Galois action on prime ideals Suppose $A$ is a $k$-algebra, with $k$ a field, and let $\overline{A} := A \otimes_k \overline{k}$, where $\overline{k}$ is an algebraic closure of $k$. Let $\frak{p}$ be any prime ideal in $A$, and let $S(\frak{p})$ be the set of prime ideals in $\overline{A}$ over $\frak{p}$. My question is the following: does $\mathrm{Aut}(\overline{k}/k)$ act transitively on $S(\frak{p})$? I know the answer is "yes" if "prime ideal" is replaced by "maximal ideal," but am interested in the more general setting. 
(I have also seen variations of this question in the context of, e.g., Dedekind domains, but I found nothing yet about my specific setting.)
 A: I think this is true. I assume you mean unital algebras. Consider the more general question: given an algebraic extension $L/K,$ is it true that for all unital $K$-algebras $A$ and all prime ideals ${\frak p}\subset A,$ the group $\mathrm{Aut}(L/K)$ acts transitively on the primes lying above ${\frak p}$ in $A\otimes_K L$? Let's call this property of $L/K$ "transitivity".
A finite Galois extension is transitive. This follows from properties of integral extensions. See Action of finite group of automorphisms on Spec A. As D_S mentioned, you can use that the $\mathrm{Aut}(L/K)$-invariant elements of $A\otimes_K L$ are just $A$; see for example the proof of Theorem 2.14 here: http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/galoisdescent.pdf
If $K$ has characteristic $p>0$ then $K^{1/p}/K$ is transitive. More specifically, there is a unique prime lying above ${\frak p}$; it's ${\frak p}^{1/p}:=\{x^{1/p}\mid x\in {\frak p}\},$ the preimage of ${\frak p}$ under the Frobenius isomorphism $K^{1/p}\to K.$
If we have a tower $M\supset L\supset K$ where both extensions are transitive, then $M\supset K$ is transitive (transitivity of transitivity!). Given ${\frak q},{\frak q}'\subset A\otimes_K M$ lying above ${\frak p},$ some element of $\mathrm{Aut}(L/K)$ takes ${\frak q}\cap A\otimes_K L$ to ${\frak q}'\cap A\otimes_K L.$ We can extend this to an element $\sigma\in\mathrm{Aut}(M/K),$ then compose with an element of $\mathrm{Aut}(M/L)$ sending $\sigma({\frak q})$ to ${\frak q}'.$
By Zorn's lemma, given any algebraic extension $M/K$ there is a maximal transitive extension $L/K$ with $L\subseteq M.$ This implies that separable closures and perfect closures are transitive. Using the fact that the algebraic closure is the separable closure of the perfect closure gives the result.
