Functoriality of the maximal unramified extension

Question

Let $K$ be a finite extension of $\mathbb{Q}_p$ and $L$ a finite Galois extension of $K$. We have $$\text{Gal}(K^{nr}/K)\cong\hat{\mathbb{Z}}\cong \text{Gal}(L^{nr}/L) $$ where $K^{nr},L^{nr}$ denotes the maximal unramified extension of $K$ and $L$ respectively. I would like to understand the map $$\text{Gal}(L^{nr}/L)\to \text{Gal}(K^{nr}/K).$$ I think, well maybe hope, it is induced by a morphism $$\mathbb{Z}\to \mathbb{Z}.$$ Would that make sense?

Answer 1

$K^{nr} = \bigcup_{p\ \nmid\ n} K(\zeta_n)$

$L^{nr} = \bigcup_{p\ \nmid\ n} L(\zeta_n)$

$Gal(K^{nr}/K) \cong Gal(\overline{k}/k)$ where $k=O_K/(\pi_K),\overline{k}=O_{K^{nr}}/(\pi_K)$, the isomorphism is the reduction map $O_{K^{nr}}\to O_{K^{nr}}/(\pi_K)$.

The map $ Gal(\overline{k}/\ell)\to Gal(\overline{k}/k)$ is a bit obvious, right ?

$ Gal(\overline{k}/\ell)$ is topologically generated by the Frobenius $\phi_\ell(a) = a^{|\ell|}=\phi_k^f(a)$ where $f=[\ell:k]$ and $\phi_k(a)=a^{|k|}$.

The isomorphism $ Gal(\overline{k}/k)\to \hat{\Bbb{Z}}$ sends $\phi_k$ to $1$.

So the map $\hat{\Bbb{Z}}\to \hat{\Bbb{Z}}$ corresponding to $ Gal(\overline{k}/\ell)\to Gal(\overline{k}/k)$ is just $c\to fc$.

No need that $L/K$ is Galois.

On the $L^{nr}$ side the Frobenius lifts uniquely to the automorphism defined by $\sigma_L(\zeta_n)=\zeta_n^{|\ell|}$.