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: http://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.

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.