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Feb 24, 2017 at 3:55 comment added R. van Dobben de Bruyn And for those of us lucky enough to have non-algebraically closed fields in our lives: the assumption should be that $f$ has geometrically connected fibres, rather than merely connected. An easy counterexample is a field extension.
Oct 17, 2016 at 12:41 comment added David Holmes I think the reference to III.10.3 is incorrect, or at least it points to something irrelevant in the edition I have to hand. Also, the statement `if $Y$ is in addition normal, then $f_∗\mathcal O_X=\mathcal O_S$ holds' needs $X$ reduced even in characteristic zero, see comment above.
Jan 3, 2013 at 1:40 comment added minimax Is the reference to Hartshorne III.10.3 correct?
Apr 28, 2011 at 18:10 comment added Karl Schwede Another quick comment, one can do even better than normality. Let's work over an algebraically closed field of characteristic zero. With J.C. Ottem's notation, assume that $f$ has connected fibers and $Y$ is seminormal. Doing Stein factorization as above, one has $g : Z \to Y$ a finite map, which is birational (at least in characteristic zero). Furthermore, one can show that $g$ also has connected fibers because $f'$ is surjective (since it is proper and dominant). Thus $g$ is an isomorphism by the defining property of seminormality (over algebraicaly closed field of char. 0).
Apr 28, 2011 at 18:08 comment added J.C. Ottem Yes, that's true. In that case $g$ is bijective, but not necessarily an isomorphism.
Apr 28, 2011 at 18:04 comment added Karl Schwede A minor correciton, if $Y$ is normal and $f$ has connected fibers, then $f_* \mathcal{O}_X = \mathcal{O}_Y$ holds in characteristic zero. In characteristic $p$, this is not the case (for example the Frobenius morphism).
Apr 28, 2011 at 17:10 history edited J.C. Ottem CC BY-SA 3.0
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Apr 28, 2011 at 16:48 history edited J.C. Ottem CC BY-SA 3.0
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Apr 28, 2011 at 16:34 history answered J.C. Ottem CC BY-SA 3.0