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I already asked this on math.stackexchange.com, but didn't receive an answer.

Suppose $f: X \to Y$ is a (possibly proper) morphism of complex manifolds (resp. smooth varieties) such that all fibers of $f$ are of constant dimension $n = \dim X - \dim Y$. We may also suppose that $f$ has connected fibers, I don't know if this is relevant. Also suppose that $Z \subset Y$ is an analytic subset of codimension $\operatorname{codim}(Z, Y) \geq 2$, such that $f$ is smooth over $f^{-1}(Y \setminus Z)$.

Question: Is it true that the fibers of $f$ are generically reduced?

Context: I encountered this problem when reading the paper Characteristic Foliation on the Discriminant Hypersurface of a Holomorphic Lagrangian Fibration by J.-M. Hwang and K. Oguiso. There one has $\dim X = 2n = 2 \dim Y$, and they argue as follows: Suppose $z \in Z$, and let $x \in f^{-1}(z)$ be a point which is a smooth point in the reduction $f^{-1}(z)_{\text{red}}$. Let $W \subset X$ be a smooth $n$-dimensional submanifold, which is meets $f^{-1}(z)_{\text{red}}$ at $x$ transversally. Then they claim that (possibly after shrinking $Y$), the restricted map $f|_W: W \to Y$ is unramified over $Y \setminus Z$ (I don't know why that is true), and hence by the purity of branched loci, $f$ is unramified over $Y$. So in particular, $f|_W^{-1}(z) = f^{-1}(z) \cap W$ is smooth at $x$, so $f^{-1}(z)$ is reduced at $x$.

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    $\begingroup$ Probably you know this, but the case $n=0$ is fine by Zariski-Nagata purity. $\endgroup$
    – Piotr Achinger
    Commented Feb 8, 2021 at 10:23
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    $\begingroup$ de Jong's talk at WAGON sites.math.washington.edu/~jarod/wagon.html seems to be exactly about this question (in the AG setting). He said that when $X$ and $Y$ are smooth (as in your setup) then the assumption in fact implies that $f$ is smooth. He said that this is a result of Dolgachev and gave a reference to arxiv.org/pdf/1003.5872.pdf Theorem 3.5 (4), (5) is the desired result, I think. $\endgroup$
    – SashaP
    Commented Feb 8, 2021 at 16:44
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    $\begingroup$ @SashaP: No. This holds only when the fibers are 1-dimensional. There is an example (due to Mumford I believe) with $\dim X=4$, $\dim Y=2$, and only one fiber singular. However in the example the fiber is reduced, so this does not answer the question of the post. $\endgroup$
    – abx
    Commented Feb 8, 2021 at 21:00
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    $\begingroup$ @SashaP Thanks for your remark, but I don't see how the estimate by Dolgachev solves my problem. The estimate is $\operatorname{codim}(\operatorname{Sing}(f)) \leq 1 + n$, where $\operatorname{Sing}(f) \subset X$ is the set of points where $f$ is not smooth. But since $f$ is smooth over $f^{-1}(Y \setminus Z)$, I only have $\operatorname{codim}(\operatorname{Sing}(f)) \geq 2$, so I cannot conclude that $f$ is smooth. $\endgroup$
    – red_trumpet
    Commented Feb 9, 2021 at 10:28
  • $\begingroup$ @red_trumpet Right, I've confused purity on the source of the morphism with that on the target. $\endgroup$
    – SashaP
    Commented Feb 10, 2021 at 1:52

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