Let $A\rightarrow B$ be a local morphism of complete noetherian rings making $B$ a formally smooth $A$-algebra. Does the induced morphism $\textrm{Spec}(B)\to\textrm{Spec}(A)$ have geometrically regular fibers ?

Noting $k$ the residue field of $A$, the answer is yes if $k$ is of characteristic $0$, by resolution of singularities. If $K$ is the residue field of $B$, the answer is also yes if the extension $K/k$ is finite by EGA IV 2, paragraph 7, theorem (7.5.1). The result subsists if the extension is of finite type, even if cannot find reference for this, but does the result subsist if $B\otimes_A k$ is the separable closure of $k$ ?

  • $\begingroup$ The answer to the final question is affirmative, but for reasons specific to that situation. Namely, if $B \otimes_A k$ is a separable closure of $k$ then $B$ has to be the completion of a strict henselization $A^{\rm{sh}}$ of $A$, so it suffices to show that $A^{\rm{sh}}$ is excellent. But $A$ is excellent since it is complete local noetherian (EGA IV$_3$, 7.8.3(iii)), and strict henselization preserves excellence of local rings (though EGA IV$_4$, 18.7.6 handles henselization, curiously EGA does not handle the analogue for strict henselization; see p. 17 of Freitag-Kiehl for a proof). $\endgroup$
    – user74230
    Mar 9, 2015 at 10:14

1 Answer 1


Yes. More generally, if $A \to B$ is formally smooth and $A$ is quasi-excellent, then $A \to B$ has geometrically regular fibers. This is the main theorem in "M. André, Localisation de la lissité formelle". The proof is difficult, even under the additional assumption that $A$ and $B$ are complete.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.