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Let $R=\mathbb{C}[x,y]$. I have the following situation: $\mathbb{C} \subseteq D \subseteq R$ is affine (= finitely generated as a $\mathbb{C}$-algebra), noetherian, has field of fractions $\mathbb{C}(x,y)$, and $R$ is separable over $D$ (= $R$ is a projective $R \otimes_D R$-module).

It would be great if one can show (at least) one of the following:
(1) Without any further assumptions, $D$ is regular (= a noetherian ring such that every localization at a maximal ideal is a regular local ring).
(2) Assuming integrality of $D \subseteq R$ (but not assuming flatness) implies regularity of $D$.
(3) Assuming flatness of $D \subseteq R$ (but not assuming integrality) implies regularity of $D$.

I am most interested to show (1); however, I am afraid there is not enough information to show (1), so showing (2) would be great too. (3) may be of some interest.

Thus far I can show only a special case of (3), namely: If $D \subseteq R$ is faithfully flat, then $D$ is regular. Reason: Just apply Proposition 8 in Bourbaki's book, 10, page 59, which says the following: If $A$ and $B$ are noetherian rings with $A \to B$ faithfully flat, then regularity of $B$ implies regularity of $A$.

Sorry for not posting (yet) the specific case I have in mind; it's just that I first wish to know if the more general case is provable (maybe it can be found in a book or a paper?).

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  • $\begingroup$ If I understand well you want a criterion to decide when such $D$ is regular. For $n=1$ flatness is good enough? $\endgroup$ Commented May 18, 2015 at 21:34
  • $\begingroup$ Yes, I want a criterion to decide the regularity of $D$. For $n=1$ your criterion sounds good-- Do you have a proof/reference for it? What happens when $n=2$? $\endgroup$ Commented May 18, 2015 at 21:50
  • $\begingroup$ For $n=1$, $D$ contains a monic polynomial, so $D\subset\mathbb C[X]$ is integral. $\endgroup$ Commented May 18, 2015 at 22:12
  • $\begingroup$ If the extension is integral, then flatness is enough. $\endgroup$ Commented May 18, 2015 at 22:23
  • $\begingroup$ Yes, because integral+flat imply faithfully flat, and then there is an exercise, for example in Matsumura's book, that claims that if $A \subseteq B$ is faithfully flat and $A$ and $B$ have the same field of fractions, then $A=B$. $\endgroup$ Commented May 18, 2015 at 22:27

1 Answer 1

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Unfortunately, none of the three things you want may be true.

First an example for 2, which will say 1 is false too. Take $D=\mathbb{C}[x^2-1,x(x^2-1), y]$.

For 3, take $D=\mathbb{C}[y^2x-1,(y^2x-2)^2x, y]$.

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  • $\begingroup$ Thanks! How one shows that $\mathbb{C}[x,y]$ is separable over $D$? $\endgroup$ Commented Jun 28, 2015 at 0:57
  • $\begingroup$ It is a calculation that is better done on paper. Most important thing to notice is that if $M$ is a maximal ideal of $D$, $M\mathbb{C}[x,y]$ is a product of (two) maximal ideals in the first case and a maximal ideal in the second. What in algebraic geometry is called an unramified map. $\endgroup$ Commented Jun 28, 2015 at 2:49
  • $\begingroup$ Thank you very much for the hints! $\endgroup$ Commented Jun 28, 2015 at 2:50

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