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Let $A \subseteq B$ be commutative noetherian rings. I have found the following claim: "Separability implies smoothness" with the following explanation: "The natural thing is to prove that a separable algebra is a regular homomorphism (and so smooth if it is of finite type). In general ...".

Remark: I am only interested in finitely generated algebras, so did not quote the rest of the explanation.

However, it seems to me, at least with my definitions, that this claim is false: If $A \subseteq B$ is separable, then (assuming the claim is true) $A \subseteq B$ is smooth, then by a result of Grothendieck, $A \subseteq B$ is flat, but this seems too much (especially in view of Examples on pages 95-97 and Corollary 9; I do not see why separability should imply flatness of $B$ over $A$. Notice that actually, this is what the explanation suggests: it says that separability implies $A \to B$ is a regular homomorphism, and by definition, a regular homomorphism is flat).

I will really appreciate if one can either post a sketch of proof/tell me where I can find a proof, or a counterexample.

Actually, I have already posted this question here, but got no comments thus far.

EDIT: I work with the definition of separability of this book, and with the definition of smoothness of this paper.

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  • $\begingroup$ There is something confusing about the premise of this question, as follows. Separability means that the surjective map $\pi: B\otimes_A B \rightarrow B$ makes $B$ a projective $B\otimes_A B$-module, but $\pi$ is surjective, so by injectivity of flat local maps it follows that $I := \ker(\pi)$ satisfies $I_{\mathfrak{p}}=0$ for all primes $\mathfrak{p} \supset I$ of $B \otimes_A B$. Hence, $\Omega^1_{B/A}=0$, so smooth $A$-algebras with positive relative dimension (e.g., $B=A[T]$!) are never separable. Have I misunderstood something? If not, then what is the motivation for this question? $\endgroup$
    – grghxy
    Commented Jul 12, 2015 at 21:53
  • $\begingroup$ Thanks. The motivation is a comment on MO which claims that "separablity implies smoothness"; the (beginning of the) explanation in the comment says: "The natural thing is to prove that a separable algebra is a regular homomorphism (and so smooth if it is of finite type). In general ...". $\endgroup$
    – user237522
    Commented Jul 12, 2015 at 22:05
  • $\begingroup$ @user237522 Where is the comment "separablity implies smoothness", can you give the link and thanks. $\endgroup$
    – Henry.L
    Commented Jul 12, 2015 at 22:42
  • $\begingroup$ Thanks. The comment is in: mathoverflow.net/questions/209379/…. However, I feel uncomfortable to attach the link, since it involves the person who wrote that comment (and if it's not true, maybe he wishes to delete it). Anyway, I can delete the link (after getting an answer). $\endgroup$
    – user237522
    Commented Jul 12, 2015 at 22:54
  • $\begingroup$ @user237522: I suspect that the person who made the earlier comment must have had in mind another definition of "separable algebra" or was confused, as my argument shows that the definition of separability that you are using is never satisfied for smooth algebra which is not etale. $\endgroup$
    – grghxy
    Commented Jul 12, 2015 at 23:13

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