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


1 Answer 1


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 $B$.

  • $\begingroup$ 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

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