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As we know there are a lot of principle of connectedness in algebraic geometry. Here is a useful and interesting one:

Suppose T is an integral curve over k. X-->T is a flat family of closed subvarieties in $P_k^n$. If there is a non-empty open subset U in T such that at every closed point t in U, the fiber X_t is connected. Then show every fiber X_t is connected for any t in T.

In consideration of uppercontinuous property, this says that if the parameter space is a curve, then if $h^0(X_t, O_{X_t})$ is locally constant on some open set, then it's locally constant everywhere!(If we further require k is algebraically closed here).

I thing this property is interesting and useful, but I can't prove it, and every reference I can find traces back to Hartshorne's exercise III.11.4. If anyone can give me a proof, I would be very grateful!

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EDIT: This is a small correction on the previous argument that I posted here, prompted by a comment of Steven Sam.

The question is stable under any base change. Let $\nu : \widetilde{T} \to T$ be the normalization of $T$. Let $\tilde{f} : \widetilde{X} \to \widetilde{T}$ be the base change of $f : X \to T$ via $\nu$. Then $\tilde{f}$ is again flat and projective and $\widetilde{T}$ is smooth. The sheaf $\tilde{f}\_* \mathcal{O}\_{\widetilde{X}}$ is torsion free since $\tilde{f} : \widetilde{X} \to \widetilde{T}$ is flat and projective. Since $\widetilde{T}$ is a smooth curve this implies that $\tilde{f}\_*\mathcal{O}\_{\widetilde{X}}$ is locally free. Since $f$ is assumed to have connected fibers over $U$, we get that $\tilde{f}$ has connected fibers on $U$ and so $\tilde{f}\_* \mathcal{O}\_{\widetilde{X}}$ has rank one on $\nu^{-1}(U)$ and so it must have rank one everywhere. Thus $\tilde{f}\_*\mathcal{O}\_{\widetilde{X}}$ is the sheaf of sections of a line bundle on $\widetilde{T}$. On the other hand the space of global sections of $\tilde{f}\_* \mathcal{O}\_{\widetilde{X}}$ is equal to the space of global sections of $\mathcal{O}\_{\widetilde{X}}$ and so $\tilde{f}\_* \mathcal{O}\_{\widetilde{X}}$ has a nowhere vanishing section corresponding to the section $1$ of $\mathcal{O}\_{\widetilde{X}}$. This shows that $\tilde{f}\_* \mathcal{O}\_{\widetilde{X}}$ is the trivial line bundle, i.e. $\tilde{f}\_* \mathcal{O}\_{\widetilde{X}} \cong \mathcal{O}\_{\widetilde{T}}$. Now this implies that all fibers of $\tilde{f}$ are connected and hence the fibers of $f$ are connected.

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I took the liberty of fixing your latex: I wanted to read it :P – Mariano Suárez-Alvarez Dec 16 '09 at 5:14
Thanks! I couldn't figure out what I was doing wrong. It is good to know how to fix those subscripts. – Tony Pantev Dec 16 '09 at 5:18
I am not sure why the connectedness of generic fiber can imply that $f_* \mathcal{O}_X$ has rank one generically. Consider this example, take A to be a DVR, hence spec A only has one closed point and one generic point. If we choose B to be a finite extension of A such that the unique prime of A ramifies in B, then Spec B-->Spec A has generic fiber but don't have rank 1. Hence this proposition seems fail to hold for finite proper morphisms. Projectivity may play a more important role rather than just implying $f_* \mathcal{O}_X$ is a coherent sheaf on T, I think. – Taisong Jing Dec 16 '09 at 5:56
In my last comment, I meant Spec B--> Spec A has CONNECTED generic fiber, sorry for missing that information. – Taisong Jing Dec 16 '09 at 5:58
Taisong, in your setup you assumed that your morphism maps you to an integral curve over a field and that in $T$ you have a non-empty open set $U$ over which all fibers of $f$ arre connected. Restricting f over $U$ gives a projectve morphism with connected fibers so the push-forward of the structure sheaf is the structure sheafd. – Tony Pantev Dec 16 '09 at 13:29

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