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The base scheme is an algebraically closed field.

Let $X\to \mathbf{P}^1$ be an arithmetic surface over $\mathbf{P}^1$ and let $P$ be a section of $X\to \mathbf{P}^1$. Let $D$ be an effective (edited) divisor on $X$. Is it trivial that the intersection product $D\cdot P \geq \mathrm{ord}_P(D) (P\cdot P)$? Here $\mathrm{ord}_P(D)$ is the order of $D$ at $P$, i.e., the coefficient of $D$ at $P$.

Here's my idea of a proof.

If $E$ is an integral divisor on $X$ and the image of $P$ is not contained in the support of $E$, then $E\cdot P \geq 0$. Therefore, we have that $D\cdot P \geq \mathrm{ord}_P(D) (P\cdot P)$.

Can we replace $P$ by any integral horizontal divisor? Can we replace $\mathbf{P}^1$ by a Dedekind scheme?

Remark. An arithmetic surface over $\mathbf{P}^1$ is a flat projective morphism $X\to \mathbf{P}^1$ with $X$ an integral regular and $2$-dimensional scheme.

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  • $\begingroup$ What do $ord_P(D)$ and $(P,P)$ mean? $\endgroup$
    – naf
    Commented Oct 9, 2011 at 9:36
  • $\begingroup$ $(P,P)$ 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 $\mathrm{ord}_P(D)$. $\endgroup$
    – Taicho
    Commented Oct 9, 2011 at 9:37
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    $\begingroup$ 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. $\endgroup$
    – naf
    Commented Oct 9, 2011 at 11:01
  • $\begingroup$ So the self-intersection of $P$ is well-defined in this case because $\mathbf{P}^1$ is proper over the base field, right? $\endgroup$
    – Taicho
    Commented Oct 9, 2011 at 11:29
  • $\begingroup$ Yes, that's right. So you could replace $\mathbf{P}^1$ by any smooth projective curve. $\endgroup$
    – naf
    Commented Oct 9, 2011 at 11:51

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