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This question popped up as I was thinking about intersection theory on fibered surfaces, which is slightly different from the situation I'm about to describe, and as a result I may have gotten some facts wrong. Anyway, here goes:

Let $X$ be a regular projective surface. If $D$ be an ample effective divisor on $X$, corresponding to an ample invertible sheaf $\mathcal{O}_X(D)$ on $X$. Then, for every closed curve $C\subset X$, $\mathcal{O}_X(D)|_C$ is ample on $C$, and hence has positive degree as an invertible sheaf on $C$. In terms of intersection theory, this means that $D\cdot C > 0$ for every closed curve $C\subset X$.

But...isn't the intersection pairing negative semidefinite? Hence, $D\cdot D\le 0$. At first, this seems like it would break the ampleness of $D$, and hence regular projective surfaces $X$ can't have any ample divisors, but this is plainly false since you can just take $X = \mathbb{P}^2$, and $D$ to be any prime divisor.

Where is the problem(s) with my logic??

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    The intersection pairing is certainly not negative semidefinite. Think of intersection theory on $\Bbb P^2$. Some curves may have negative self-intersection but a typical divisor won't. – Angina Seng Apr 23 '17 at 05:35
  • @LordSharktheUnknown Hmm, on $\mathbb{P}^2$, is it true that the intersection number of a curve of degree $d$ and a curve of degree $e$ is $d\cdot e$? So, the self intersection of a curve of degree $d$ is $d^2$? Interesting...what about fibered surfaces causes the intersection pairing to be negative semidefinite? –  Apr 23 '17 at 06:42
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    Fibered or not, the intersection pairing is not negative semi-definite. Where did you read that? – Mohan Apr 23 '17 at 13:12
  • @Mohan Hmm, well on fibered surfaces usually the intersection pairing is only defined when one of the divisors is vertical, and so you can only talk about the self intersection of vertical divisors, and restricted to vertical divisors, the intersection pairing is negative semidefinite. (This is Qing Liu's book, chapter 9.1) –  Apr 23 '17 at 21:26
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    I do not know what you mean. If $X$ is a regular projective surface, there is an intersection pairing on it, whether it is fibered or not. If fibered over a projective curve, it still has horizontal divisors and by definition of projective surface, has an ample divisor which intersects all curves positively. – Mohan Apr 23 '17 at 23:14
  • @Mohan What I meant was that in the context of fibered surfaces as treated in Qing Liu, his definition of a fibered surface is an integral surface projective over a Dedekind scheme. He doesn't require the surface itself to be projective over a field, and hence you can have arithmetic surfaces like curves over $\mathbb{Z}$, where now the base curve is not necessarily proper, and hence the there is some issue with defining intersection numbers of two non-vertical divisors (I would guess the issue is that the pairing doesn't factor through $Pic(X)\times Pic(X)$... –  Apr 24 '17 at 17:07
  • @Mohan So, in the context of fibered surfaces, if $S$ is the base Dedekind scheme, because the pairing is insensitive to addition by a principal divisor, and because you can always compute the pairing locally on $S$, the self-intersection of any vertical fiber $X_s$ is 0. Because the intersection of any two distinct prime divisors is nonnegative, this implies that if $X_s$ is connected and has multiple irreducible components, then some of them must have negative self-intersection. Similarly, if a regular projective surface over $k$ were to have a principal divisor which is connected with ... –  Apr 24 '17 at 17:13
  • @Mohan multiple irreducible components, then by the same reasoning, some of those components would have to have negative self-intersection...right? So if this doesn't happen, does that mean that every connected principal divisor in a regular projective surface is prime? –  Apr 24 '17 at 17:15
  • Are you interested in surfaces fibered over Dedekind domains or over projective curves? For the arithmetic case, the intersection pairing is more complicated, by counting what happens at infinity and will lead you to Arakelov theory. So, do you have a specific question? – Mohan Apr 24 '17 at 17:55
  • @Mohan Here's an updated question: https://math.stackexchange.com/questions/2250217/ample-invertible-sheaves-on-projective-fibered-surfaces-over-projective-curves –  Apr 24 '17 at 18:52

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