2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Just a guess -- does $\mathbb{Z}[2i] \to \mathbb{Z}[i]$ give a counterexample to the new lemma? ( $\endgroup$ – hunter Jul 22 '20 at 15:11
  • 1
    $\begingroup$ Well this is a blowup, so it is not flat at the exceptional locus. $\endgroup$ – RandomMathUser Jul 22 '20 at 17:31
0
$\begingroup$

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 $X'=S$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.