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.)

  • $\begingroup$ Couldn't we replace $A$ by $A_{\mathfrak{p}}$ and assume $\mathfrak{p}$ is maximal? $\endgroup$
    – Mohan
    Dec 29, 2017 at 16:16
  • $\begingroup$ @ Mohan: could you eloborate on that ? $\endgroup$
    – Boccherini
    Jan 2, 2018 at 16:09
  • $\begingroup$ First, I would like to know what you tried after my comment. $\endgroup$
    – Mohan
    Jan 2, 2018 at 16:16
  • $\begingroup$ Can you first show that $A = \{ x \in \overline{A} : \sigma(x) = x \textrm{ for all } \sigma \in Aut(\overline{k}/k) \}$? $\endgroup$
    – D_S
    Jan 3, 2018 at 4:27
  • $\begingroup$ @ Mohan: not much; you seem to imply that the action of $\mathrm{Aut}(\overline{k}/k)$ on the maximal ideals in $A_{\frak{p}} \otimes_k \overline{k}$ over the maximal ideal $A_{\frak{p}}\frak{p}$ of $A_{\frak{p}}$ is closely related to its action on $S(\frak{p})$ ? $\endgroup$
    – Boccherini
    Jan 3, 2018 at 13:36

1 Answer 1


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.

  • $\begingroup$ @ DAP: in your 4th paragraph, I guess you mean $\frak{q}, \frak{q}'$ $\subset (A\otimes_K M)$ and $\frak{q}$ $\cap (A \otimes_K L)$ to $\frak{q}'$ $\cap (A \otimes_K L)$. $\endgroup$
    – Boccherini
    Jan 22, 2018 at 15:28
  • $\begingroup$ @Boccherini: fixed $\endgroup$
    – Dap
    Jan 22, 2018 at 21:20
  • $\begingroup$ @ DAP: could you explain why separable and perfect closures are transitive (cf. the last paragraph of your answer) ? Also, is there an implication for general algebraic extensions (cf. your setting) ? $\endgroup$
    – Boccherini
    Jan 24, 2018 at 13:18

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