# Flatness of $\Omega_{B/K}$ over $B$.

Let $K$ be a field of characteristic zero. Assume that $K \subset A \subseteq B$ are noetherian integral domains, with $A$ regular (= all its localizations at maximal ideals are regular local rings). Assume also that $B$ is a flat $A$-module.

I do not mind to further assume integrality of $B$ over $A$ if that helps-- in that case $B$ is faithfully flat over $A$ (since integral+flat implies faithfully flat).

My question: Is it possible to show that $B$ is regular? I suspect there exists a counterexample.

Corollary on page 201 of this paper says that $B$ is regular iff $\Omega_{B/K}$ is flat over $B$.

I wonder if there is a connection between $A$-flatness of $\Omega_{A/K}$ (= which is known, since $A$ is regular) and $B$-flatness of $\Omega_{B/K}$.

Recall: $\Omega_{B/K}$ is the $B$-module of $K$-differentials = $\ker g/(\ker g)^2$, where $g: B\otimes_K B \to B$ is defined by $g(b_1 \otimes_K b_2)=b_1b_2$.

If I am not wrong, the converse of what I ask is true, namely: regularity of $B$ implies regularity of $A$, by Proposition 8, page 59 of Bourbaki's book "Commutative Algebra", chapter 10.

## 1 Answer

I don't think this is necessarily true.

Let $A=K[t]$ and $B=K[x,y,t]/(xy-t^2)$. Then $B$ is flat over $A$ (this is because the Hilbert polynomial is constant in this family). Of course they are flat as $K$-modules as well, since $K$-modules are vector spaces, and vector spaces are free.

Then all the conditions are satisfied, but $B$ is singular at the origin.

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Here's another example, which satisfies all conditions, including integral and faithfully flat, thanks to Jake Levinson's comment below:

$$k[x] \to k[x,t]/(x^2-t^3)$$

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Read the below musings with skepticism:

I don't know if B is regular if we further impose integrality or faithful flatness. I suspect it is true, however, for the following reason: a ring being integral (and finite type) over another ring says that the corresponding maps of varities are covering maps, and since covering maps are local diffeomorphisms, smoothness of A should imply smoothness of B.

• Nice answer, thanks. However, I am still curious if adding integrality does imply regularity of $B$. Unfortunately, I am not so familiar with the material appearing in your last three lines, though it seems to hint a positive answer. Should I post the flat+integral case as a separate question? (with mentioning your plausible positive answer)? – user237522 Jun 12 '15 at 14:41
• You could change your example to $B = K[x,t]/(x^2-t^3)$, which is flat and integral over $K[t]$ but not regular. And faithfully flat. – Jake Levinson Jun 12 '15 at 17:52
• @Fredrik: it is not true that being integral means the corresponding map is smooth. Being integral means it is finite; the condition you are thinking of is etale -- in particular unramified. – Jake Levinson Jun 12 '15 at 17:57
• @JakeLevinson Thank you for the comment! I'll add that example as well. – Fredrik Meyer Jun 13 '15 at 9:16
• Thank you very much Jake Levinson and Fredrik Meyer! You have showed that f.g. (as modules) and faithful flatness are not enough to conclude regularity of $B$; what additional conditions are needed in order to get regularity of $B$? Maybe separability of $B$ over $A$? If the ring extension is separable (and by assumption the rings are noetherian), then the ring extension is unramified, and I have assumed it's flat, so it's etale; – user237522 Jun 14 '15 at 2:40