# Global sections in proper flat families

I was trying to unravel comments under the question When is it true that the ring of global regular functions on a projective variety is just the base ring?. It is apparently being claimed that if a proper morphism of finite presentation between schemes $$X\rightarrow S$$ is flat and has geometrically connected & reduced fibers then the natural map $$\mathcal{O}_S(S)\rightarrow \mathcal{O}_X(X)$$ is an isomorphism. Is it true? By Grothendieck's coherence theorem the map is finite but I am not sure what to do next.

Maybe I should also mention that in my book empty space is connected (this probably does not change anything). Actually, flat morphism locally of finite presentation is open https://stacks.math.columbia.edu/tag/01UA so as long as we assume that $$S$$ is connected the fibers will be non-empty.

• Sketch: 1) For $S = \operatorname{Spec} k$ for $k$ a field, use flat base change to $\bar k$ and the assumption on the geometric fiber. 2) In general, the flatness assumption + cohomology and base change/Grauert tell you that the formation of $f_* \mathcal{O}_X$ commutes with arbitrary base change $S'\to S$, 3) combine 1+2 and see that $\mathcal{O}_S\to f_* \mathcal{O}_X$ is an isomorphism on all fibers, and hence an isomorphism. Jul 31, 2019 at 20:38

This follows from Stein factorization: https://stacks.math.columbia.edu/tag/03H2

One only has to observe that if the fibers are geometrically connected & reduced, then in the Stein factorization $$X\to S'\to S$$, where $$S'\to S$$ is finite, that morphism also has to have geometrically connected & reduced fibers, which means that it must be an isomorphism (each scheme theoretic fiber is a geometrically connected & reduced zero dimensional scheme, i.e., a reduced point).

• then fin. pres. assumption is useless?
– user143832
Jul 31, 2019 at 20:51
• sure, I mean if the base is not Noetherian then a proper morphism is not necessarily of finite presentation, stacks.math.columbia.edu/tag/05LC (and I am not sure how the standard results work then).
– user143832
Jul 31, 2019 at 21:01
• in fact, it seems to me that the example in the linked page starting with "Let $M=R\oplus\dots$" is a counterexample to the statement in absence of fin. pres. assumption (it has two at least two $R$-direct summands in the global sections).
– user143832
Jul 31, 2019 at 21:05
• I am not sure what example you are referring to, I did not find a line with "$M=R\oplus\dots$" there. However, two things: i) on the stacks project, the proof actually considers this issue by writing $f$ as a limit of finitely presented morphisms, and ii) as I mentioned I am not sure what your example is, but direct sums tend to have non-connected fibers... Aug 1, 2019 at 7:29
• sorry, on this page: stacks.math.columbia.edu/tag/05LB Ctrl+F "the symmetric algebra", you will see the example. The fibers are projective spaces (so geometrically integral).
– user143832
Aug 1, 2019 at 7:53