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I have just read: Direct proof of non-flatness and wondered what is exactly the claim that Alex Youcis is referring to: "...but are you aware of the fact that flatness preserves normality. In your case $A$ is non-normal and $B$ is normal, so $B/A$ can't be flat".

Can one please explain what exactly "flatness preserves normality" means?

Given $A \subseteq B$ (commutative rings) with $B$ flat over $A$, does "flatness preserves normality" mean: (1) If $B$ is normal, then $A$ is normal. (2) If $A$ is normal, then $B$ is normal. (3) Something else?

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user237522
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    This should have been a comment under that question asking the author of the claim what he wanted to say. – user26857 Jun 16 '15 at 06:34
  • I guess you are right; I first thought to directly ask Alex Youcis what he meant, but then thought that maybe his claim "flatness preserves normailty" is something well-known, like "localization preserves flatness", so it deserves to be asked separately. – user237522 Jun 16 '15 at 12:57
  • Truly, I thought that Alex Youcis meant to claim something in the spirit of Theorem 1.7 (i) implies (ii) of Adjamagbo's paper: "On separable algebras over a UFD and the Jacobian Conjecture in any characteristic" (appears in van den Essen's book "Automorphisms of affine spaces"). – user237522 Jun 16 '15 at 13:08

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The simplest claim that would make Alex's point is that a flat integral extension of rings is faithfully flat, and that the only faithfully flat extensions of domains with the same field of fractions are identity maps. In other words if $B$ is integrally closed and integral and flat over $A$, then $B=A$, and in particular $A$ is also integrally closed.

That handles the situation of $k[x^2,x^3]\subset k[x]$. But the stronger claim (1) is false, as you can see by taking $B$ to be the field of fractions of any non-integrally closed $A$. (2) is also false, since if $A$ is a field then every $B$ is flat.

EDIT: The one other thing you might wonder is whether an integral extension of an integrally closed ring is integrally closed, and again the answer is no: $\mathbb{Z}[X^2,X^3]/\langle X^5+2\rangle$ is an integral extension of $\mathbb{Z}$ which is not integrally closed, since it doesn't contain $X$.

Kevin Carlson
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  • Thanks. I know the claims you are talking about (flat+integral implies faithfully flat, etc., and truly, before reading your answer I have noticed the impossibility of (1) you mention). It's just that to me "flatness preserves normality" sounds like it should mean something else, but in the special situation it appeared, it probably means what you explained (about faithful flatness etc.). As for your edit: Are there conditions that make an integral extension of an integrally closed ring, integrally closed? – user237522 Jun 16 '15 at 12:37
  • @user237522 I'm not sure. – Kevin Carlson Jun 16 '15 at 21:34
  • Instead of the scary ring $\mathbb{Z}[X^2,X^3]/\langle X^5+2\rangle$ I suggest $\mathbb Z[\sqrt{-3}]$. – user26857 Jun 17 '15 at 08:04