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I believe it is something silly, but I'm a newbie, so why on a surface the intersection of an effective divisor and a divisor from ample bundle is non-negative? In fact, I need that an intersection of an effective divisor with a hyperplane section is non-negative.

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    $\begingroup$ The accepted answer is correct, but there is a much more down-to-earth explanation: if $L$ is a hyperplane section on $X$, then there is an embedding $i:X \rightarrow \mathbf P^n$ such that $L=i^*H$ where $H$ is the hyperplane class on $\mathbf P^n$. But then for $D$ an effective divisor, one can calculate $D \cdot L$ in $\mathbf P^n$ instead: $D \cdot L = i_* D \cdot H$, which is just the degree of the 1-cycle $i_*D$. If $L$ is ample but not very ample, replace it with a positive multiple which is very ample, and apply the same argument. $\endgroup$ – user64687 May 24 '15 at 19:09
  • $\begingroup$ When you say $D\cdot L=i_*D\cdot H$ are you using something like the projection formula? Where do you use the fact that D is effective? My intuition is that effective means an "actual" curve on the surface so when you do the pushforward we just see the curve in $\mathbb{P}^n$ and then the degree, which is nonnegative, is just the intersection with an hyperplane. $\endgroup$ – SC30 Nov 27 '17 at 0:06
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This follows immediately from Nakai-Moishezon's criterion: a Cartier divisor $D$ on a proper scheme $X$ is ample if and only if, for every subscheme $Y$ of $X$, one has $D^{\dim(Y)} \cdot Y > 0$. You can find this theorem for surfaces in Hartshorne's Algebraic Geometry or, more general, in Debarre's Higher-Dimensional Algebraic Geometry (in this last one, check also Kleiman's criterion as an alternative).

In your case, I think you have a projective surface $S \subset \mathbb{P}^{n}$ and the (ample) divisor is that associate to the restriction of $\mathcal{O}_{\mathbb{P}^{n}}(1)$ to $S$, that is, the hyperplane section. The hyperplane class $H$, associate to $\mathcal{O}_{\mathbb{P}^{n}}(1)$, is clearly ample (you can show this directly from the definition of ample). Since the pullback of an ample divisor under a finite morphism of noetherian schemes is ample, we have the hyperplane section on $S$ ample (in your case, the morphism is the inclusion).

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