TITLE: $f_L:X\to\mathbb P^r$ if finite $\Longleftrightarrow L$ is...?
QUESTION [9 upvotes]: Suppose we have a line bundle $L$ on an algebraic variety $X$. Let us assume $L$ is globally generated, and let $f_L:X\to \mathbb P^r$ denote the corresponding morphism.

Question. Under which condition(s) on $L$ is the morphism $f_L$
  finite?

For example, if $X$ is a curve then to give a finite degree $d$ morphism $X\to \mathbb P^1$ (so we fixed $r=1$) is the same as to give a degree $d$ line bundle (plus two independent sections). But I think this readily fails if, for instance, we drop the assumption $r=1$, or the assumption that $X$ is a curve.
However, $L$ determines $f_L$ (yes, up to the choice of a basis for the generating sections), so there should be some condition to put directly on $L$ to translate the finiteness of $f_L$. I know this is a very basic question but I am unable to find these conditions.
I only noticed that they should be weaker than "being very ample" (because a closed immersion is finite). Moreover, a finite morphism is quasi-finite, so the linear system $|L|$ should be one whose divisors are zero-dimensional subschemes of $X$.
Thank you!

REPLY [5 votes]: The morphism $f_L$ is finite if and only if $L$ is ample. 

First as $O_{\mathbb P^r}(1)$ is ample, if $f_L$ is finite then $L=f_L^*O_{\mathbb P^r}(1)$ is ample (can you find a reference for this fact ?). 
Conversely, if $L$ is ample, for any closed point $y\in \mathbb P^r$, 
chose a hyperplane $H$ not passing through $y$, then $L\simeq O(f^{-1}(H))$ is trivial on $f^{-1}(y)$. This means the structure sheaf of $f^{-1}(y)$ is ample. So $f^{-1}(y)$ is affine and proper, hence finite. As $f$ is proper and quasi-finite, it is finite.