Let $K$ be an algebraically closed field and let $k$ be a subfield of $K$ such that the field extension $K \mid k$ is algebraic. Let $B$ be the polynomial ring $K [x_1, \ldots, x_n]$ and let $A$ be the subring $k [x_1, \ldots, x_n]$. It is clear that any $k$-subalgebra of $K$ is a field, so it follows that the intersection of $A$ and any maximal ideal of $B$ is a maximal ideal of $A$. Moreover, every maximal ideal of $A$ arises in this way.
Question. What are the maximal ideals of $A$, described in terms of the maximal ideals of $B$?
It is well-known that the maximal ideals of $B$ are all of the form $(x_1 - a_1, \ldots, x_n - a_n)$ for some $n$-tuple $(a_1, \ldots, a_n)$ of elements of $K$. Thus, for each maximal ideal $\mathfrak{m}$ of $A$, we may define the set $$P(\mathfrak{m}) = \{ (a_1, \ldots, a_n) \in K^n : (x_1 - a_1, \ldots, x_n - a_n) \cap A = \mathfrak{m} \}$$ and this set is finite (because it is a 0-dimensional closed subset of $\mathbb{A}^n_K$). Consider $G = \mathrm{Aut}(K \mid k)$. Of course, $G$ acts on $B$, and every element of $A$ is invariant under the action of $G$, so it follows that $P (\mathfrak{m})$ is closed under the action of $G$ in $K^n$ as well, i.e. $P (\mathfrak{m})$ is the union of $G$-orbits.
Conjecture. $G$ acts transitively on $P (\mathfrak{m})$.
This is certainly true if $k$ is a perfect field and $n = 1$ – after all, in that case, $\mathfrak{m}$ is a principal ideal generated by a polynomial in one variable irreducible over $k$. In fact this is true even without the assumption that $k$ is perfect: we know that the residue field $\kappa (\mathfrak{m}) = A / \mathfrak{m}$ is a finite field extension of $k$ and that the right $G$-set of $k$-algebra embeddings $\kappa (\mathfrak{m}) \to K$ is isomorphic to the right $G$-set of zeros in $K$ for the generator of $\mathfrak{m}$; but the homomorphism extension property of $\kappa (\mathfrak{m}) \to K$ implies that any two embeddings can be connected by an isomorphism of $K$, so $G$ acts transitively on the set of $k$-algebra embeddings $\kappa (\mathfrak{m}) \to K$.
What about the general case where $n > 1$? I had always taken for granted that the indicated conjecture was true, but I have not been able to find a proof in the usual places. By descent theory, one can show that $$\operatorname{Spec} B \otimes_k B \rightrightarrows \operatorname{Spec} B \to \operatorname{Spec} A$$ is a coequaliser diagram in the category of schemes, so $\operatorname{Spec} A$ is a scheme-theoretic quotient of $\operatorname{Spec} B$ by an internal equivalence relation, but it is not clear to me what this internal equivalence relation is when $n > 0$. (For $n = 0$ and $k$ perfect, something quite remarkable happens: $\operatorname{Spec} K \otimes_k K$ has a canonical continuous free transitive action of $\mathrm{Gal}(K \mid k)$!)