How the Frobenius behaves under the change of base field Let $k'/k$ be an extension of finite fields, $X$ be a scheme over $k'$ and thus over $k$. Then $X\otimes_{k} \bar{k}$ is $n = [k':k]$ disjoint union of $X\otimes_{k'} \bar{k}$. How the Frobenius of the former can be expressed in the latter?
Milne's book on etale cohomology claims that:
for a sheaf $E$ on $X_{et}$, denote $W = H_c^r(X\otimes_k \bar k,\bar E)$, $V = H_c^r(X\otimes_{k'} \bar {k},\bar E)$, then $(W,F) = (V^n,F_n')$, where
\begin{equation}
  F_n'(v_1,\dots, v_n) = (F'v_n,v_1,\dots,v_{n-1}),
\end{equation}
and $F'$ is the Frobenius map on $X/k'$.
I don't know why this is true. I want to express $\mathrm{id}_X \otimes_k F_{\bar k/k}$ on
$X\otimes_{k} \bar{k}$ in terms of $X\otimes_{k'} \bar{k}$ under the identification
\begin{align*}
    X \times_{k'} G_{k'}
    = X\times_{k'} k' \times_k k'
    = X\times_k k'
\end{align*}
where $G = \operatorname{Gal}(k'/k)$. But it is just hard to make it clear why a single $F'$ shows up.
 A: The situation is base-changed (by $X \rightarrow \mathrm{Spec}\,k’$) from the case $X=\mathrm{Spec}\,k’$ so we may as well assume this.
The question then can be written as follows: describe the action of $\mathrm{id} \otimes F_{k}$ acting on $k’ \otimes_k \overline{k}$, in terms of the $F_{k’}$-action on $\overline{k}$.
Now, we can write $k’$ as $k[X]/(P)$ for some irreducible polynomial $P \in k[X]$ of degree $d$, and we can factor $P$ in $\overline{k}$ as $P(X)=\prod_{i=0}^{d-1}{(X-F_k^i(x))}$ for some $x \in \overline{k}$.
This produces the $d$ $\overline{k}$-morphisms $k’ \otimes_k \overline{k} \rightarrow \overline{k}$: we note them $e_i$ for $0 \leq i < d$ and $e_i(Q(X) mod P)=Q(F_k^i(x))$.
So we have an isomorphism of $k$-algebras $(e_0,\ldots,e_{d—1}): k’ \otimes \overline{k} \rightarrow \overline{k}^d$; under this isomorphism, $F_k$ acts on the target by rotating the indices and applying itself to every coordinate, and $\overline{k} \subset \overline{k}^d$ is the diagonal embedding.
But consider the following map $I: (x_i)_{0 \leq i < d} \in \overline{k}^d \longmapsto (F_k^{-i}(x_i))_i \in \overline{k}^d$. If the target space is seen as a $\overline{k}$-algebra through the embedding $f: x \in \overline{k} \longmapsto (F_k^{-i}(x))_i$, this is a $\overline{k}$-isomorphism. Moreover, we have $I(F_k((x_i)_i))=I((F_k(x_{i-1}))_i)=(F_k^{1-i}(x_{i-1})_i=(F_k(x_{d-1}),x_0,\ldots,F_k^{2-d}(x_{d-2}))$.
Now, if $I((x_i)_i)=(y_0,\ldots,y_{d-1})$, then $F_k(x_{d-1})=F_{k’}(y_{d-1})$, and for $i \geq 1$, $F_k^{1-i}(x_{i-1})=y_{i-1}$, so that $I(F_k((x_i)_i))=(F_{k’}y_{d-1},y_0,\ldots,y_{d-2})$.
In other words, as $\overline{k}$-algebras with semi-linear $F_k$-actions, $k’ \otimes_k \overline{k}$ is isomorphic to $\overline{k}^d$ with the above embedding ($F_k^{-i}$ on the $i$-th coordinate, $0 \leq i < d$) and $F_k(y_0,\ldots,y_{d-1})=(F_{k’}y_{d-1},y_0,\ldots,y_{d-2})$.
Then it should be smooth sailing from there if you’re careful enough with the machinery.
