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Assume $A=K[x,y]\subset K[x,y][w]=B$, $K$ is a field of characteristic zero, $w$ is integral over $A$ (so $B$ is a f.g. $A$-module), but $w$ is not in the field of fractions of $A$, and $B$ is an integral domain.

It can be shown (see Bourbaki "Commutative algebra" chapter 10, Corollary on page 58) that $B$ is a projective $A$-module, hence (by Quillen-Suslin) free $A$-module (maybe freeness follows more immediately from integrality).

My question: Is $B$ separable over $A$? ($B$ is separable over $A$ if $B$ is a projective $B\otimes_A B$-module under $\mu:B\otimes_A B \to B$, defined by $\mu(b_1 \otimes_A b_2)=b_1b_2$).

I have already asked this question here: https://math.stackexchange.com/questions/1321950/is-a-specific-ring-extension-b-of-kx-y-integrally-closed-separable/1323861#1323861 (where additional remarks on it can be found) but since I have not got an answer to it, I post it here also.

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  • $\begingroup$ The first answer in math.stackexchange should suffice since separable implies smooth (take in mind that if $B$ is smooth over $A$, then it is smooth over $K$ and so a regular ring). For more generality, see propositions 2.3 and 2.5 in sciencedirect.com/science/article/pii/0021869392900057 $\endgroup$
    – Vinteuil
    Commented Jun 16, 2015 at 8:43
  • $\begingroup$ @Vinteuil Thank you very much! What I have missed is "separable implies smooth", and then the example in math.stackexchange cannot be separable, since separability would imply (after some considerations you have made) that $B$ is regular but it is not, as Brevik explained. (I have now noticed that I am not sure I know how to prove that $B=A[w]$, with $w^2=x^2y$ is not integrally closed; I guess it's not difficult to prove this). $\endgroup$
    – user237522
    Commented Jun 17, 2015 at 2:41
  • $\begingroup$ @Vinteuil Please can you tell me where can I find a proof of your claim "separable implies smooth"? (Sorry if this implication may be considered trivial). Perhaps I should try Matsumura's book "Commutative Algebra"? $\endgroup$
    – user237522
    Commented Jun 18, 2015 at 3:19
  • $\begingroup$ 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 pd$_{B\otimes_AB}B<\infty$ for a flat algebra implies regularity. This result was obtained in gdz.sub.uni-goettingen.de/dms/load/img/?PPN=GDZPPN002067722 . The separable case was known before, but I do not know any reference, except Cartan-Eilenberg when $A$ is a field (theorem 7.10 in chapter IX, page 179). Cartan-Eilenberg was written before Serre introduced flatness, so maybe it is easy to modify $A$ field by $A \to B$ flat in their proof. $\endgroup$
    – Vinteuil
    Commented Jun 18, 2015 at 7:41
  • $\begingroup$ @Vinteuil Thank you very much for your explanation! It is very helpful to me. $\endgroup$
    – user237522
    Commented Jun 18, 2015 at 23:01

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