# Galois groups of maximal unramified extensions

I have been workin throught Cassels-Frohlich Algebraic number theory. Im looking at local fields and their cohomology, in particular the Brauer group. On page 133 of casses-frohlich they make the folloing claim that I cant quite follow, its most likely something simple I'm missing but heres the thing:

If we let $L/K$ be a finite extension of degree $n$ and let $L_{nr},K_{nr}$ be the maximal unramified extensions of $L$ and $K$ respectively. They define $\Gamma_{K}=Gal(K_{nr}/K)$ and similarly for $\Gamma_{L}$. Now they want to make a map $Res: H^2(\Gamma_{K},K_{nr}^{*}) \longrightarrow H^2(\Gamma_{L},L_{nr}^{*})$ and to do so we need a map $\Gamma_{L} \longrightarrow \Gamma_{K}$, and they claim there is an inclusion map $\Gamma_{L} \longrightarrow \Gamma_{K}$. But I dont see how this can be an inclusion map? I know $L_{nr}=LK_{nr}$ but does this mean there is a natural inclusion? if so what am I forgetting?

Thank you

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I guess you're assuming the residue fields of $K$ and $L$ are finite? Otherwise it's not clear to me that $K_{nr}L=L_{nr}$. Assuming this is the case, it is exactly what you need for injectivity of the restriction map $\mathrm{Gal}(L_{nr}/L)\rightarrow\mathrm{Gal}(K_{nr}/K)$. What does it mean for something to be in the kernel? If $\sigma$ is in the kernel, then its restriction to $K_{nr}$ is the identity. But by definition $\sigma$ fixes $L$ pointwise, so it must fix the composite $LK_{nr}=L_{nr}$. So $\sigma$ is the identity.