Recovering the Zariski topology from the Zariski topology over an extension Suppose $A$ is a $k$-algebra, with $k$ a field, and let $\ell$ be a field extension of $k$. Is there an easy way to see/recover $\mathrm{Spec}(A)$ in/from $\mathrm{Spec}(A \otimes_k \ell)$, using the action of $\mathrm{Aut}(\ell/k)$ ? Important special case: $\ell$ is an algebraic closure of $k$. 
To illustrate my question, let $\ell$ be an algebraic closure of $k$. Then a maximal ideal $\frak{m}$ of $A$ corresponds to an $\mathrm{Aut}(\ell/k)$-orbit of maximal ideals in $A \otimes_k \ell$ (the maximal ideals in $A \otimes_k \ell$ ''over $\frak{m}$''). Is such a thing true for the other prime ideals (and for any field extension $\ell$, not ''just'' algebraic closures) ? Is there a similar way (so using the $\mathrm{Aut}(\ell/k)$-action as above) to describe the closed sets of $\mathrm{Spec}(A)$ ?
 A: Let $A$ be a finitely generated $k$-algebra, and let $K$ be the algebraic
closure of $k$. I explain how to recover $spec(A)$ from $spec(A\otimes K)$.
Equivalently, I explain how to recover $spm(A)$ from $spm(A\otimes K)$ (max specs).
Consider the map $spm(A\otimes K)\rightarrow spm(A)$.
(a) The map is surjective.
Every maximal ideal $\mathfrak{m}$ of $A$ is the kernel
of a $k$-algebra homomorphism $A\rightarrow K$,
which extends to a $K$-algebra homomorphism $A_{K}\rightarrow K$, 
whose kernel is a maximal ideal lying over $\mathfrak{m}$.
(b) The map is continuous. Let $S=Z(f_{1},\ldots,f_{s})$ be
closed in $spm(A)$. Then $\pi^{-1}(S)=Z(f_{1},\ldots,f_{s})$ in
$spm(A_{K})$.
(c) The map is closed. Let $T=Z(f_{1},\ldots
,f_{s})$ be a closed subset of $spm(A_{K})$. Choose a basis
$(e_{j})_{j\in J}$ for $K$ as a $k$-vector space, and write
$f_{i}=\sum_{j}e_{j}f_{ij}$ (finite sum) with $f_{ij}\in A$. Every maximal
ideal of $A_{K}$ is the kernel of a $K$-algebra
homomorphism $A_{K}\rightarrow K$, and the $f_{i}$
map to zero under such a homomorphism if and only if every $f_{ij}$ does.
Therefore $T=Z(\ldots,f_{ij},\ldots)$ in $spm(A_{K})$, and it
follows that $\pi(T)=Z(\ldots,f_{ij},\ldots)$ in $spm(A)$.
Thus, the map is a quotient map of topological spaces, and the fibres are the
orbits of $Aut(K/k)$.
