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Let a surjective morphism from a smooth projective surface to a projective surface contract an irreducible curve $C$. Pull back an ample divisor $H$ from the target to obtain $L$. The projection formula for algebraic cycles gives $L\cdot C=0$ and $L^2=\mathrm{deg}(f)H^2>0$. The Hodge index theorem for algebraic surfaces makes the intersection form negative definite on the orthogonal complement of this positive-square class. Since an ample divisor on the source meets $C$ positively, its numerical class is nonzero, and $C^2<0$.

If $D$ is nonzero numerically with $D^2=0$ and $D\cdot A>0$ for ample $A$, then $D\cdot E=0$ implies $E^2\le 0$, with equality only when $[E]$ is proportional to $[D]$. To see this, use the Hodge index theorem for algebraic surfaces to split off the positive $A$ direction and diagonalize the remaining negative definite form. Orthogonality to a nonzero isotropic vector then leaves a negative semidefinite form whose radical is precisely that vector's span.

Embed $Y$ in projective space and let $H$ be its hyperplane Cartier divisor. Since $f$ is surjective between integral surfaces, it is generically finite, with degree $r>0$. Its pullback of a Cartier divisor $L=f^{\ast }H$ is a nef divisor, and the projection formula for algebraic cycles gives $L^2=r{H}^2>0$. The same formula gives $L\cdot C=H\cdot f_{\ast }C=0$, because the curve is contracted to a point.

The Hodge index theorem for algebraic surfaces makes the intersection pairing on the Picard group of a surface have signature $(1,\rho -1)$ on $N^1(X{)}_{\mathbb{R}}$. The orthogonal complement of any positive-square vector, including $L$, is negative definite. The numerical class of $C$ is nonzero, since an ample divisor on $X$ intersects every nonzero effective divisor positively. Thus the contracted curve has negative self-intersection conclusion applies:

The strict statement as printed needs the qualification if $F\ne 0$. Part (a) already shows $F^2=0$ in the present situation, and the next part proves $F=0$; an unconditional $F^2<0$ would contradict that conclusion.

To prove the needed conditional statement, suppose $F\ne 0$. A fixed part has $h^0(X,F)=1$. Indeed, a different $F^{\prime }\in |F|$ has smaller multiplicity along some component of $F$; otherwise $F^{\prime }-F$ would be a nonzero effective linearly trivial divisor. Adding a member of the movable part avoiding that component contradicts fixedness.

The Hodge index theorem for algebraic surfaces, together with $D\cdot F=0$ and $D^2=0$, gives $F^2\le 0$. If equality held, Riemann–Roch theorem for algebraic surfaces would give $\chi (X,F)=2$, and Serre duality would give $h^2(X,F)=h^0(X,-F)=0$. Hence $h^0(X,F)=2+h^1(X,F)\ge 2$, a contradiction. ThereforeThis is the fixed-part elimination on a K3 surface argument.

Choose an ample divisor $A$. The Hodge index theorem for algebraic surfaces implies $D\cdot A>0$: if it were zero, $D^2=0$ would force the numerical class of $D$ to be zero. Consequently $-D$ cannot be effective, because every nonzero effective divisor has positive intersection with $A$, while $-D\cdot A<0$.

The Riemann–Roch theorem for algebraic surfaces and the preceding computation give $\chi (X,{\mathcal{O}}_X(D))=2+D^2/2=2$. By Serre duality, $h^2(X,D)=h^0(X,-D)=0$. Thus