4
$\begingroup$

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.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
7
$\begingroup$

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.

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ I learn algebraic geometry in analytic way.Fact 3 is the thing I missed.In fact before I ask the question, I expected somtheing like Fact 3 to be valid.But I failed to find similar statements in my textbook.Thank you for the clear exposition. $\endgroup$
    – Jun Li
    Sep 1, 2011 at 6:35
  • $\begingroup$ Fact 3 is in fact not very hard to prove: Using the Leray spectral sequence and the projection formula we have $H^i(X,(f^*M)^n \otimes F)=H^i(P,M^n\otimes f_*M)$. $\endgroup$
    – J.C. Ottem
    Sep 1, 2011 at 12:44
  • $\begingroup$ Thank you for the proof of Fact 3.I'm learning something about spectral sequence right now.Its a great pleasure to hear of an application. $\endgroup$
    – Jun Li
    Sep 1, 2011 at 14:46
  • $\begingroup$ Note that Ottem's way to prove Fact 3 uses Serre's cohomological characterization of ampleness which works only if X and Y are proper over a field (or in fact over any noetherian ring). This is of course the case in your original question. The more general version I stated is maybe slightly less standard than I claimed. Its proof can be found in the Section on ample line bundles in EGA2 or - and I can't resist to give this reference - in my book with Görtz, Prop. 13.83. $\endgroup$
    – Torsten Wedhorn
    Sep 1, 2011 at 16:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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