[Math] Number of irreducible and connected components constant in flat families

ag.algebraic-geometry

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?

Best Answer

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.

Related Question