I am reading Neukirchs ANT and I'm having some difficulties with abstract Galois theory and abstract valuation theory. I know that my question is quite specific and requires more information than what I can give here. I'm hoping that someone read the book as well and remembers what happens here. In the following we have a Galois extension $L/K$, which just means that $G_L$ is normal in $G_K$. Then he passes to the maximal unramified extension of $L$, denoted $\widetilde{L}$. Then he considers the Galois group $G(\widetilde{L}/K)$ but how do we know that the extension $\widetilde{L}/K$ is Galois? Does it hold in general that the maximal unramified extension of a Galois extension is Galois? Are all unramified extensions of a Galois extension Galois?
In the abstract theory $\widetilde{L}/K$ is Galois because:
$\widetilde{K}/K$ is a Galois extension because $G_{\widetilde{K}} = I_K$ is the kernel of $d_K: G_K \to \hat{\mathbb{Z}}$ and hence a normal subgroup of $G_K$. $G_{\widetilde{L}} = G_{\widetilde{K}L} = G_{\widetilde{K}} \cap G_L$ is a normal subgroup of $G_K$ as the intersection of two normal subgroups.
Does it hold in general that the maximal unramified extension of a Galois extension is Galois?
Yes. Suppose that $L/K$ is Galois, and $M/L$ is unramified. Let $N/K$ be the Galois closure of $M/K$. It has the property that it is the compositum of $\sigma M$ for $\sigma \in \mathrm{Gal}(N/K)$ (or equivalently $\sigma M$ for all $\sigma \in \mathrm{Gal}(\overline{K}/K)$. Since $\sigma$ is an automorphism, $\sigma M/\sigma L$ is also unramified. Since $L/K$ is Galois, $\sigma L = L$. Thus $\sigma M/L$ is unramified for all $\sigma$. So now you just have to prove that the compositum of unramified extensions (all the $\sigma M$ over $K$) is unramified. (A well-known exercise).
Are all unramified extensions of a Galois extension Galois?
Not in general. One can easily come up with theoretical constructions but let's give an actual example. Let $K$ be the degree $3$ subfield of $\mathbf{Q}(\zeta_{163})$, which is $\mathbf{Q}[x]/(x^3 - x^2 - 54x + 169)$. The class group is $(\mathbf{Z}/2 \mathbf{Z})^2$. There is a corresponding maximal abelian extension $L/K$ which (by a similar argument to the first part) is Galois over $\mathbf{Q}$, and in fact $\mathrm{Gal}(L/\mathbf{Q}) = A_4$, and $L$ is the splitting field of $x^4 - x^3 - 7x^2 + 2x + 9$. But now clearly $L/K$ contains subfields of degree $2$ which can't be Galois over $\mathbf{Q}$ since $A_4$ has no order six quotient.