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Let $A=K[x,y] \subset K[x,y][w]=B$, $K$ 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.

(If $w$ was in the field of fractions of $A$, then being integral over $A$ and $A$ being integrally closed, would imply that $w \in A$; so I want to dismiss of this trivial case $A=B$).

One can show (just use Exercise 13 on page 164 of Kaplansky's book "Commutative rings") that $B$ is Gorenstein. (But it is NOT known if $B$ is a regular ring. Recall that: regular $\subset$ complete intersection $\subset$ Gorenstein $\subset$ Cohen-Macaulay). From Bourbaki's book "Commutative Algebra", chapter 10, page 58, $B$ is a projective $A$-module, hence from Quillen-Suslin theorem, $B$ is a free $A$-module. (I think that in my specific case it is immediate, without Bourbaki and Quillen-Suslin, to show that $B$ is free over $A$.)

My question(s):

(i) I would like to know if it is possible to show that $B$ is integrally closed. (A somewhat relevant but different question can be found in Regular Ring is Integrally Closed?).

(ii) 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 $\mu(b_1 \otimes_A b_2)=b_1b_2$.)

What I have tried thus far:

(1) Serre's criterion for normality; can be found in Matsumura's book "Commutative Ring Theory" on page 183, Theorem 23.8. (see also corollary to Theorem 11.5 on page 82). I do have $(S_2)$, since $B$ is Cohen-Macaulay, but I do not know how to show $(R_1)$.

(2) Showing that $B$ is regular is more specific than showing it is integrally closed (since a regular ring is normal, so a regular domain is integrally closed). I have tried to view $B=A[z]/(h(z))$ and then show that $h(z)$ satisfies conditions that guarantee that the quotient is regular, but have not yet succeeded (probably since I do not have enough information about $h(z)$ except it is monic).

(3) A paper of Seibt "A generalized Jacobian criterion for regularity", Corollary on page 201. However, I do not know how to show that the $B$-module of $K$-differentials is a flat $B$-module. Is there a connection between flatness of $B$ over $K$ (=which is known. $A$ is a flat $K$-module, since it's free) and flatness of the $B$-module of $K$-differentials over $B$?

(4) A paper of Wang "A Jacobian criterion for separability": I have tried to use Corollary 8, but it is not applicable in my case, since I do not have enough information about the derivative of $h$ evaluated at $w$.

-Please notice that I have already asked a similar, less specific, question: When does the regularity of $A$ implies the regularity of $A[w]$?

-Notice the following "answer" to (i): If we assume that the integral closure of $B$ is flat over $B$, then $B$ is integrally closed, see Integral closure $\tilde{A}$ is flat over $A$, then $A$ is integrally closed. I do not like such an answer, since I do not know if the integral closure of $B$ is flat over $B$.

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If I understand the setup correctly, it seems to me that $B$ need not be integrally closed; the simplest example is the "Whitney umbrella" obtained by adjoining $w$ so that $w^2=x^2y$. $B$ fails to be regular in codimension $1$, as can be seen using the Jacobian.

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  • $\begingroup$ Thank you very much! an interesting counterexample to (i). I wonder if $B=K[x,y,T]/(T^2-x^2y)$ is separable over $K[x,y]$? (Maybe it's easy to answer this, I do not know). $\endgroup$ – user237522 Jun 14 '15 at 1:01
  • $\begingroup$ @user237522 Sorry, I'm not too familiar with the notion of separability in this context. Seems as though it should be a mild condition. $\endgroup$ – John Brevik Jun 14 '15 at 11:44
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$B=K[x,y,z]/(h(z))$ with $h(z)=z^2-x^2y$ is not separable over $A=k[x,y]$ for $\frac{\partial h}{\partial z}=2z$ is not invertible in $B$: if the characteristic of $K$ is two then this is obvious; otherwise suppose $z$ is invertible, so there is $f(x,y,z)$ such that $zf(x,y,z)-1\in(z^2-x^2y)$. By sending $z$ to $0$ we get $1\in (x^2y)$, a contradiction.

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  • $\begingroup$ Thank you! I guess you have used Corollary 8 of Wang's paper, which I have mentioned in (4) above. I understand all your arguments except the last one (I am probably missing something trivial). $\endgroup$ – user237522 Jun 15 '15 at 21:53
  • $\begingroup$ Yes, thanks. I am still not comfortable with this last argument. BTW, in the "specific" example I have in mind, the invertible elements of $B$ are exactly $K^*$, so $z$ is not invertible. Also, in my example, it is not possible to have $h(Z)=Z^2-x^2y$; I do not know exactly how $h(Z)$ looks like, although given an $h(Z)$, I may be able to say if it can serve as $h(Z)$ or not. (Perhaps I should someday post the specific example I have in mind). $\endgroup$ – user237522 Jun 15 '15 at 22:39

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