Let $k\_0$ be a field of characteristic 0, and let $k$ is a fixed algebraic closure of $k\_0$.
Write $G={\rm Gal}(k/k\_0)$.


Let $A\_0$ be a finitely generated commutative $k\_0$-algebra with a unit.
Then $A:= A\_0 \otimes\_{k\_0} k$ is a finitely generated commutative $k$-algebra with a unit.
The Galois group $G $ acts on $A$ via the action on $k$, and this action is $k$-semilinear:
$^s(\lambda f)=\,^s\lambda\cdot\,^s f$ for any $s\in G $, $\lambda\in k$, and $f\in A$.
The Galois action is continuous:
for any element $f\in A$ the stabilizer ${\rm Stab}\_G (f)$ is open in $G $.


Conversely, let $A$ be a finitely generated commutative $k$-algebra with a unit
endowed with a $k$-semilinear $G $-action such that the stabilizer in $G $
of any element $f \in A$ is open.
Set $A\_0=A^G $ (the algebra of $G $-invariants).



> 
> **Question.**
>  Is it true that $A\_0$ is finitely generated over $k\_0$ and that $A\cong A\_0\otimes\_{k\_0} k$ ?
> 
> 
> 


I would be grateful for any reference (or proof) for this elementary assertion.



Infinite Galois descent for finitely generated commutative algebras over a field
As an alternative to my first answer, here is an easier (but less powerful) approach that uses the finite generation in a more essential way:


Let $k \to \ell$ be a Galois field extension with Galois group $G$. Let $B$ be an $\ell$-algebra with an $\ell$-semilinear action $G \times B \to B$ with open stabilisers. Assume $B$ is finitely generated over $\ell$; say $B = \ell[x\_1,\ldots,x\_n]/I$. Let $U\_i$ be the stabiliser of $x\_i$, and set $U = U\_1 \cap \ldots \cap U\_n$.


Then any $u \in U$ acts on $B$ by
$$\sum a\_I x^I \mapsto \sum u(a\_I) x^I.\label{Eq action}\tag{1}$$
Shrinking $U$ if necessary, we may assume that $U$ is open and normal, and that $I$ is defined over $\ell^U$, i.e. there is an ideal $I\_U \subseteq \ell^U[x\_1,\ldots,x\_n]$ such that $B\_U = \ell^U[x\_1,\ldots,x\_n]/I\_U$ satisfies $B\_U \otimes\_{\ell^U} \ell \cong B$. Both sides have $U$-action given by (\ref{Eq action}), so in particular we see that
$$B^U = B\_U = \left\{\sum a\_Ix^I \in B\ \bigg|\ a\_I \in \ell^U \text{ for all } I\right\}.$$
This in particular implies that the natural map $B^U \otimes\_{\ell^U} \ell \to B$ is an isomorphism.


Since $(B^U)^{G/U} = B^G$, this reduces us to Galois descent for the finite Galois extension $k \to \ell^U$ with Galois group $G/U$, which is classical. $\square$


