K3 ample linear systems from Serre extensions and instability

If global generation fails, choose a point Z of length one where evaluation fails. If global generation holds but very ampleness fails, choose a length-two subscheme Z where evaluation is not surjective, minimal for that property. A length-two subscheme of a smooth surface is either two distinct reduced points or the tangent infinitesimal scheme with local ring $C[ϵ]\mathrm{/}{ϵ}^2$; both are local complete intersections. Every proper length-one subscheme imposes an independent condition in the second case. This follows from the proper globally generated morphism: an ample generated line has finite morphism, and failure to be a closed immersion is exactly failure to separate two points or a tangent vector, by the finite local Nakayama criterion.

We construct the Serre extension and check local freeness rather than assuming Cayley–Bacharach. The sequence 0 -> $I_Z$ L -> L -> L|Z -> 0 and the dual finite resolutions give an exact segment $Ex{t}^1(I_{Z}L,O_S)$ -> $H0(omeg{a}_{Z}tensor{L}^{-1}\mathrm{\mid }Z)$ -> $H2(L^{-1})$. Its right-hand arrow is dual to evaluation H0(L)->H0(L|Z), under the local residue trace. Moreover $H1(L^{-1})=0$ by [F1] and Serre duality. Thus a nonzero functional annihilating the evaluation image lifts to an extension 0 -> $O_S$ -> V -> $I_Z$ L -> 0. In length one the image is zero and its nonzero functional generates the local extension module. For two reduced points the evaluation image is a line with both coordinates nonzero, so a nonzero annihilating functional has both coordinates nonzero. For the tangent ring choose a local generator for L and write the image as C u, with $u=u0+u1$ epsilon a unit, since L is generated. The residue trace tr(a0+a1 epsilon)=a1 identifies the dual ring with the ring. An annihilator lambda(b)=tr(a b), lambda(u)=0, has $a1=-a0$ u1/u0; nonzero lambda therefore has a0 nonzero and a is a unit. In ALL cases the extension class is a unit at every point of Z.

Locally $I_Z=(f,g)$ has the two-generator Koszul relation. Its extension V has generators v0,v1,v2 with one relation a v0+g v1−f $v2=0$, where a lifts the local residue/extension class. Since a is a unit modulo (f,g), it is a unit in the local ring; eliminate v0 to see V is free of rank two. Away from Z the extension is an extension of line bundles and is locally free as well. Chern additivity gives c1(V)=L and $c2(V)=ell=length(Z)\le 2$. The use of finite-resolution duality in step2.1 is justified by dualizing these explicit Koszul complexes; the residue evaluation is their last coefficient map, so no duality theorem for arbitrary singular sheaves was imported.

V cannot be slope-stable with respect to L. A stable rank-two bundle has only scalar endomorphisms: choose an eigenvalue of an endomorphism, whose characteristic coefficients are constant on the connected projective surface. If its difference from that scalar were nonzero, its positive-rank kernel and image would both be proper subsheaves of strictly smaller slope, contradicting degree additivity in 0 -> kernel -> V -> image -> 0. Thus h0(End V)=1. Since $K_S$ is trivial, Serre duality gives h2(End V)=1, so $\chi (\mathrm{End}V)\le 2$. But GRR and Chern additivity give $\chi (\mathrm{End}V)=8+d-4\mathrm{\ell }\ge 4$, a contradiction. This proves nonstability, without any discriminant theorem.

Choose a rank-one subsheaf of V of slope at least that of V and saturate it. It is a line bundle M: its quotient is torsion-free, and the depth lemma together with [F3] makes this saturated kernel reflexive and locally free. Write its quotient as $I_W$ N, where $N=L-M$ and W has finite length. Put $a=L$ N and $q=N^2$. The inclusion M->V cannot factor through the $O_S$ in the Serre extension, since L $M\ge d\mathrm{/}2>0$ implies H0(−M)=0. Its nonzero projection to $I_Z$ L is a section of $I_Z$ N. Therefore N has a nonzero EFFECTIVE divisor containing Z; in particular a is a positive integer. Chern additivity gives $ell=M$ N+length(W)=a−q+length(W). Hence $q\ge a-ell$, and Hodge index gives $q\le a^2\mathrm{/}d$.

In the strictly destabilizing case L $M>d\mathrm{/}2$, we have $a<d\mathrm{/}2$ and therefore $q<a\mathrm{/}2$. For $ell=1$, $q\ge a-1$ makes $a<2$, so $a=1$; evenness and $0\le q<1\mathrm{/}2$ give $q=0$. For $ell=2$, $a\ge 4$ would give $a-2\ge a\mathrm{/}2$, impossible; $a=3$ would require even $q\ge 2$ but $q<3\mathrm{/}2$, also impossible. Thus $a=1$ or 2, and the inequalities and evenness again give $q=0$. We have obtained an effective square-zero divisor of L-degree at most ell.

In the equal-slope case $a=d\mathrm{/}2$, the inequalities become $d\mathrm{/}2-ell\le q\le d\mathrm{/}4$, with q even. For $ell=1$ and $d\ge 4$ there is no such even integer (at $d=4$ the only possible value would be one). For $ell=2$ they force $d\le 8$. At $d=4$ one gets $q=0$ and $a=2$; at $d=6$ there is no even value; at $d=8$ one gets $q=2$ and equality in Hodge index. Equality gives N numerically L/2, and [F2] then gives $L=2N$ as actual line bundles, with $N^2=2$. Thus the only new case is the stated imprimitive degree-eight exception.

An effective square-zero divisor D of L-degree one or two yields a nef square-zero divisor of no larger degree. If the degree is one it is irreducible, so it is nef. In degree two it is either irreducible (again nef), twice a degree-one component of square zero, or the sum of two degree-one components. For any such component C, adjunction gives $C^2\ge -2$ and Hodge gives $C^2\le 1\mathrm{/}d<1$, so $C^2$ is either −2 or zero. A square-zero component already gives the claim. If both squares are −2, then $D^2=0$ forces their mutual intersection to be two. Thus D has zero intersection with its two components, and nonnegative intersection with every other curve, hence is nef. This exhaustive degree argument does not use a K3 linear-system classification.

A nonzero nef square-zero divisor D is a multiple of an elliptic pencil. RR and Serre duality give $h^0(D)\ge 2$. Write its complete system as movable part P plus fixed part B. Nefness and $D^2=0$ imply D $P=D$ $B=0$. Two general members of |P| have nonnegative intersection, so $P^2\ge 0$. The Lorentzian Hodge inequality forces P numerically proportional to D, whence $P^2=P$ $B=B^2=0$. If B were nonzero, RR would give $h^0(B)\ge 2$; varying B and multiplying by a member of the movable system avoiding a changed component contradicts that B is fixed. Thus there is no fixed part. Two general members now have total intersection zero and no common curves, so there are no base points. The map given by |D| has curve image; take its connected-fibre Stein factor f:S->B0. The injection $H1(B0,O_{B0})->H1(S,O_S)=0$ makes $B0=P1$. Thus $D=f\ast O_{P1}(k)$, and the connected fibre class $E=f\ast O(1)$ is nef and primitive: if E were a multiple $m>1$ of a line class, that class is nef square zero and the same pencil argument would give $h^0(E)\ge m+1$, while connected fibres give h0(E)=2. A general fibre is smooth by Bertini, connected and of arithmetic genus one by adjunction, hence an elliptic curve. Its L-degree does not exceed that of D.

Steps1.1–9.1 prove the obstruction assertions. In particular if Pic(S)=ZL with L primitive and $L^2\ge 4$, no nonzero square-zero class exists and L cannot be twice another line bundle. There is neither a length-one nor a length-two obstruction, so L is globally generated and very ample by the criterion in step1.1. This conclusion has NOT been applied to a K3 section before its Picard hypothesis is supplied.

Finally suppose $d\ge 6$ and E,E' are distinct primitive nef elliptic pencil classes with $e=L$ $E\le 2$ and $e^{\mathrm{\prime }}=L{E}^{\mathrm{\prime }}\le 2$. Their intersection is positive; zero would force proportionality by Hodge index and hence equality by primitivity. It cannot be one: then $(E-E^{\mathrm{\prime }}{)}^2=-2$, RR makes one of E−E' or E'−E effective, and its intersection with the appropriate nef pencil is −1, impossible. Thus $E{E}^{\mathrm{\prime }}\ge 2$. On the other hand write $E=(e\mathrm{/}d)L+v$ and E'=(e'/d)L+v' in the L-orthogonal negative-definite space. Since $E^2=E^{\mathrm{\prime }2}=0$, Cauchy–Schwarz gives E $E^{\mathrm{\prime }}\le 2e{e}^{\mathrm{\prime }}\mathrm{/}d\le 8\mathrm{/}d<2$. Contradiction. This proves uniqueness and completes the lemma.