Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 139 / 2 / vi

First derive the intersection constraints. Since $D$ is nef and $F,M$ are effective, $D\cdot F,D\cdot M\ge 0$; their sum is $D^2=0$, so both vanish. Since the movable divisor $M$ is nef, $M^2,M\cdot F\ge 0$, and $M\cdot D=0$ forces both to vanish. In particularFor every component $E$ of $F$, nefness and $D\cdot F=0$ imply $D\cdot E=0$. The isotropic orthogonality consequence of the Hodge index theorem gives $E^2\le 0$. If $E^2=0$, Riemann–Roch theorem for algebraic surfaces and Serre duality give $h^0(X,E)\ge 2$, since $-E$ cannot be effective. Choose $E^{\prime }\in |E|$ different from $E$. It cannot contain $E$: otherwise $E^{\prime }-E$ would be a nonzero effective numerically trivial divisor, contradicting its positive intersection with an ample divisor. Choose a member of $|M|$ avoiding $E$. Replacing one copy of $E$ in $F$ by $E^{\prime }$ now produces a member of $|D|$ with smaller multiplicity along $E$, contradicting the definition of the fixed part. Therefore

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.