# A rational curve on projective complex surfaces.

Let $X$ be a smooth projective complex surface and $C$ a smooth curve on it. How can one conclude that $(C,C)=-1$ provided that $(C,C)+(K_X,C)<0$?

By adjunction formula, $(C,C)+(K_X,C)=2g(C)-2$, we have $C\cong \mathbb{P}^1$, but I don't see why its self-intersection number is $-1$.

-

Your statement of the adjunction formula is incorrect. In fact, $$K_C\simeq K_X|_C\otimes O_X(C)|_C$$ as invertible sheaves on $C$. Taking degrees we get $$2g(C)-2 = (K_X, C)+(C,C).$$ When $(K_X, C)<0$ and $(C, C)<0$, this implies that $g(C)=0$ and $(K_X, C)=(C,C)=-1$.
Note that there may exist $C\simeq \mathbb P^1$ with $(C,C)=-2$ and $(K_X,C)=0$. In this case $(C,C)\pm (K_X,C)< 0$ and $(C,C)\ne -1$.
You are right; my adjunction formula was wrong. $g(C)=0$ is automatic from the assumption. But we cannot conclude that $(C,C)=-1$. Maybe the problem is wrong. – M. K. Nov 18 '12 at 18:36