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?



Proper morphisms with geometrically reduced and connected fibers
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.


