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If $f:X \to Y$ is a flat and proper surjective morphism between smooth schemes over an algebraically closed field, and $f$ has connected fibers, does it imply that $$f`_*\mathcal O_X = \mathcal O_Y?$$

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According to the answers below, I think we should agree that $f$ is supposed to be surjective. Could you edit? –  Qfwfq Apr 23 '10 at 11:13

4 Answers 4

up vote 8 down vote accepted

This follows from Zariski's main theorem if the characteristic is zero and it is false in positive characteristics: consider the the morphism $\mathbb{A}^1 \to \mathbb{A}^1$ given by $x \mapsto x^p$ where $p$ is the characteristic. The statement would also be true in char p if you assume that the general fibre is reduced.

(Note that it suffices to assume that X is integral and Y is normal. $f$ should of course also be surjective.)

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Can you please explain how Zariski's main theorem implies the result. The schemes I use also proper and the general fiber is indeed reduced in the char p case. –  Athena Apr 27 '10 at 11:50
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We have a factorisation $X \to Spec \ f_* \mathcal{O}_X \to Y$ with $Spec f_* \mathcal{O}_X$ an integral scheme which is finite over $Y$. Since $k$ is algebraically closed and the fibres of $f$ are connected it follows that the morphism $Spec \ f_* \mathcal{O}_X \to Y$ also has connected fibres. It is therefore birational if $k$ is of char $0$ (since the extension of function fields is separable) and is birational in char p if the fibres are also reduced. Since $Y$ is normal it follows that the map is an isomorphism. (So it seems that ZMT is not really necessary...) –  ulrich Apr 28 '10 at 5:15

If $X$ is smooth over $K$ and if $K$ is of characteristic $p>0$, then the relative Frobenius $F:X\rightarrow X\times_{F_X} K=:X'$ is faithfully flat $K$-morphism and finite, hence proper. Moreover it's a homeomorphism, so the fibers are connected. But it's not hard to see that $F_*\mathcal{O}_{X}$ is locally free of rank $p^n$, if $X$ is of dimension $n$.

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Since $f : X \to Y$ is a projective morphism, we can use Stein factorization and write $f = h \circ g$ with

  • $g : X \to Z$ having connected fibers and $g_*\mathcal{O}_X = \mathcal{O}_Z$

  • $h : Z \to Y$ a finite morphism.

Since $f$ has connected fiber and surjective, $h$ then must be surjective and has degree 1. It follows that $h$ is an isomorphism. Hence, $f_*\mathcal{O}_X = \mathcal{O}_Y$.

I think we only need to assume that $X$ and $Y$ are noetherian, normal and over $\mathbb{C}$.

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Over $\mathbb{C}$, I would reason as follows.

For any $V$ open in $Y$, $\mathcal{O}_Y(V) \cong \mathcal{O}_X(U)$, where $U:=f^{-1}(V)$, via the pullback

$f^*:\mathcal{O}_Y (V) \rightarrow \mathcal{O}_X (U)$

$f^*:h \mapsto h \circ f$, thinking of $h$ as a morphism $h:Y \rightarrow \mathbb{A}^1$.

The above pullback is in general injective, but in this case it's even surjective because every regular function $g$ on $X$ has constant value on the fibers (by base change, in this case the (reduction of the) fibers are connected proper varieties), hence is the pullback of a function $h$ on $Y$.

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