If we have the result as the title, then I can solve my real question. The original question was stated as follows.


In a paper I found the following lemma：


Let $S$ be a nonsingular projective surface, $R\in PicS$ a divisor with $R^2>0$ . Let 


$(E\_{i})$ be the family of distinct curves such that $R\cdot E\_{i}=0$.


Then the $E\_{i}$ are numerically independent.


The proof just says that the result follows from Hodge Index Theorem. But I cant see how. HIT just assert that the intersection matrix $(E\_i.E\_j)$ is negative semidefinite. Why $E\_i$ can't be numerically dependent?


Any hint is welcome. Thanks a lot.



Does negative semidefinite intersection matrix of bunch of curves implies zero intersection with other curves?
The answer already given here seems to assume the E\_i are pairwise disjoint, which I didn't take to be the case from your question.


To get the result in general, I think you need to use the additional fact that an effective divisor can never be numerically trivial (since it will have positive intersection with an ample divisor.) Then take a hypothetical numerical relation among the E\_i. By the above, it must have both positive and negative coefficients. Group them to get a relation like


$\sum a\_i E\_i = \sum b\_j E\_j $ (modulo numerical equivalence)


with the a's and b's positive and all of the E\_i and E\_j distinct from each other. If you call the left and right hand sides of this equation A and B, then you have that $A^2<0$ since it's a nonzero element and the intersection pairing is negative definite, but $A \cdot B \geq 0$ since all the curves intersect properly. This contradicts $A\equiv B$. 


