Question

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?

Answer 1

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....

Answer 2

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.