Hartshorne Exercise V.1.9.
(a) If $H$ is an ample divisor on a surface $X$, and $D$ is any divisor, show that
$$(D^2)(H^2)\le(D.H)^2.$$
(b) Now let $X$ be a product of two curves $X=C\times C’$. Let $l= C\times\text{pt}$, and $m=\text{pt}\times C’$. For any divisor D on $X$,let $a=D.l$,$b=D.m$. Then we say $D$ has type $(a,b)$. If D has type $(a,b)$, with $a,b\in\Z$, show that
$$D^2\le2ab,$$
and equality holds if and only if $D=bl +am$. [Hint: Show that $H=l+m$ is ample, let $E=l-m$, let $D’= (H^2)(E^2)D – (E^2)(D.H)H – (H^2)(D.E)E$, and apply (1.9). This inequality is due to Castelnuovo and Severi. See Grothendieck [2].]
(a) By Nakai-Moshizon $H^2>0$. Consider the divisor $D’=H^2D-(D.H)H$, and observe that, by the Hodge Index theorem,
$$(D’)^2=(H^2)^2D^2+(D.H)^2H^2-2H^2(D.H)^2=H^2(H^2D^2-(D.H)^2)<0.$$
(b) First, if $p\in C$ is any point then $h^0(\OO_C(np))>1$ and $np$ is a basepoint free divisor on $C$ for large $n$, hence there is another effective divisor $P\sim np$ on $C$ such that $P\cap np=\varnothing$. But $P$ is just a sum of points on $C$, so it follows that we can move the divisors $nl$ and $nm$ on $X$ to linearly equivalent ones that don’t meet the originals. This shows that $H^2=2$. Any curve $F$ on $X$ has image in $C$ a point or all of $C$. If the image is all of $C$ then $F$ meets the fiber $m$ once. Otherwise $F$ is contained in a fiber over $C$, but every such fiber is a copy of $C’$ so actually $F$ is the whole fiber and thus meets the line $l$ once. By Nakai-Moishizon, $H$ is ample, and $H^2=2, E^2=-2,H.E=0, D’.H=0$. By the Hodge index theorem, $D’^2<0$ and then it just takes some simple algebra to conclude that,
$D^2\le 2ab$ with equality iff $D\equiv bl+am$.
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