Question

Let $X$ be a smooth projective manifold, $L$ is a holomorphic vector bundle over $X$ such that $mL$ ($m$ is some positive integer) is base point free. If $\iota_m$ is the map into the projective space associated to the complete system and each fiber of this map has dimension zero. Then $L$ is ample.

I see this statement when reading a paper. It seems well-known. I'm afraid that it may be very easy but I can't prove it myself.

Answer 1

This is indeed rather standard (where I assume that you mean that $L$ is a line bundle). It follows from the following general facts:

1. Let $m$ be a positive integer. Then a line bundle $L$ is ample if and only if $mL$ is ample.

2. Let $X$ be a proper (e.g. a projective) scheme over a field $k$ (or in fact over any base scheme $S$), let $P$ be a separated scheme over $k$ (e.g., $P$ a projective space), and let $i: X \to P$ be a morphism of $k$-schemes with finite fibers. Then $i$ is finite (and in particular affine).

3. Let $f: X \to Y$ be an affine (or even quasi-affine) morphism of schemes of finite type over a field (or over any base scheme) and let $M$ be an ample line bundle on $Y$. Then $f^*M$ is ample.

Thus in your case it suffices to show that $mL$ is ample by Fact 1. Because of Fact 2, the morphism $i_m$ is finite and hence affine. By definition $mL$ is the pullback of the ample line bundle $O(1)$ via $i_m$. Therefore $mL$ is ample be Fact 3.

Facts 1 - 3 should be found in most of the text books on Algebraic Geometry. Fact 1 and Fact 3 are basic and not too difficult facts on ample line bundles. Fact 2 follows from Zariski's main theorem.