Let $D$ be a divisor on smooth rational surface $X$ (over $\mathbb{C}$) such that the linear system $|nD|$ has dimension 1 (fore some $n$), has no fixed components and base points. How can I show that $D^2=0$?
Thanks!
Asked
Active
Viewed 688 times
4
-
2Please show your working so far :) – Shaun Feb 21 '14 at 19:08
-
You may want to start with Riemann-Roch for surfaces – Michael Feb 21 '14 at 19:21
-
@Michael so we have $h^0(S)+h^0(K-S)=h^1(S)+\frac{1}{2}S(S-K)+1$ for rational surface and divisor S on it, so in my case it will be just $h^0(K-nD)=h^1(nD)+\frac{1}{2}nD(nD-K)$, am I right? But it still doesn't help to evaluate $D^2$. – Andrew Feb 21 '14 at 19:34
1 Answers
6
I don't use the fact that $X$ is rational, but $|nD|$ being base point free and dimension $1$ gives a morphism $f:X \to P^1$, with $\mathcal{O}_X(nD) \simeq f^*(\mathcal{O}_{P^1}(1))$, i.e. $nD \sim f^*(p)$, where $p \in P^1$ is a point. (Geometrically, $f$ is contracting each divisor in $|nD|$ to a point in $P^1$). Then $p \cdot p=0$ (since in $P^1$ any two points are rationally equivalent and so we can replace one of the $p$'s by another point $q$, and then $p\cdot q =0$.) Then since $f^*$ preserves intersection numbers, $(nD)^2=f^*(p)^2=f^*(p^2)=0$.
ykm
- 1,571
- 8
- 16
-
Dear yogesh, your answer is indeed completely correct. As you suspect, rationality of $X$ does not play a role. – Feb 22 '14 at 16:38
-