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Oct 9, 2011 at 11:51 comment added naf Yes, that's right. So you could replace $\mathbf{P}^1$ by any smooth projective curve.
Oct 9, 2011 at 11:29 comment added Taicho So the self-intersection of $P$ is well-defined in this case because $\mathbf{P}^1$ is proper over the base field, right?
Oct 9, 2011 at 11:22 history edited Taicho CC BY-SA 3.0
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Oct 9, 2011 at 11:01 comment added naf Your idea (and the statement itself) is correct only if $D$ is an effective divisor. In this case you can replace $P$ by any integral divisor. If the base is not proper then one cannot define intersection numbers for horizontal divisors.
Oct 9, 2011 at 9:37 comment added Taicho $(P,P)$ is the self-intersection of $P$. So I should have denoted that by $P\cdot P$. The order of $D$ at $P$ is denoted by $\mathrm{ord}_P(D)$.
Oct 9, 2011 at 9:36 history edited Taicho CC BY-SA 3.0
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Oct 9, 2011 at 9:36 comment added naf What do $ord_P(D)$ and $(P,P)$ mean?
Oct 9, 2011 at 8:50 history asked Taicho CC BY-SA 3.0