# If $\pi : X \to Y$ is a flat, proper $O$-connected morphism of locally Noetherian schemes, then is $h^0(X_q, O_{X_q}) = 1$?

If $\pi : X \to Y$ is a flat, proper $O$-connected morphism of locally Noetherian schemes, then is $h^0(X_q, O_{X_q}) = 1$ (the dimension of the space of global sections of the fibers)?

This seems to be used in a proof I am reading, but I can't see why it is true.

$O$-connected means that the natural map $O_Y \to \pi_* O_X$ is an isomorphism.

For a point $p$ with residue field $k(q)$ I would like to say something like, $k(q) = O_Y \otimes k(q) \cong (\pi_* O_X) \otimes k(q) \cong \pi'_*(O_X \otimes k(q))$ (where this last $\pi'$ is projection onto $k(q)$, hence gives global sections of the fibers), but $Spec k(q) \to Y$ is generally not flat, so I don't know how to argue for the last isomorphism.

(Or maybe there is some way to guarantee that the fibers are all geometrically connected and geometrically reduced?)

For reference, the proof I am reading is 28.1.11 in Ravi Vakil's notes.

• @Hoot Oh you are right... thanks. I was getting confused, because I felt that closed subschemes where not usually flat. But of course everything is flat over a field... – Lorenzo Apr 25 '16 at 13:37
• @Hoot Wait... no I am confused still. Of course $A$ (say Spec A is $Y$ or something) is flat over $A/m$, but the question is whether the morphism $Spec (A \to A/m)$ is flat, and that isn't clear. The morphism should have been from $Spec k(q) \to Y$, because I am trying to pullback along it. It was a typo. – Lorenzo Apr 25 '16 at 14:01
• Did you ever get a resolution on this? Hartshorne has a similar statement in III.12 working over a $k = \bar k$ where I think we all agree everything works. Vexing. – Hoot Jun 7 '16 at 4:11
• @Hoot I agree with you that the theorem on formal functions can be used to prove that the fibers are connected of $\pi$ are connected. Is it possible to upgrade this to the fibers being reduced? Is it possible to upgrade the theorem on formal functions to the analogous statements about the geometric fibers? So the short answer to your question is: I don't know. – Lorenzo Jun 7 '16 at 22:09
• @AreaMan Almost a year later, did you figure this out? I am stuck at the same point. – Asvin Apr 1 '17 at 0:01