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A) Let $f:F\rightarrow S$ be a flat proper morphism of schemes with geometrically normal fibers. Then supposedly the number of $\textbf{connected}$ components of the geometric fibers is constant. Why is this? Without some kind of vanishing of cohomology or information on the base, I don't see why this is true.

B) Furthermore, supposedly if $F$ is now a flat proper morphism with reduced, connected, nodal curves as geometric fibers, then there is a Zariski open subset of $S$ on which the fibers all have the same number of $\textbf{irreducible}$ components. Why is this?

Finally, how far can these results be generalized? For example, is B) true for any flat proper morphism?

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    $\begingroup$ For A): this is because in the Stein factorization of $f$, the quasi-finite morphism is étale. See EGA III, 2, 7.8.6 or Illusie's article in the book "FGA explained" Prop. 8.5.16. You have to assume that $S$ is noetherian, though. For B): EGA IV, 9, Prop. 9.7.8. It is the number of geometric irreducible components, which remains the same in a Zariski open set, though (I think it is false otherwise, even in the situation you consider). $\endgroup$
    – Damian Rössler
    Commented Nov 29, 2012 at 16:45
  • $\begingroup$ You didn't want to put this in an answer and get the points? :) As usual Damian, your response has helped me a lot! Thanks $\endgroup$
    – HNuer
    Commented Nov 29, 2012 at 16:53
  • $\begingroup$ You might like our paper arxiv.org/abs/math/0602626 in which $F$ has a finite map to $Y \times S$, $Y$ is proper, and the fibers are only required to be reduced. We prove results about completing such families at holes in $S$, you could say. $\endgroup$
    – Allen Knutson
    Commented Nov 29, 2012 at 18:18
  • $\begingroup$ See stacks.math.columbia.edu/tag/0E0N and stacks.math.columbia.edu/tag/055A. $\endgroup$
    – Aoi Koshigaya
    Commented Nov 10, 2020 at 1:23

2 Answers 2

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A) You can easily reduce to the case that $S$ is the spectrum of a DVR $R$. Furthermore, by passing to a finite extension of $R$, you can assume that the components of the closed fiber are geometrically connected. Say that there are $d$ of them; then by semicontinuity the generic geometric fiber has at most $d$ components. Take a Stein factorization $X \to T \to S$; then $T$ has $d$ points over the closed point of $S$. Then $T$ is flat over $S$; this implies that the number of connected components of the generic geometric fiber of $X$ over $S$, that equals the degree of the generic fiber of $T$ over $S$, is at least $d$. This proves the equality.

B) The number of connected components of a geometric fiber is the dimension of H^0 of the structure sheaf of the fiber; the result follows from semicontinuity.

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  • $\begingroup$ Not to be pedantic, but did you switch A) and B)? Also for your argument in B), $\dim H^0$ of the fiber being the number of connected components doesn't necessarily apply if the fibers have embedded components, which I believe normality rules out, so would reduced geometric fibers be enough there? Also I don't see how semicontinuity finishes it. It implies there is an open set of $S$ upon which the minimal number of components is achieved, but why does this imply it is all of $S$? $\endgroup$
    – HNuer
    Commented Nov 29, 2012 at 18:54
  • $\begingroup$ I did not switch A and B, but you are right that I was not answering your question. I'll clarify tomorrow. $\endgroup$
    – Angelo
    Commented Nov 29, 2012 at 21:00
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Part (A) is precisely the content of Theorem 4.7, (iii) in

Deligne, Pierre; Mumford, D., The irreducibility of the space of curves of a given genus, Publ. Math., Inst. Hautes Étud. Sci. 36, 75-109 (1969). ZBL0181.48803.

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