# Number of irreducible and connected components constant in flat families

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?

• 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). – Damian Rössler Nov 29 '12 at 16:45
• 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 – HNuer Nov 29 '12 at 16:53
• 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. – Allen Knutson Nov 29 '12 at 18:18

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.
• 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 $S$? – HNuer Nov 29 '12 at 18:54