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Let $(R,m)$ and $(S,n)$ be two local rings, $R \subseteq S$, $R$ regular, $S$ a finitely generated and flat $R$-algebra, and $mS=n$.

In comments to this question it was claimed that in such situation necessarily $R \subseteq S$ is unramified; however, I am not able to prove this. If I am not wrong (but I may be wrong), we need to show that $A/m \subseteq B/n$ is a (finite?) separable field extension.

Could one please help or just refer to the relevant known results that help prove that $R \subseteq S$ is unramified?

Also see this question, where the above question was discussed but not solved.

Thank you very much!

Edit: Assume (in addition to the above conditions) that $A$ and $B$ are $k$-algebras, where $k$ is a perfect field. Then:

(a) If $S$ is a finitely generated $R$-module, then $R/m \subseteq S/n$ is a field extension of finite degree (hence algebraic). Then the field extension is separable.

(b) If $S$ is an algebraic $R$-algebra, then $R/m \subseteq S/n$ is an algebraic field extension. Then again the field extension is separable. (A special case when $S$ is algebraic over $R$ is when $S=R[w]$, where $w \in S$ is algebraic over $R$).

Also, I now see that there is no need to assume regularity of $R$.

Summarizing, we have:

Let $(R,m)$ and $(S,n)$ be two local rings which are $k$-algebras, $k$ is a perfect field, $R \subseteq S$, $S$ is an algebraic $R$-algebra (= every element of $S$ satisfies a polynomial over $R$), and $mS=n$. Then $R \subseteq S$ is unramified.

Please, am I missing something or am I right? Any comments are welcome.

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  • $\begingroup$ This is not quite right; think about the case where $R$ and $S$ are fields. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Apr 22, 2021 at 1:54
  • $\begingroup$ Take an imperfect field $k$, $a \in k - k^p$. Then $k \to k(a^{1/p})$ satisfies all the assumptions in your question but is not separable. $\endgroup$
    – David Benjamin Lim
    Commented Apr 22, 2021 at 3:07
  • $\begingroup$ Thank you very much, R. van Dobben de Bruyn and David Benjamin. $\endgroup$
    – user237522
    Commented Apr 22, 2021 at 5:32
  • $\begingroup$ @R.vanDobbendeBruyn, please, if the fields $R$ and $S$ are perfect and $R \subseteq S$ is algebraic, then the extension is separable (hence, since $0S=0$, the extension is unramified) or am I still missing something? Also, is the original claim (= first two lines in the question) true? It is assumed that $S$ is finitely generated as an $R$-algebra. (Is this enough or should $S$ be finitely generated as an $R$-module?). $\endgroup$
    – user237522
    Commented Apr 22, 2021 at 6:13
  • $\begingroup$ I gave you a counterexample to the claim in the second paragraph. Note that the linked question was talking about characteristic $0$, where all field extensions are separable. If $R$ and $S$ are essentially of finite type over a perfect field, there is still a chance, but you did not put this assumption anywhere. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Apr 22, 2021 at 15:18

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