Galois action on $K\otimes_{\mathbb{Q}} \mathbb{Q}_\ell$

Question

Let $K/\mathbb{Q}$ be a finite extension and let $\ell$ be a prime number.

For each prime $\lambda$ of $K$ lying over $\ell,$ choose an embedding $\sigma_\lambda:K\hookrightarrow K_\lambda$ of $K$ into the $\lambda$-adic completion of $K.$

There is a natural isomorphism $$K\otimes_{\mathbb{Q}} \mathbb{Q}_\ell \longrightarrow \prod_{\lambda\mid\ell} K_\lambda\,; \; \; \; x\otimes y \longmapsto (\sigma_\lambda(x)y \, : \, \lambda\mid\ell).$$

If we now assume, in addition, that $K/\mathbb{Q}$ is Galois with group $G,$ then $G$ has a natural action on $K\otimes_\mathbb{Q} \mathbb{Q}_\ell$ (acting on the first factor).

Is there a "nice" description of resulting action on the product $\prod_{\lambda\mid\ell} K_\lambda$?

Answer 1

For example, for unramified $\ell$, the Galois group permutes the product factors, with decomposition groups being the isotropy groups. (We know that decomposition groups surject to local Galois groups...)

Answer 2

$G=Gal(K/Q), D(P_1) = \{ \sigma\in G, \sigma(P_1)=P_1\}$ $$\ell O_K=\prod_{j=1}^g P_j^e, \qquad G= \bigcup_{j=1}^g \sigma_j D(P_1), \qquad P_j = \sigma_j(P_1)$$ Take $\pi_1\in P_1,\not \in P_1^2$ and for $j\ne 1$, $\pi_1\equiv 1\bmod P_j$. Then $$(1-\pi_1)^{\ell^n}\equiv 0\bmod P_j^{\ell^n}, \qquad (1-\pi_1)^{\ell^n}\equiv 1\bmod P_1^{\ell^n}$$ Let $$r_1 = \lim_{n\to \infty} ( (1-\pi_1)\otimes 1)^{\ell^n} \in K\otimes_\Bbb{Q}\Bbb{Q}_\ell, \qquad {\sum_{j=1}^g \sigma_j(r_1)=1, \sigma_j(r_1)^2= \sigma_j(r_1), \\ \sigma_i(r_1)\sigma_j(r_1)=0}$$

Take a normal basis $$K_{P_1} = \sum_{m=1}^{ef} h_m(b) \Bbb{Q}_\ell$$ Where the $h_m$ are the elements of $D(P_1)$ and $b$ is an element of $K$.

$\sigma_j(b)$ generates a normal basis of $K_{P_j}$.

Then $$K\otimes_\Bbb{Q}\Bbb{Q}_\ell =\sum_{j=1}^g \underbrace{\sigma_j(r_1) \sum_{m=1}^{ef} \sigma_jh_m \sigma_j^{-1}(\sigma_j(b)\otimes 1) \Bbb{Q}_\ell}_{\cong K_{P_j}}$$ Because $r_1$ is fixed by $D(P_1)$ the Galois action on the RHS is clear.