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I am trying to understand the proof of the following statement in http://virtualmath1.stanford.edu/~conrad/mordellsem/Notes/L13.pdf

Lemma 7.1. Let $f:X\to S$ be a proper flat surjective map to a noetherian scheme $S$, and assume that $f$ has geometrically connected and smooth generic fibers. Then all fibers are geometrically connected.

Proof. We may and do assume that $S$ is reduced and irreducible (by base change to irreducible components of $S$, equipped with the reduced structure). For a non-generic point $s\in S$ there is a discrete valuation on the function field of $S$ that dominates $O_{S,s}$ [EGA,II, 7.1.7], so by base change to such a ring we can assume that $S=\operatorname{Spec} R$ for a discrete valuation ring $R$. Let $K=\operatorname{Frac}(R)$. By $R$-flatness of $X$ and smoothness and geometric connectedness of the generic fiber, the $R$-finite $H^0(X,O_X)$ injects into $H^0(X_K,O_{X_K})=K$. Thus, $R=H^0(X,O_X)$ by the normality of $R$. That is, $X\to\operatorname{Spec} R$ is its own Stein factorization. But Stein factorizations always have geometrically connected fibers [EGA, III1, 4.3.4].

So I am not sure where the smoothness condition is needed exactly, it looks like we only need that the generic fiber is geometrically reduced. Indeed, the injection $H^0(X,O_X) \to H^0(X_K,O_{X_K})$ only require the flatness of $X$: the restriction to $X_K$ is injective if $X_K$ contains all associated primes, but since $X$ is flat over $R$ all associated primes live in the generic fiber. And $H^0(X_K,O_{X_K})=K$ only requires $X_K$ to be geometrically connected and geometrically reduced.

So it looks like the following statement is true:

Lemma. Let $f:X\to S$ be a proper flat surjective map to a noetherian scheme $S$, and assume that $f$ has geometrically connected and reduced generic fibers. Then all fibers are geometrically connected.

Am I missing something?

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  • $\begingroup$ Just a guess -- does $\mathbb{Z}[2i] \to \mathbb{Z}[i]$ give a counterexample to the new lemma? ( $\endgroup$
    – hunter
    Jul 22, 2020 at 15:11
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    $\begingroup$ Well this is a blowup, so it is not flat at the exceptional locus. $\endgroup$ Jul 22, 2020 at 17:31

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So in fact we even have this:

Lemma. Let $f:X→S$ be a proper flat surjective map, S loc. noeth and assume that f has geometrically connected fibers. Then all fibers are geometrically connected.

Indeed, by EGA.4.15.5.9 the number of geometrically connected components is lower semi-continuous, so if it is one on the generic fibers it is one (by surjectivity) everywhere. This extends to the non loc. noeth case by the usual approximation techniques when $f$ is locally of finite presentation.

So we don't even need geometrically reduced fibers. What they give is that in this case $f$ is Stein, ie $f_\ast O_X = O_S$. Indeed in this case if $X'=\mathrm{Spec} f_\ast O_X$, then $X'$ is flat over $S$, so finite étale over $S$ since $X \to X' \to S$ is the Stein factorisation. The fibers of $X' \to X$ are geometrically connected and reduced over $S$, so $X'_s = \mathrm{Spec}\ k(s)$ and $X'=S$.

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