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I am confused with the following observation: Let $f : X:=\mathrm{Spec}(K) \rightarrow \mathrm{Spec}(k) =: Y$ be a scheme morphism corresponding to a non-trivial finite field extension ( hence $f$ is proper and $X$, $Y$ are noetherian). Its Stein factorization is $ X \overset { \mathrm{id}_X} \longrightarrow X \overset {f} \longrightarrow Y$. But it's clear that $f$ is not a birational, even if $char=0$. We don't have

$f_{*} \mathcal{O}_X = \mathcal{O}_Y$

in this example ( $char = 0 $ or not) either, although we have $f$ has connected fiber and $Y$ is normal.


When I read the following discussion, I got a question.

When will the pushforward of a structure sheaf still be a structure sheaf?

Let $f: X \rightarrow Y$ be a proper morphism of noetherian shcemes and $ X \overset {f'} \rightarrow Z \overset {g} \rightarrow Y$ be its Stein factorization. In J.C. Ottem response, he mentioned that if the fibers of $f$ are connected, then $g$ must be birational ( I think that we assume both $X$ and $Y$ are integral schemes), and from this, one gets $g_{*} \mathcal{O}_Z = \mathcal{O}_Y$, hence we have

$f_{*} \mathcal{O}_X = \mathcal{O}_Y$

He also give the reference, which is Hartshorne III.10.3, for which I think the right one is Hartshorne III.11.3, but 11.3 is telling that the isomorphism gives connected fibers.

So I would like to know :

(1) How to see that $g$ is birational? The correct reference?

(2) In which part we need the characteristic 0 condition?

(3) Could we replace the characteristic 0 condition to another condition, e.g $f$ has integral fibers?

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    $\begingroup$ 1), 2) The fact that $g$ is birational follows from generic smoothness in characteristic zero and the reference is Hartshorne III.10.7 (and not III.10.3). I don't know if there is an analogue to the characteristic $p$ case. $\endgroup$ – J.C. Ottem Sep 7 '11 at 7:43
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    $\begingroup$ You need to assume that the fibres are geometrically connected in order to get $g$ to be birational (in characteristic $0$). Over a field of characteristic $p>0$ it suffices to assume that the fibres are geometrically connected and geometrically reduced. This follows from the elementary fact that if $k$ is a field and $A$ is a finite $k$-algebra then $Spec(A)$ is geometrically connected and reduced iff $A = k$ (as a $k$-algebra). $\endgroup$ – ulrich Sep 7 '11 at 11:27
  • $\begingroup$ Sorry, in my first comment I misread what was said. I got $f$ and $f'$ confused. $\endgroup$ – Karl Schwede Sep 7 '11 at 14:28
  • $\begingroup$ ulrich, a dumb question when you say that ``$A$ is a finite $k$-algebra'', you mean a finite as a $k$-vector space, not finitely generated right? $\endgroup$ – Karl Schwede Sep 7 '11 at 17:14
  • $\begingroup$ @ Karl, yes and it should be finite as a $k$-vector space here to get the conclusion also. And I think that finite algebra is the common terminology for algebras which are finite as modules (over base ring), see p.30 of Atiyah & Macdonald's book "Introduction to Commutative Algebra" $\endgroup$ – user565739 Sep 7 '11 at 18:09

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