ if q = p, where $\delta$ = 1 (resp. 0) if K contains (resp. does not contain) a primitive q-th root of 1. For $H^2$ it’s harder, but one knows that its dimension is equal to $\delta$ . One can even make the relation explicit when there is one. A self contained reference is H. Koch’s book « Galois theory of p-extensions ».
Now take K = $\mathbf Q_{p}$. The above results imply that in all cases, G($\mathbf Q_{p}$)$\cong$ $\mathbf Z_{q}$. This means that a q-group H is realizable as a Galois group over $\mathbf Q_{p}$ iff H is cyclic. Note that for p $\neq$ q, whatever K is, we are stuck to H with at most two generators. This is rather meager, but we can add « degrees of liberty » in posing a different problem with p = q : given a p-group H, is there a local p-adic field K such that H can be realized as a Galois group over K ? The answer is « yes » : just take K not containing a primitive p-th root of 1; the above results show that G(K) is pro-p-free on N + 1 generators, hence admits H as a quotient if N is large enough .
