Etale cohomology of a number field I am asking a question that I should feel ashamed of! 
Suppose $K$ is a number field and $K/\mathbb{Q}$ is a finite Galois extension of rank $n$, say
\begin{equation}
K=Q[x]/f(x)
\end{equation}
where the degree of the polynomial $f(x)$ is $n$. Let's consider $X:=\text{Spec}\,K$ to be a $\text{Spec}\,\mathbb{Q}$-variety. The variety $X_{\overline{\mathbb{Q}}}$ is just,
\begin{equation}
X_{\overline{\mathbb{Q}}}:=\text{Spec}\,K \otimes_{\mathbb{Q}}\overline{\mathbb{Q}}=\text{Spec}\,\overline{\mathbb{Q}}[x]/f(x)
\end{equation}
which is a variety defined over $\overline{\mathbb{Q}}$ consists of $n$ points that correspond to the $n$ roots of $f(x)$. 
Question 1: The Galois group $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ has a right action on the variety $X_{\overline{\mathbb{Q}}}$, what is this action? I guess its action on the $n$ points is like the Galois group $\text{Gal}(K/\mathbb{Q})$ permutes the $n$ roots? 
Now the etale cohomology group
\begin{equation}
H^0_{et}(X_{\overline{\mathbb{Q}}},\mathbb{Q}_\ell)
\end{equation}
is a representation of the Galois group $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$. 
Question 2 This group is an $n$ dimensional $\mathbb{Q}_\ell$ vector space, I feel it is just the representation that factors through  $\text{Gal}(K/\mathbb{Q})$, but I could not make this precise, could anyone give a rigourous construction?
 A: Here is another way to think about it (since it works for any field, not just $\mathbb{Q}$, I will write $k$ instead) :
If $K/k$ is Galois, 
$$
\begin{array}{rcl}K\otimes_k \overline{k}&\longrightarrow &\displaystyle\prod_{\sigma:K\rightarrow\overline{k}}\overline{k}\\u\otimes x&\longmapsto&(\sigma\mapsto\sigma(u)x)\end{array}$$
is an isomorphism. (This is the same as the one I wrote in my comment which uses a primitive element, its minimal polynomial $P$, the other roots of $P$ and a fixed embedding of $K$ into $\overline{k}$. This way you are free of many choices...).
The action of $\operatorname{Gal}(\overline{k}/k)$ on the left hand side is $\tau(u\otimes x)=u\otimes \tau (x)$. Under the isomorphism, it translates as $\tau((x_\sigma)_\sigma)=(\tau (x_{\tau^{-1}\sigma}))_\sigma$. (Check it)
Taking spectrum, one has 
$$X_{\overline{k}}=\operatorname{Spec}(K\otimes_k\overline{k})=\operatorname{Spec}(\prod_{\sigma:K\rightarrow\overline{k}}\overline{k})=\coprod_{\sigma:K\rightarrow\overline{k}}\operatorname{Spec}\overline{k}=\operatorname{Hom}(K,\overline{k})\times \operatorname{Spec}\overline{k} $$
Thus, $\operatorname{Gal}(\overline{k}/k)$ acts on the right on points as $(\sigma,p)\tau=(\tau^{-1}\sigma,p)$. 
Of course the subgroup $\operatorname{Gal}(\overline{k}/K)\subset \operatorname{Gal}(\overline{k}/k)$ does not depends on the embedding, and acts trivially on the underlying set of $X_{\overline{k}}$ (but not on $X_\overline{k}$).

What is the action on cohomology ? Well if $S$ is a set of points with algebraically closed residue fiedls, $H^0(S,\mathbb{Q}_l)=\mathbb{Q}_l^S$, and if $G$ acts on the underlying set of $S$ (on the right), it also acts (on the left) on $H^0(S,\mathbb{Q}_l)$ by functioriality.
In this case 
$$ H^0(X_\overline{k},\mathbb{Q}_l)= \prod_{\sigma:K\rightarrow\overline{k}} H^0(\overline{k},\mathbb{Q}_l)=\prod_{\sigma:K\rightarrow\overline{k}}\mathbb{Q}_l$$
And the action of $\operatorname{Gal}(\overline{k}/k)$ is just $\tau(x_\sigma)_\sigma=(x_{\tau^{-1}\sigma})_\sigma$ (it is indeed a left action though it does not looks like it). Clearly it does factor through its quotient $\operatorname{Gal}(K/k)$.
So we have shown that the cohomology of $H^0(X_\overline{k},\mathbb{Q}_l)$ is just the regular representation of $\operatorname{Gal}(K/k)$.
