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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.

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  • $\begingroup$ 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. $\endgroup$
    – Piotr Achinger
    Commented Jul 31, 2019 at 20:38

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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).

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  • $\begingroup$ then fin. pres. assumption is useless? $\endgroup$
    – user143832
    Commented Jul 31, 2019 at 20:51
  • $\begingroup$ 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). $\endgroup$
    – user143832
    Commented Jul 31, 2019 at 21:01
  • $\begingroup$ 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). $\endgroup$
    – user143832
    Commented Jul 31, 2019 at 21:05
  • $\begingroup$ 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... $\endgroup$
    – Sándor Kovács
    Commented Aug 1, 2019 at 7:29
  • $\begingroup$ 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). $\endgroup$
    – user143832
    Commented Aug 1, 2019 at 7:53

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