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

Write $F={\sum }_C{\min }_{G\in |H|}{\mathrm{mult}}_C(G)C$. This is the fixed part of a linear system. The movable part of a linear system is $|M|$, with $M=H-F$; multiplication by the defining section of $F$ identifies its sections with those of $H$. In particular $|H|=F+|M|$, and no curve occurs in every member of $|M|$. The occurrence of $|F|$ in the definition request is understood as $|H|$.

For any integral curve $C$, choose a member $G\in |M|$ not containing $C$. Their intersection number is a sum of nonnegative local intersection multiplicities, so $M\cdot C=G\cdot C\ge 0$. Therefore the movable divisor $M$ is nef. If $M=0$, the same assertion is immediate.

The relevant invariant of a divisor is its Iitaka dimension, also called its Kodaira dimension. If two linearly independent global sections $s,t$ occur in $H^0(X,nH)$, their ratio is a nonconstant rational function. The sections $s^q,s^{q-1}t,\dots ,t^q$ are linearly independent, because a nonconstant rational function over an algebraically closed field is transcendental over that field. Thus some complete linear system of a divisor has positive-dimensional image, contradicting $\kappa (H)=0$.

Consequently $h^0(X,nH)=1$ for every $n>0$; the nonemptiness of $|H|$ supplies a nonzero section in every multiple. Its unique divisor is entirely fixed. Hence every movable part $M_n$ is zero.

Because $|D|$ has no fixed components, choose two members $G_1,G_2$ with no common component. This uses the infinitude of the algebraically closed field and avoidance of finitely many proper linear subspaces of the section space. On a smooth algebraic surface, their intersection number is the sum of local intersection multiplicities. Since $G_1\cdot G_2=D^2=0$, they are disjoint. A basepoint of $|D|$ would belong to both, so $|D|$ is basepoint-free.

The resulting morphism has nonconstant image since $\dim |D|\ge 1$. Its image cannot be a surface: a generically finite morphism defined by $D$ would make $D^2$ the positive product of its degree and the degree of its image. Thus the image is a curve. Apply Stein factorization to obtain $g:X\to C$ with connected fibers and $g_{\ast }{\mathcal{O}}_X={\mathcal{O}}_C$, where $C$ is a smooth projective curve. There is a positive-degree line bundle $L$ on $C$ with ${\mathcal{O}}_X(D)=g^{\ast }L$. The projection formula for sheaves and Riemann-Roch theorem give $h^0(X,mD)=h^0(C,L^m)=m\mathrm{deg}L+O(1)$. Therefore

Each $M_n$ is a nef divisor. If $M_n^2>0$, the volume of a nef divisor formula gives quadratic growth of $h^0(X,q{M}_n)$. Multiplication by the section of $q{F}_n$ injects this space into $H^0(X,qnH)$, so the birational linear system criterion for bigness makes the Iitaka dimension of $H$ two.

Conversely, if $\kappa (H)=2$, some $|nH|$, and hence its movable part $|M_n|$, has a two-dimensional image. If $M_n^2=0$, the previous part would make $|M_n|$ basepoint-free with image a curve; this is impossible. Since $M_n$ is nef, $M_n^2\ge 0$, hence $M_n^2>0$.

If all $M_n^2=0$ and one $|M_{n^{\prime }}|$ has positive dimension, the preceding part supplies a curve image. The just-proved equivalence excludes dimension two, so $\kappa (H)=1$. Conversely $\kappa (H)=1$ excludes every positive $M_n^2$ and requires a positive-dimensional linear system. Thus

A connected integral projective variety has only constant regular functions, so $h^0(X,{\mathcal{O}}_X)=1$. By Serre duality and $K_X\sim 0$, $h^2(X,{\mathcal{O}}_X)=h^0(X,{\mathcal{O}}_X(K_X))=1$. The assumed first sheaf cohomology vanishes. Therefore the Euler characteristic of a coherent sheaf isThese hypotheses characterize the K3 surface setting used below.

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

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.

The intersection calculation in part (vi)(a) gives $F^2=0$. Part (vi)(b), with its necessary $F\ne 0$ qualification, excludes every nonzero fixed part. Thus $F=0$ and $|D|$ is movable. Part (v) gives $\dim |D|\ge 1$, and $D^2=0$ by hypothesis. Applying part (ii) provesThus a nontrivial nef divisor of square zero on a K3 surface defines a fibration over a curve; the key step is fixed-part elimination on a K3 surface.