0
$\begingroup$

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?

$\endgroup$
5
  • 1
    $\begingroup$ 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, 2011 at 7:43
  • 3
    $\begingroup$ 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). $\endgroup$
    – naf
    Sep 7, 2011 at 11:27
  • $\begingroup$ Sorry, in my first comment I misread what was said. I got $f$ and $f'$ confused. $\endgroup$ Sep 7, 2011 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? $\endgroup$ Sep 7, 2011 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, 2011 at 18:09

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.