1
$\begingroup$

I'm trying to prove that the sum of two ample divisors on a projective complex algebraic surface S is it self an ample divisor.
To do this i need to verify that the index intersection between two ample divisors is positive.
Is it true that if A and B are two ample divisors on S the index intersection AB is a positive integer ?

$\endgroup$

1 Answer 1

3
$\begingroup$

I must say I've never heard the term "index intersection" before. Maybe this is an issue of language: the usual English phrase is "intersection number".

Anyway, yes, this is true. Here's the proof:

  1. Intersection numbers are bilinear in both arguments, so we can assume $A$ and $B$ are very ample.
  2. A very ample divisor is effective (by definition).
  3. A very ample divisor has positive interesection number with any effective divisor (easy exercise).
$\endgroup$
4
  • $\begingroup$ But on my book a divisor is very ample if the associated morphism from my surface to the n-dimensional complex projective space is an embedding.(i'm not assuming that a very ample divisor is effective) is it a consequence of the previous definition? $\endgroup$
    – dario
    Sep 19, 2014 at 14:20
  • $\begingroup$ I've prooved the index intersection between an ample divisor and an effective divisor is positive. The proble is to change the effective divisor with another ample divisor. $\endgroup$
    – dario
    Sep 19, 2014 at 14:24
  • $\begingroup$ @dario: an equivalent definition of very ample is this: $A$ is very ample if there is an emedding $X \hookrightarrow \mathbf P^n$ such that $A= H_{|X}$, where $H$ is the hyperplane class on $\mathbf P^n$. If you look into how the "associated morphism" is defined, you'll see that's the same thing. $\endgroup$
    – user64687
    Sep 19, 2014 at 14:25
  • $\begingroup$ ok now i understand- Thanks $\endgroup$
    – dario
    Sep 19, 2014 at 14:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .