3
$\begingroup$

In [1], Huybrechts and Mauri argue that a holomorphic Lagrangian fibration $f: X \to B$ with smooth base $B$ is flat. This is an application of so called miracle flatness [2, Thm 23.1], because Lagrangian fibrations have equidimensional fibers. Then they write

Remark 1.18. Note that the conclusion that $f$ is flat really needs the base to be smooth. In fact, by miracle flatness, $f$ is flat if and only if $B$ is smooth.

I see that smoothness of $B$ is needed to apply miracle flatness, but I don't see the converse. Why does flatness necessarily fail if $B$ is not smooth?

[1] Huybrechts, Mauri, Lagrangian fibrations, arXiv, 2022

[2] Matsumura, Commutative Ring Theory, 1986

$\endgroup$

1 Answer 1

2
$\begingroup$

The answer lies in Theorem 23.7 from Matsumura's Commutative Ring Theory:

Theorem 23.7. Let $(A, \mathfrak m, k)$ and $(B, \mathfrak n, k')$ be local Noetherian local rings, and $A \to B$ a local homomorphism [...]. We assume that $B$ is flat over $A$. (i) If $B$ is regular then so is $A$.

As $X$ is always assumed to be smooth, the flatness of $f$ will imply the smoothness of the base $B$.

$\endgroup$
2
  • $\begingroup$ Maybe I am confused, but does this imply that the regularity satisfies fpqc descent for Noetherian ring (I have not heard results of this kind before)? Namely, given a faithfully flat map $R\to S$ of Noetherian rings. If $S$ is regular, then so is $R$. $\endgroup$
    – Z. M
    Mar 25 at 12:46
  • $\begingroup$ @Z.M Good point, I think you are correct. Though I'm not too experienced with descent. $\endgroup$ Mar 25 at 13:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.