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Let $f:X \rightarrow Y$ be a morphism between two smooth projective varieties $X,Y$ which are defined over an algebraically closed field $k$. I am looking for some criteria which guaranties the projectivity of $f$.

For instance if $f$ is finite it is projective. Here we don't need the projectivity of the varieties $X,Y$.

Is the morphism $f$ projective if

Question1: The fibers of $f$ are finite?

Question 2: $f$ is one-to-one?

Question 3: $f$ is onto?

Does the assumption $k=\mathbb{C}$ make the questions easier?

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    $\begingroup$ A morphism between projective varieties is always projective. $\endgroup$
    – Angelo
    Commented Oct 7, 2010 at 10:12
  • $\begingroup$ Thanks Angelo. I suspected this to be true. But I couldn't prove it myself. $\endgroup$
    – Passenger
    Commented Oct 7, 2010 at 10:24
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    $\begingroup$ If you want a reference you can have a look at Goertz-Wedhorn AGI Summary 13.71 $\endgroup$
    – babubba
    Commented Oct 7, 2010 at 10:26

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Here is an attempt to prove Angelo's comment (it seems too simple to use a reference for it):

$X,Y$ defined over $S$. If they are both proper over $S$, then so is $f$ by Hartshorne, II.4.8(e). In particular $f$ is separated and universally closed.

If $X$ is projective over $S$ then for some $n$ there exists $\iota: X\to \mathbb P^n_{S}$, a universally closed separated immersion.

The morphism $\nu:X\to\mathbb P^n_S\times_S Y=:\mathbb P^n_Y$ defined by $x \mapsto (x,f(x))\in \mathbb P^n_S\times_SY$ is the composition of the base extension of $f$ by the projection $\pi:\mathbb P^n_Y\to Y$; $f_{\pi}$, the embedding $\iota$ base extended by the identity of $X$; $X\times_SX\to \mathbb P^n_S\times_S X$ and the diagonal morphism of $X$; $\Delta_X: X\to X\times_S X$. I.e., $\nu=f_{\pi}\circ (\iota\times_S{\rm id}_X)\circ \Delta_X: X\to \mathbb P^n_S\times_S Y$. Actually, this might be a better definition than the one with "coordinates". Since $f$ is separated and universally closed and $\iota$ is universally closed, it follows that $\nu$ is closed. It is obviously an embedding. Now $f=\pi\circ\nu$ and hence it is projective.

Well, may be it was not that simple, and Angelo might tell me that it is wrong....

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  • $\begingroup$ Sure, this is one way to do it. $\endgroup$
    – Angelo
    Commented Oct 7, 2010 at 19:39
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A morphism $X \to Y$ factorizes as an embedding $X \to X\times Y$ (the graph of $f$) followed by the projection $X\times Y \to Y$. The first is a closed embedding (and hence is projective) if $Y$ is separable and the second is projective if $X$ is projective.

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  • $\begingroup$ Krampusz, I am sorry, when I was typing my answer I didn't see yours. Should I delete mine? $\endgroup$
    – Sasha
    Commented Oct 8, 2010 at 5:29
  • $\begingroup$ The reason is evident --- your answer is much longer and people are usually lazy to read long answers. By the way, I don't think it is a good idea to delete the accepted answer. $\endgroup$
    – Sasha
    Commented Oct 9, 2010 at 17:36

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