Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $$f`_*\mathcal O_X = \mathcal O_Y?$$

share|improve this question
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.)

share|improve this answer
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
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 $n$.

share|improve this answer

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 $\mathbb{C}$.

share|improve this answer

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 $Y$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.