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Let $D$ be a divisor on an algebraic complex smooth projective surface $S$. Assume that the complete linear system $|D|$ is not empty and has no fixed component. Is it true that $D^2 \geq 0$, where $D^2$ is the self-intersection of $D$?

Francesco Genovese
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    Yes! Since $|D|$ has no fixed component, it contains two divisors, say $D_1$ and $D_2$, which intersect in a finite number of points (possibly zero). But then $D^2=D_1 \cdot D_2 \geq 0$. –  Aug 01 '13 at 20:53
  • I'm not convinced. What if $D_1$ and $D_2$ share a common component? After all, the intersection of $D_1$ and $D_2$ could be bigger than the base locus of $|D|$. – Francesco Genovese Aug 01 '13 at 21:17
  • I'm not saying that arbitrary choices of the $D_i$ will work: rather, because $|D|$ has no fixed component, we can find two divisors in $|D|$ that do no share any component. –  Aug 01 '13 at 21:21
  • Could you explain me how this is possible, in detail? I'm sorry if I am a bit pedantic, but I'm quite a newbie in AG! – Francesco Genovese Aug 01 '13 at 21:30
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    The point is that divisors in |D| containing some given divisor E form a linear subspace, and the hypothesis of no fixed component says that for any $E$, this is a proper subspace. Now choose any divisor $D_1 \in D$; it has finitely many components, so by the previous sentence the set of divisors in $|D|$ not containing any of the components of $D_1$ is a nonempty open set of $|D|$. Choosing $D_2$ to be any divisor in this open set, we are done. –  Aug 01 '13 at 21:52

1 Answers1

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Let $D_1\in\left| D\right|$. We know that $D^2 = D\cdot D_1$. If $D,D_1$ share no common component, then we win, since the intersection in this case is defined by the sum over the multiplicities of the local rings at the intersection points. Otherwise, suppose that $D = D'+E$ and $D_1 = D_1' +E$ with nontrivial component $E$. By hypothesis, there exists $D_2$ in the linear series that does not have the component $E$, and $D^2 = D\cdot D_2 = D'\cdot D_2 + E\cdot D_2$ where $E\cdot D_2\ge 0$. Now suppose that $D' = D''+E'$ and $D_2 = D_2'+E'$ share a component $E'$. Repeating the argument we find $D_3$ that does not contain $E'$, and then $D'\cdot D_2 = D'\cdot D_3 = D''\cdot D_3 + E'\cdot D_3$ where $E'\cdot D_3\ge 0$. Continuing on, we continue removing components from $D$ until we no longer need to (or can), thus obtaining the result.

Andrew
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