# Proper flat morphism with geometrically connected and smooth generic fiber

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?

• Just a guess -- does $\mathbb{Z}[2i] \to \mathbb{Z}[i]$ give a counterexample to the new lemma? ( – hunter Jul 22 '20 at 15:11
• Well this is a blowup, so it is not flat at the exceptional locus. – RandomMathUser Jul 22 '20 at 17:31

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 $$X'=S$$.