# Proper morphisms with geometrically reduced and connected fibers

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?

• What is your argument for showing $S'\to S$ has reduced fibers without flatness of $f$? Jul 20, 2020 at 21:36
• 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. Jul 21, 2020 at 0:29
• I miscomputed the pushforward of the structure sheaf. The example in my previous comment is wrong. Jul 21, 2020 at 9:42
• @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. Jul 21, 2020 at 15:27

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.
• 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. Jul 22, 2020 at 14:26
• @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. Sep 24, 2021 at 15:59