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Let $f: X \to S$ be a proper morphism ($S$ locally noetherian), and $X \to S' \to S$ its Stein factorisation. By Zariski's Main Theorem the number of geometric connected components of the fibers of $f$ can be read from the cardinal of the fibers of the finite $S' \to S$. In particular if all fibers of $f$ are geometrically connected, then $S' \to S$ is radicial.

I expect that if furthermore the fibers of $f$ are geometrically reduced (and $f$ is surjective and $S$ reduced in order to remove trivial counterexamples), then $S'=S$ that is $f$ is an $\mathcal{O}$-morphism (viz. $f_*\mathcal{O}_X = \mathcal{O}_S$). Strangely I only find this fact when $f$ is furthermore assumed to be flat, for instance: https://stacks.math.columbia.edu/tag/0E0L.

Here is an outline of a demonstration (suggested by a friend): we want to show that $S' \to S$ is an isomorphism. Since it is a surjection by assumption on $f$, it suffices to show that it is an immersion. By our assumptions on $f$, $S' \to S$ has geometrically connected and reduced fibers. We way assume that $S=\textrm{Spec} A$ and $S'=\textrm{Spec} B$, with $A \to B$ finite. Let $C$ be the cokernel of $A \to B$ (seen as a $A$-module). If $p$ is a prime ideal in $A$, $B \otimes_A \overline{k}(p) = \overline{k}(p)$ (since it is connected and reduced over $\overline{k}(p)$), so $C \otimes_A \overline{k}(p)=0$, so $C=0$.

Is the above proof indeed correct? Does the hypotheses already imply that $f$ is flat? Is there a reference to this result somewhere in the literature, presumably in EGA?

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    $\begingroup$ What is your argument for showing $S'\to S$ has reduced fibers without flatness of $f$? $\endgroup$
    – Mohan
    Commented Jul 20, 2020 at 21:36
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    $\begingroup$ That is false when $S$ is a cuspidal plane cubic, when $X$ is the union of a twisted cubic and a tangent line in $3$-space, and the morphism is linear projection from a general point on the line. $\endgroup$
    – Jason Starr
    Commented Jul 21, 2020 at 0:29
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    $\begingroup$ I miscomputed the pushforward of the structure sheaf. The example in my previous comment is wrong. $\endgroup$
    – Jason Starr
    Commented Jul 21, 2020 at 9:42
  • $\begingroup$ @Mohan: you are right, unlike for geometric connectedness which is a topological condition so obvious, it is not obvious that $S' \to S$ has geometrically reduced fibers. Still I would expect that if it were not the case then $X \to S$ would have non geometrically reduced fibers too, but my intuition is probably wrong. $\endgroup$
    – RandomMathUser
    Commented Jul 21, 2020 at 15:27

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Here is a standard example. Take $\mathbb{P}^1\subset\mathbb{P}^3$ of large degree and let $S$ be the cone, with the vertex $p$, only singular point. Let $f:X\to S$ the blow up of $p$. One can check that $X$ is smooth and thus the Stein factorization $S'$ is the normalization of $S$. The fiber over $p$ in $X$ is smooth irreducible (scheme-theoretically), but the fiber in $S'$ is not reduced.

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    $\begingroup$ Thanks for the counterexample. Now this makes me wonder if there is a nice condition, less strong than flatness, that ensures that $S'$ has reduced geometric fibers. $\endgroup$
    – RandomMathUser
    Commented Jul 22, 2020 at 14:26
  • $\begingroup$ How does one see that the fiber in S' is non-reduced? (or that S is not normal) $\endgroup$
    – Nikolas Kuhn
    Commented Sep 24, 2021 at 15:41
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    $\begingroup$ @NikolasKuhn One has a natural map $\oplus H^0(O_{\mathbb{P}^3}(n))\to \oplus H^0(O_{\mathbb{P}^1}(nd))=B$, where $d$ is the degree and let $A$ be the image of the first ring in $B$. Then, $A$ is a proper subring of $B$ since $d$ is large and $B$ is an integral birational extension of $A$. One checks that spectrum of $B$ is $S'$ and that of $A$ is $S$. Also, $B$ is normal. Hope you can fix the rest. $\endgroup$
    – Mohan
    Commented Sep 24, 2021 at 15:59

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