Resolutions of Fat Point Ideals Involving Eight General Points of P2

Ž .Journal of Algebra 244, 684�705 2001doi:10.1006�jabr.2001.8931, available online at http:��www.idealibrary.com on

Resolutions of Fat Point Ideals Involving EightGeneral Points of P 2

Stephanie Fitchett

Honors College, Florida Atlantic Uni�ersity, Jupiter, Florida 33458E-mail: [email protected]

Brian Harbourne1

Department of Mathematics and Statistics, Uni�ersity of Nebraska,Lincoln, Nebraska 68588

E-mail: [email protected]

and

Sandeep Holay

Department of Mathematics, Southeast Community College, Lincoln, Nebraska 68508E-mail: [email protected]

Communicated by Craig Huneke

Received January 2, 2001

The main result, Theorem 1, provides an algorithm for determining the minimalfree resolution of fat point subschemes of P2 involving up to eight general pointsof arbitrary multiplicities. The resolutions obtained hold for any algebraicallyclosed field, independent of the characteristic. The algorithm works by giving aformula in nice cases and a reduction to the nice cases otherwise. The algorithm,which does not involve Grobner bases, is very fast. Partial information is also¨obtained in certain cases with n � 8. � 2001 Academic Press

Key Words: minimal free resolution; rational surface; fat point.

1 This author benefited from a National Science Foundation grant.

6840021-8693�01 $35.00Copyright � 2001 by Academic PressAll rights of reproduction in any form reserved.

RESOLUTIONS OF FAT POINT IDEALS 685

1. INTRODUCTION

Determining the Hilbert function and minimal free resolution of idealsdefining n general fat points of P 2 is a difficult problem that has attractedthe attention of numerous researchers over the years. For n � 9, the

� �problem remains unsolved in general. For n � 9, Nagata N resolved the� �problem for Hilbert functions, while for n � 5, Catalisano C resolved the

Ž .problem for minimal free resolutions, extended using different methods� � � �by Fitchett F2 to n � 6 and by Harbourne H6 to n � 7. In this paper,

we now extend this work to n � 8. Our results are based on studying linearsystems on blow ups of P 2 at the n points. For n � 8, these surfaces areDel Pezzo surfaces, and hence the semigroup of classes of effectivedivisors is finitely generated. It will be very difficult to extend our ap-proach to the case of n � 9; for one thing, the surfaces obtained for n � 9are no longer Del Pezzos, for another, the semigroup of effective divisorclasses is no longer finitely generated.

We now recall the notion of fat points. Consider n distinct pointsp , . . . , p of P 2. Given nonnegative integers m , a fat point subscheme1 n iZ � m p � �� m p is that subscheme defined by the homogeneous1 1 n n

m1 m n � 2 �ideal I � I �� I in the homogeneous coordinate ring R � k PZ 1 n2 Ž .of P over any algebraically closed field k , where I is the idealj

generated by all forms vanishing at p .jBecause Z has codimension 2 and is arithmetically Cohen�Macaulay,

the minimal free graded resolution of I is of the form 0 � F � F � IZ 1 0 Z� 0. To determine F , it then suffices to compute for each t the0

Ž . Ž . Ž .dimension � Z of the cokernel of the multiplication map � Z : It�1 t Z tŽ . � �� i R � I , since F is simply � R �i . If we know the Hilbert1 Z t�1 0 i� 0

Ž Ž . Ž .function h of I i.e., the dimension h t � dim I for each degree tZ Z Z Z tŽ . .of the homogeneous component I of I of degree t , then exactness ofZ t Z

the resolution allows us to determine the Hilbert function of F and thus1Ž . Ž .since F is free F itself. More explicitly, since in general � Z �1 1 t

Ž Ž .. � �si Ž Ž ..dim Tor I , k and F � � R �i , where s � dim Tor I , k ,0 Z t 1 i 1 Z ii� 0we can tensor the Koszul complex for k by I to compute these Tor’s. WeZ

3 Ž .find that � � s � � h i , where � is the difference operator, defined byi i ZŽ . Ž . Ž .�h i � h i � h i � 1 .Z Z Z

Instead of working with I and its components, by a standard translation,one can instead work on the surface X obtained by blowing up the ndistinct points p , . . . , p of P 2, in which case one then has the correspond-1 ning birational morphism � : X � P 2. Throughout this paper, this is whatX denotes.

We now recall this standard translation. With respect to � : X � P 2, let2 �1Ž .L be the total transform to X of a line on P , and let E � � p fori i

1 � i � n. We call the divisors L, E , . . . , E an exceptional configuration;1 n

FITCHETT, HARBOURNE, AND HOLAY686

� � � � � � Ž .their classes L , E , . . . , E Cl X form a basis for the divisor class1 nŽ . Žgroup Cl X of X, with intersections given by E � E � �� where �i j i j i j

. 2denotes Kronecker’s delta , L � E � 0 for all i, and L � 1. For Z � m pi 1 1Ž .� �� m p , under natural identifications we have I �n n Z t

0Ž Ž .. Ž . 0Ž Ž ..H X, OO F ; hence h t � h X, OO F , where F � tL � m EX t Z X t t 1 1Ž . Ž . Ž .� �� m E . Moreover, the map � Z : I R � I given byn n t Z t 1 Z t�1

multiplication corresponds under these identifications to the natural map0Ž Ž .. 0Ž Ž .. 0Ž Ž . Ž ..� : H X , OO F H X , OO L � H X , OO F OO L �F X t X X t Xt0Ž Ž ..H X, OO F . Thus, although in this paper we are concerned with �X t�1 F

for an arbitrary divisor F, our results have an immediate application tocomputing resolutions of fat point ideals. We also note that as long as

0Ž Ž ..h X, OO H is known for arbitrary divisors H, when computing theXdimension of the cokernel of � for some divisor F, it is just as good toFcompute the dimension of the kernel of � or the rank of � , dependingF Fon convenience.

In addition to assuming that we have a birational morphism X � P 2

Ži.e., that X is basic, that is, obtained by blowing up n � 0 points p ofi2 .P , we assume that there is a particular smooth, irreducible anticanonical

Ž .divisor D on X i.e., the points blown up lie on a smooth cubic , and thatXX satisfies the following properties:

Ž .A1 The only integral curves besides possibly D of negative self-XŽintersection on X are exceptional curves i.e., smooth rational curves of

.self-intersection �1 .Ž . 1Ž Ž ..A2 h X, OO F � 0 for any effective, numerically effective divi-X

Žsor F. F being numerically effecti�e means that F � H � 0 for every.effective divisor H.

In these circumstances, we say that X is a good surface.2 Ž .For example, any blow up X of P at n � 8 general points is good: A2

� � Ž .holds by Theorem 8 of H3 , and A1 holds by adjunction since theŽanticanonical class �K is ample see the first two paragraphs of theX

� �.proof of Theorem 1 of H3 . For another example, say that a basic surfaceX with a smooth, irreducible anticanonical divisor D is injecti�e if theX

Ž . Ž . Žcanonical map Pic X � Pic D is injective. Although we sometimesXuse D to denote an arbitrary divisor, throughout this paper we use D toX

.denote a smooth, irreducible anticanonical divisor. Such a surface X isŽ . � � Ž .good: A2 holds by Theorem III.1 of H4 , while A1 holds by adjunction,

since any integral curve C of negative self-intersection which is neither anexceptional curve nor D must by adjunction have C � D � 0, and henceX X

Ž . Ž .C � 0 since Pic X � Pic D is injective.XŽ . Ž .We note that the condition that Pic X � Pic D is injective holds inX

all characteristics if the points blown up are sufficiently general points of a


smooth plane cubic curve, if the ground field k is sufficiently large. Itcannot hold, however, if k is the algebraic closure of a finite field. On theother hand, assuming that the points p are general points of a smoothi

plane cubic curve, our result for the rank of � for a particular F or forF

the minimal free resolution of the ideal I for a particular Z holds overZ

any algebraically closed field k. This is because our result for a particularF or Z holds for some set of n K-points of a smooth plane cubic k-curve

2C � P , where K is the algebraic closure of a sufficiently large extensionnK of k, but the k-points of C are a dense subset of the K-points.

We denote the set of reduced, irreducible curves C on X with0Ž Ž .. � � Ž .h X, OO C � 1 by � . By Theorem III.1 of H4 , A1 implies that �X X X

includes the set of exceptional curves on X, and D if D2 � 0, butX X

nothing else.� � ŽMoreover, as shown in Section 4 of H6 mutatis mutandis, since �K X

� �.need not always be the class of an effective divisor in H6 , given that X isgood, one can in a completely effective manner determine the fixedcomponents and dimension of any complete linear system on X. Ourmain interest, then, is also to determine the rank of � for an arbitraryF

divisor F.Ž . Ž .Assuming A1 and A2 , our approach is to reduce the problem of

computing the rank of � for an arbitrary divisor F to that of doing so forF� �a special class of divisors. A similar approach was taken in H6 , in which a

reduction was made to ample F. Unfortunately, the class of ample divisorsŽis still too coarse to get a nice answer in general note that the exceptional

Ž .Ž . .case in Theorem 1 c ii is ample , even for n � 8 points, so here wepresent a refined reduction to a particular special set of divisors F.Although for arbitrary n we cannot always handle the resulting divisors bypresent methods, for n � 8, using ad hoc methods we always can, therebyachieving a complete determination. In terms of fat points, this gives acomplete determination of the minimal free resolution for the ideal of anyfat points subscheme involving up to eight general points of P 2, essentiallyby reduction to nice cases where a formula applies.

To state our main result, we introduce some terminology and notation.0Ž . 0Ž Ž ..For convenience, we often write simply h X, F for h X, OO F . Also,X

we say that a divisor F on X is monotone provided that F � E � F � E �1 2�� F � E . We now define quantities and for each curve C in � .n X

For C � E for any i, let � � 0. Otherwise, let m be the maxi-i C C C

mum of C � E , . . . , C � E ; define to be the maximum of m and of1 n C CŽ . � Ž .C � L � m ; define to be the minimum of m and of C � L � m ;C C C C

and if C is a smooth rational curve, then define to be � , andC Cotherwise define to be the maximum of � and 2. For example, if C isC C

anticanonical, then � � 1 and � � 2.C C C


Here now is our main result, Theorem 1. For general n, it determinesŽ .the rank of � for certain sufficiently nice divisors F, which is sometimesF

enough to determine resolutions, as shown in Section 4. For n � 8,Theorem 1 gives a reduction for arbitrary F to the nice case, thereby inprinciple providing an algorithm for computing the rank of � for anyFdivisor F. Briefly, the algorithm works as follows. By applying Lemma 4

Ž . 0Ž . Žwith Lemma 3 a , we can compute h X, F for any divisor F. See Section� � .4 for examples, or H1, H4, and H6 for more comprehensive background.

Ž . Ž .By Lemma 3, � is finite for n � 8, so cases a and b of Theorem 1 areXŽ . Ž .easy to implement. If cases a or c occur, then we are done, while case

Ž . 0Ž .b need be applied only if h X, F � 0 since � is clearly injective ifF0Ž . Ž . 0Ž . Ž .h X, F � 0. In case b with h X, F � 0, repeated applications of b

0Ž .gives a divisor F for which either h X, F � 0 or which satisfies theŽ . Ž . Žconditions either of a or c . It is not hard to implement the algorithm

� �explicitly; see H9 , which includes an explicit Macaulay script implement-.ing this algorithm.

THEOREM 1. Let X be a good surface with exceptional configurationL, E , . . . , E . Let F be a monotone di�isor on X.1 n

Ž . Ža If F � C � for all C � , then � has maximal rank i.e., isC X F. Ž Ž .. Ž 0Ž .either injecti�e or surjecti�e , hence dim cok � � max 0, h X, F � L �F

0Ž ..3h X, F .Ž . Ž Ž ..b If F � C � for some C � , then dim ker � �C X FŽ Ž ..dim ker � .F� C

Ž . Ž . Ž .c If n � 8 but neither case a nor case b holds, then eitherŽ . Ž . Ž .i F � L � E � E � 0, in which case the dimension of cok �1 2 F

1Ž Ž .. 1Ž Ž ..is h X, F � L � E � h X, F � L � E ,1 2

Ž . � � � � � �ii F is 3L � E � �� E � r 8 L � 3E � �� 3E � E1 7 1 7 8Ž Ž .. Ž Ž ..for some r � 1, in which case dim cok � � r and dim ker � � r � 1,F F

orŽ . Ž Ž .. Žiii � has maximal rank, and hence dim cok � � max 0, F � LF F

2 .� F � K � F .X

Ž .Proof. Part a is Corollary 5. The proof here depends on a criterionfor � to be surjective, which involves vanishing of certain h1’s, in theF

Ž . Ž . Ž .form of q* F and l* F defined later; see Lemma 2 b . Being defined in1 Ž . Ž .terms of h ’s involving F, intuitively we can expect q* F and l* F to

vanish if F is sufficiently nice, which is precisely what our hypothesesŽ .guarantee. Part b is Proposition 6. This part generalizes the fact that �F

and � have kernels of the same dimension if F is effective and C is anF� Cintegral curve with F � C � 0, which is obvious since in this situation C is a

� � Ž .fixed component of F . Most of the work here is in proving c , which


follows from Proposition 13 and Proposition 15. The proof of this partŽ . Ž .amounts to an analysis of all possibilities not dealt with by a or b .

2. THE NICE CASE AND THE REDUCTION

Ž .We regard case a of Theorem 1 as the nice case, because there weŽ .have the rank of � directly. Case b of Theorem 1 is the reduction case.F

Ž . Ž .In this section we prove cases a and b of Theorem 1. Given a basicsurface X with exceptional configuration L, E , . . . , E and a monotone1 n

0Ž . Ž . 1Ž .divisor F on X, we denote h X, F � E by q F , h X, F � E by1 1Ž . 0Ž Ž .. Ž . 1Ž Ž .. Ž .q* F , h X, F � L � E by l F , and h X, F � L � E by l* F .1 1

LEMMA 2. Let X be obtained by blowing up distinct points of P 2, withexceptional configuration L, E , . . . , E . Let F be a monotone di�isor on X.1 nThen

Ž . Ž Ž .. Ž . Ž . Ž .a dim ker � � q F � l F ; in particular, � is injecti�e if q FF FŽ .� 0 � l F .

Ž . 1Ž . Ž Ž .. Ž .b If F is effecti�e and h X, F � 0, then dim cok � � q* F �FŽ . Ž . Ž .l* F ; in particular, � is surjecti�e, if also l* F � 0 � q* F .F

Ž . � �Proof. a See Lemma 4.1 of H8 , which assumes that F � E � 0ifor all i, but it is easy to check that the result holds even if F � E � 0 forisome i.

Ž .b Clearly, L is numerically effective. Thus, F � L � 0, since F is1Ž . Ž . Ž . Žeffective. Now, from h X, F � 0 and 0 � OO F � OO F � L � OO FX X L

. 1Ž .� L � 0, we see that h X, F � L also vanishes, and we compute0Ž . 0Ž . 0Ž . Ž .h X, F � L � 3h X, F � 2 � F � L � 2h X, F . Similarly, l* F �Ž . Ž . 0Ž . Ž . Ž .l F � F � L � E � 1 � h X, F and q* F � q F � F � E � 1 �1 10Ž . Ž Ž . Ž .. Ž Ž . Ž .. 0Ž .h X , F , so l* F � l F � q* F � q F � h X , F � L �

0Ž . Ž Ž .. Ž Ž .. 0Ž .3h X, F . Therefore, dim cok � � dim ker � � h X, F � L �F F0Ž . Ž . Ž . 0Ž . 0Ž . Ž . Ž .3h X, F � l F � q F � h X, F � L � 3h X, F � l* F � q* F ,

as required.

We need to refer to the following result.

LEMMA 3. Let C be a cur�e on the blow up X of P 2 at eight generalpoints, with exceptional configuration L, E , . . . , E . Then1 8

Ž .a Up to permutation of the indices, C is an exceptional cur�e if and� � � � � � �only if C is one of the following: E , L � E � E , 2 L � E8 1 2 1

� � � �� E , 3L � 2 E � E � �� E , 4L � 2 E � 2 E � 2 E � E5 1 2 7 1 2 3 4� � � �� E , 5L � 2 E � �� 2 E � E � E , or 6L � 3E � 2 E8 1 6 7 8 1 2

�� 2 E .8


Ž .b Up to permutation of the indices, C is a smooth rational cur�e with2 � � � � �C � 0 if and only if C is one of the following: L � E , 2 L � E1 1

� � � �� E , 3L � 2 E � E � �� E , 4L � 2 E � 2 E � 2 E � E4 1 2 6 1 2 3 4� � � �� E , 4L � 3E � E � �� E , 5L � 3E � 2 E � 2 E � 2 E7 1 2 8 1 2 3 4

� � � �� E � �� E , 5L � 2 E � �� 2 E � E , 6L � 3E � 3E � 2 E5 8 1 6 7 1 2 3� �� 2 E � E � E , 7L � 3E � �� 3E � 2 E � 2 E � 2 E �6 7 8 1 4 5 6 7

� � � � �E , 7L � 4E � 3E � 2 E � �� 2 E , 8 L � 3E � �� 3E � E ,8 1 2 3 8 1 7 8� � �8 L � 4E � 3E � �� 3E � 2 E � 2 E � 2 E , 9L � 4E � 4E �1 2 5 6 7 8 1 2

� � �3E � �� 3E � 2 E , 10L � 4E � �� 4E � 3E � �� 3E , and3 7 8 1 4 5 8� �11L � 4E � �� 4E � 3E .1 7 8

Ž . ŽProof. Under the action on Cl X by the Weyl group which is gener-ated by permutation of the indices and by the action of quadratic Cremona

� �transformations centered at any three of the eight points; see H1 or� �.H2 , any class of an effective divisor F is in the orbit of a class F� of the

� �form dL � a E � �� a E , where d � 0, 3d � a � �� a � 01 1 8 8 1 8Ž .since, as mentioned in Section 1, �K is ample , d � a � a � a � 0,X 1 2 3and a � a � �� a . If F is reduced and irreducible, then F� is either1 2 8� � � �E or a � 0. In the latter case, F� is, by Lemma 1.4 of H1 , a8 8nonnegative sum of the classes of L, L � E , 2 L � E � E , 3L � E �1 1 2 1E � E , . . . , 3L � E � �� E . But any such sum has nonnegative self-2 3 1 8

� �intersection. Thus classes of exceptional curves are in the orbit of E ;8Ž .these are the classes listed in a . The only such sums with self-intersection

� � � �0 are the positive multiples of L � E ; hence only L � E itself1 1represents the class of a reduced and irreducible divisor with self-intersec-

Ž . � �tion 0. Thus the classes to be listed in b comprise the orbit of L � E1Ž .under the action of the Weyl group; it is easy to check that the list in b is

Ž .up to permutations the complete orbit.

Ž .Given a surface X, denote by EFF or just EFF the subsemigroup ofXŽ .the divisor class group Cl X of X of classes of effective divisors, and let

NEFF denote the cone of classes of numerically effective divisors.

LEMMA 4. Let X be a good surface obtained by blowing up n points of P 2.

Ž .a If 2 � n � 8, then EFF is generated by the classes of exceptionalcur�es, while for n � 8, EFF is generated by the classes of exceptional cur�esand by �K .X

Ž . Ž .b Let F Cl X . If 2 � n � 8, then F is in NEFF if and only ifF � C � 0 for e�ery exceptional cur�e C, while if 9 � n, then F is in NEFF ifand only if both �F � K � 0 and F � C � 0 for e�ery exceptional cur�e C.X

Ž . 1Ž . 2Ž . 0Ž .c EFF contains NEFF, and h X, F � h X, F � 0 and h X, FŽ 2 .� F � F � K �2 � 1 for any F NEFF.X


Ž . Ž . Ž .Proof. a By A1 , any effective divisor is up to linear equivalence anonnegative sum of D , exceptional curves and an element of NEFF. ButX

� �by Corollary 3.2 and Lemma 1.4, both of H1 , any element of NEFF isŽ .with respect to some exceptional configuration L, E , . . . , E a sum of1 n� � � � � �L , L � E , 2 L � E � E , �K and classes of exceptional curves.1 1 2 X

� � � � � � � � � � �But n � 2, so L � L � E � E � E � E , L � E � L � E1 2 1 2 1 1� � � � � � � � � � �� E � E , and 2 L � E � E � 2 L � E � E � E � E are2 2 1 2 1 2 1 2

sums of classes of exceptional curves, so NEFF is generated by the classesof exceptional curves and by the class �K of D . Moreover, forX Xn � 2, 3, 4, 5, 6 or 7, �K is, respectively, the following sums of classes ofX

� � � � � � � �exceptional curves: 3 L � E � E � 2 E � 2 E , L � E � E �1 2 1 2 1 2� � � � � � � � � � � �L � E � E � L � E � E � E � E � E , L � E � E �1 3 2 3 1 2 3 1 2� � � � � � � � � � �L � E � E � L � E � E � E � E , L � E � E � L �3 4 1 2 1 2 1 2

� � � � � � � � � �E � E � L � E � E � E , L � E � E � L � E � E � L3 4 1 5 1 1 2 3 4� � � � �� E � E , and 2 L � E � E � �� E � L � E � E .5 6 1 2 5 6 7

Ž . Ž .b This follows immediately from a , except in the case where� � �n � 8. For n � 8, we have �2 K � 3L � 2 E � E � �� E � 3LX 1 2 7

�� E � �� E � 2 E ; hence if F � E � 0 for all exceptional curves E,2 7 8then also �F � K � 0, and our result follows here as well.X

Ž . � � 2 2Ž .c By Proposition 4 of H3 , F � 0 and h X, F � 0 for anyF NEFF. Since we assume that �K is effective, we also have �F � KX X

0Ž . Ž 2 .� 0. But by Riemann�Roch, h X, F � F � F � K �2 � 1, soX0Ž . 1Ž .h X, F � 1, hence F EFF. Now h X, F � 0 for any F NEFF by

Ž .A2 , and the rest is immediate from Riemann�Roch.

Ž .Here is the proof of Theorem 1 a .

COROLLARY 5. Let X be a good surface with exceptional configurationL, E , . . . , E , and let F be a monotone di�isor on X. If F � C � for all1 n CC � , then � has maximal rank.X F

0Ž .Proof. If h X, F � 0, then clearly � is injective and so has maximalF0Ž .rank. So assume that h X, F � 0. Since both � 0 and F � C � C C

for all C � and hence for all curves C of negative self-intersection, weXsee that F is numerically effective.

If n � 5, then � is surjective simply because F is numerically effectiveFŽ � �.Theorem 3.1.2, H7 . So we may assume that n � 6.

1Ž . Ž .Since F is numerically effective, h X, F � 0 by A2 and F � E � 0iŽ .for all i, and so F � E � E � 0 for all i. If C is an exceptional curve but1 i

not E for any i, then � C � E and hence F � C � implies thati C 1 CŽ .F � E � C � 0. If 6 � n � 8, then F � E is numerically effective by1 1

Ž . 1Ž . Ž .Lemma 4, and hence q* F � h X, F � E � 0 by A2 . If n � 9, then1Ž .D � , and so in addition we have F � E � D � � 1 � 1, andX X 1 X D X


Ž . 1Žagain we see that F � E is numerically effective and q* F � h X, F �1.E � 0.1

Ž . Ž Ž ..Now consider l* F . If F � L � E � E � 0 for some i, then clearly1 ii � 1 and F � E � 0. Since F is monotone, we see that F � E � 0 for all i,1 i

Ž .so F is up to linear equivalence a nonnegative multiple of L, for which itŽ Žis easy to see that � has maximal rank. Thus we may assume F � L �F

..E � E � 0 for all i. For any other exceptional curve C, we have F � C �1 iŽ Ž .. Ž . , so F � L � E � C � � L � E � C � 0 as well. Thus, asC 1 C 1

Ž . Ž . 1Žbefore, F � L � E is numerically effective if n � 8, so l* F � h X, F1Ž .. Ž Ž� L � E � 0, while if n � 9, then we have in addition that F � L �1.. Ž . 1Ž Ž ..E � D � � 2 � 0, so again l* F � h X, F � L � E � 0. The1 X D 1X

Ž .result now follows by Lemma 2 b .

Ž .We now prove part b of Theorem 1.

PROPOSITION 6. Let F be a di�isor on a good surface X with exceptionalconfiguration L, E , . . . , E . If F � C � for some C � , then1 n C X

Ž Ž .. Ž Ž ..dim ker � � dim ker � .F F�C

0Ž Ž .. Ž . ŽProof. If h X, OO F � 0, then of course neither OO F nor OO F �X X X. Ž . Ž .C has any global sections, so both ker � and ker � vanish.F F�C

Therefore, we may assume that F is effective.� �If C is a fixed component of F for some C � , then clearly theX

Ž . Ž .canonical injection OO F � C � OO F induces an isomorphism of bothX X� �global sections and kernels of �. So now we may assume that F is fixed

component free; hence F � C � 0 for all C � .XSince � 2, if F � D � , then it must be that F � D is 0 or 1. IfD X D XX X

Ž . Ž . Ž .0, then OO F is in the kernel of Pic X � Pic D , and since this isX XŽ . Ž .injective, we see that F � 0, in which case both ker � and ker �F F�D X0Ž Ž .. 0Ž Ž ..again vanish. If 1, then h D , OO F � 1 so � : H D , OO FX D OO ŽF . X DX D XX0Ž . 0Ž Ž .. H X, L � H D , OO F � L has the same kernel as the mapX D X0Ž . 0Ž Ž ..H X, L � H D , OO L given by restriction, which is easily seen toX D X

Ž . Ž .be injective. From the exact sequence 0 � ker � � ker � �F�D FXŽ . Ž . Ž . Ž .ker � induced by 0 � OO F � D � OO F � OO F � 0 weOO ŽF . X X X DD XX Ž Ž .. Ž Ž ..see that dim ker � � dim ker � .F F�D X

Finally, if F � C � for some C � and C � D , then C is anC X XŽ .exceptional curve. As earlier, the exact sequence 0 � OO F � C �X

Ž . Ž . Ž .OO F � OO F � 0 induces an exact sequence 0 � ker � �X C F�CŽ . Ž .ker � � ker � , where � denotes the mapF F�C , C F�C , C

H 0 C , OO F OO H 0 X , L � H 0 C , OO F � L OO .Ž . Ž . Ž .Ž . Ž .X C X C

� � Ž Ž ..By Theorem 3.1 of F1 , ker is injective if F � C � , so dim ker �� C FF�C , CŽ Ž ..equals dim ker � .F�C


Ž .3. THE MAIN THEOREM, PART C

Ž .To prove part c of Theorem 1, we need some additional background,which we now develop. For the purpose of stating the next result, givensheaves FF and LL on a scheme Y, we denote the kernel of

H 0 Y , FF H 0 Y , LL � H 0 Y , LL LLŽ . Ž . Ž .

Ž . Ž . Ž .by R FF, LL and the cokernel by S FF, LL taking Y to be understood .

PROPOSITION 7. Let Y be a closed subscheme of projecti�e space, let FF

and LL be coherent shea�es on Y, and let CC be the sheaf associated to an0Ž .effecti�e Cartier di�isor C on Y. If the restriction homomorphisms H Y, FF

0Ž . 0Ž . 0Ž .� H C, FF OO and H Y, FF LL � H C, FF LL OO are surjec-C Cti�e, then we ha�e an exact sequence

0 � R FF CC�1 , LL � R FF , LL � R FF OO , LL �Ž . Ž . Ž .C

S FF CC�1 , LL � S FF , LL � S FF OO , LL � 0.Ž . Ž . Ž .C

Ž � �.Proof. This is a snake lemma argument see M .

EXAMPLE 8. The preceding result is often applied in situations in whichŽ . Ž . Ž .we have an exact sequence 0 � OO F � C � OO F � OO F � 0,X X C

where C is a curve on a surface X with an exceptional configuration1Ž .L, E , . . . , E , and F is a divisor on X with h X, F � C � 0 and1 n

1Ž .h L, F � C � L � 0. In such a situation, we can apply Proposition 7Ž . Ž . Ž .with Y � X, FF � OO F , CC � OO C and LL � OO L . For example, let XX X X

be a blow up of P 2 at eight general points, with exceptional configurationL, E , . . . , E . Recall that D is a smooth, irreducible anticanonical divisor1 8 Xon X. For future reference, we show that � and � haveD �t L 2 D �t LX X

maximal rank for all integers t. First, say that t � 0. Consider the exactŽ . Ž . Ž .sequence 0 � OO 2 D � E � OO 2 D � OO 2 � 0, where E is theX X X X E

� �exceptional curve whose class is 6L � 3E � 2 E � �� 2 E . Since1 2 82 D � E � E , clearly � is injective, and by Lemma 12, � is alsoX 1 2 D �E 2, EX

Ž Ž ..injective. Hence, applying Proposition 7, it follows that dim ker � �2 D X

0; i.e., � has maximal rank, and an easy calculation now shows that2 D XŽ Ž ..dim cok � � 0 as well. From Proposition 7 and the exact sequence2 D X

Ž . Ž . Ž .0 � OO D � OO 2 D � OO 2 D � 0, it also follows thatX X X X D XXŽ Ž ..dim ker � � 0. It is easier to see that � and � haveD D �t L 2 D �t LX X X0Ž . 0Žmaximal rank for all t � 0, since h X, D � tL � 0 and h X, 2 D �X X

.tL � 0 for t � 0, in which case � and � are clearly injective,D �t L 2 D �t LX X1Ž . 1Ž .while for t � 0 we have h X, D � tL � 0 and h X, 2 D � tL � 0, inX X1Žwhich case we apply the general fact that � is surjective if h X, H �H� L


. 1Ž . 0Ž .tL � 0 for t � 0. Indeed, h X, H � 0 ensures that H X, H � L �0Ž . Ž Ž . Ž ..H L, H � L is surjective, and hence S OO L , OO H � L �L XŽ Ž . Ž ..S OO L , OO H � L . Then applying Proposition 7 with Y � X, C � L,L L

Ž . Ž .FF � OO L , and LL � OO H � L , we see � is onto�just note thatX X H�LŽ Ž . Ž ..S OO L , OO H � L � 0, since in this situation it is easy to check thatX X

Ž Ž .. Ž Ž . Ž ..both S OO , OO H � L � 0 and S OO L , OO H � L � 0.X X L L

PROPOSITION 9. Let F be a monotone di�isor on a good surface X withŽ .exceptional configuration L, E , . . . , E . If F � L � E � D � 0, then1 n 1 2

dim ker � � h0 X , F � L � E � h0 X , F � L � E .Ž . Ž . Ž .Ž . Ž . Ž .F 1 2

1Ž .If in addition F is effecti�e and h X, F � 0, then

dim cok � � h1 X , F � L � E � h0 X , F � L � E .Ž . Ž . Ž .Ž . Ž . Ž .F 1 2

0Ž . Ž Ž ..Proof. If h X, F � 0, then certainly dim ker � � 0, and bothF0Ž Ž .. 0Ž Ž ..h X, F � L � E and h X, F � L � E are also 0. Otherwise, this1 2

Ž . � �is Proposition II.2 e and Remark II.3, both of H6 .

We recall that a nonzero element of a free Abelian group is primiti�e ifit is not a multiple greater than 1 of another element of the group.

LEMMA 10. Let F be a di�isor on the blow up X of P 2 at 2 � n � 8general points, and suppose that F � C � �1 for all exceptional cur�es C on

1Ž . � �X. Then either h X, F � 0 or F � rH � K with r � 2, where H isXprimiti�e and H is smooth, rational, and numerically effecti�e with H 2 � 0, in

1Ž .which case h X, F � r � 1.

Ž .Proof. Since F � C � �1 for all exceptional curves C, C � F � K � 0Xfor all exceptional curves C, which implies that F � K � D is numeri-X

1Ž . 1Ž .cally effective by Lemma 4. Therefore, h X, F � h X, D � K �X1Ž Ž .. 1Ž .h X, K � D � K � h X, �D , with the center equality due toX X

� � Ž � �Serre duality. By Ramanujam vanishing R see T for a characteristic p� �. 1Ž . 2 2version, or see Theorem 2.8 of H5 , h X, �D � 0 if D � 0. If D � 0,

Ž .then we still have D � �K � 0 since, for n � 8, �K is ample. ThenX X� �by an easy calculation applying Lemma 1.4 of H1 , we see that the only

� � � �possibility is for D to be in the orbit of r L � E under the action of the1Ž . Ž . � � � �Weyl group see the proof of Lemma 3 on Cl X . That is, D � r H for

� �some r, where H is primitive and H is a smooth rational curve with2 1Ž . ŽŽH � 0. Now h X, �D � r � 1 follows by induction via 0 � OO �r �X. . Ž .1 H � OO �rH � OO � 0.X H

� �We need to refer to the main result of H6 .


THEOREM 11. Let F be a monotone, numerically effecti�e di�isor on theblow up X of P 2 at se�en general points, L, E , . . . , E being the correspond-1 7ing exceptional configuration. Let t denote the number of indices i such thatF

Ž 0Ž . 0Ž .. ŽF � E � �F � K and let � max 0, h X, F � L � 3h X, F thei X F.‘‘expected,’’ maximal rank dimension of the cokernel of � . ThenF

Ž Ž .. Ž .dim cok � � max t , , unless F is, up to linear equi�alence, eitherF F F0, B, B � K � E , B � 2 K � E � E , B � 3K � E � E � E , BX 4 X 4 5 X 4 5 6� 4K � E � E � E � E , G, or G � K � E , where B � 4L � 2 EX 4 5 6 7 X 7 1� 2 E � 2 E � E � �� E and G � 5L � 2 E � �� 2 E � E , in2 3 4 7 1 6 7

Ž Ž ..which case � is injecti�e and dim cok � � .F F F

� �Proof. This is Theorem I.6.1 of H6 .

LEMMA 12. Let C be a smooth rational cur�e on the blow up X of P 2 ateight general points with exceptional configuration L, E , . . . , E , and assume1 8C � L � d. Let 0 � t and consider the map

� : H 0 X , OO t H 0 X , OO 1 � H 0 C , OO t � d ,Ž . Ž . Ž .Ž . Ž . Ž .t , C C X C

gi�en by restriction and multiplication on simple tensors.

Ž . 2 Ž .a If C � �1 i.e., C is an exceptional cur�e on X , then � hast, Cmaximal rank.

Ž . 2 Žb If C � 0, then � has maximal rank, except up to permutationt, C.of the indices in the following cases:

Ž . � � � �i C � 4L � 3E � E � �� E and t � 1.1 2 8

Ž . � � � �ii C � 8 L � 3E � �� 3E � E and t � 3.1 7 8

Ž . Ž .For both i and ii , the rank of � is one short of maximal rank.t, C

Proof. It is easy to see that the kernel of the surjective sheaf map0Ž . Ž . Ž . Ž .OO H X, L � OO d is OO �a � OO �b , for some nonnegative aC C C C

Žand b with a � b � d. If we assume a � b, then it turns out see the proof� �. � 4of Theorem 3.1 of F1 that a � min d � m , m , and thus when d � mC C C

and m differ by at most 1, a is the smaller of d � m and m and b isC C Cthe larger. Moreover, when a and b differ by at most 1, then for each t,

0Ž Ž . Ž .. 1Ž Ž . Ž ..either h C, OO t � a � OO t � b � 0 or h C, OO t � a � OO t � bC C C C� 0, and hence � has maximal rank for every t.t, C

Ž . Ž .Part a now follows by checking Lemma 3 a to see that d � m andCm always differ by at most 1 if C is an exceptional curve.C

Ž . Ž .The same proof via Lemma 3 b also works for b , except for thefollowing four cases, for which d � m and m differ by more than 1:C C4L � 3E � E � �� E , 8 L � 3E � �� 3E � E , 10L � 4E1 2 8 1 7 8 1� �� 4E � 3E � �� 3E , and 11L � 4E � �� 4E � 3E .4 5 8 1 7 8

To handle C � 4L � 3E � E � �� E , let D � 4L � 3E � E1 2 8 1 2Ž .� �� E so D � C � 1 and D � C � E , and consider the exact se-7 8


quence

0 � OO D � C � OO D � OO 1 � 0.Ž . Ž . Ž .X X C

Ž Ž .. Ž .We can compute dim cok � trivially � is bijective , and � has aE E D8 8

one-dimensional kernel and cokernel by Theorem 11, so by Proposition 7,Ž Ž .. Ž . Ž .dim ker � � 1. Since the kernel of � is OO t � a � OO t � b , we1, C t, C C C

see that

h0 C , OO 1 � a � OO 1 � b � 1,Ž . Ž .Ž .C C

which with a � b � 4 gives a � 1 and b � 3. We see that � hast, Cmaximal rank unless t � 1, in which case the rank is one short of maximal.

To handle C � 8 L � 3E � �� 3E � E , let D � 8 L � 3E1 7 8 1Ž .� �� 3E so again D � C � 1 and D � C � E . Consider the exact7 8

sequence

0 � OO 3D � C � OO 3D � OO 3 � 0.Ž . Ž . Ž .X X C

Ž . Ž .Note that 3D � C � 2 D � E . The inclusion ker � � ker � is8 2 D 2 D�E8

clearly an isomorphism since it is induced by the canonical injectionŽ . Ž .OO 2 D � OO 2 D � E , which gives an isomorphism on global sections.X X 8

But by Theorem 11, � and � have kernels of dimensions 0 and 1,2 D 3 DŽ Ž ..respectively, so by Proposition 7, dim ker � is at least one dimen-3, C

�sional. Thus a � 3, but we know that a � b � 8 and a is at least min d �4m, m � 3, so in fact a � 3 and b � 5, which gives the desired result.

To handle C � 10L � 4E � �� 4E � 3E � �� 3E , note that1 4 5 8Ž . Ž . Ž .a, b must be either 4, 6 or 5, 5 . We just need to compute the dimen-sion of the kernel of � to tell which. From the exact sequence4, C

0 � OO �2 K � OO �2 K � C � OO 4 � 0,Ž . Ž . Ž .X X X X C

Ž .and the fact that � is bijective see Example 8 , it suffices to compute�2 K XŽ Ž ..dim ker � . Let D � 6L � 3E � 2 E � �� 2 E and look at�2 K �C 1 2 8X

0 � OO �2 K � C � D � OO �2 K � C � OO 2 � 0.Ž . Ž . Ž .X X X X D

Ž .Using the injectivity of � from part a , we reduce to � . Now2, D �2 K �C�DX

let E � 6L � 2 E � 3E � 2 E � �� 2 E , look at1 2 3 8

0 � OO �2 K � C � D � E � OO �2 K � C � D � OO 2 � 0,Ž . Ž . Ž .X X X X E

and note that �2 K � C � D � E � 4L � E � E � 2 E � 2 E � EX 1 2 3 4 5� �� E . Permuting indices gives F � 4L � 2 E � 2 E � E � �� E ,8 1 2 3 8

Ž . Ž .which has l F � q F � 0. Hence � , and thus � is�2 K �C�D�E 4, CX


injective by Lemma 2; i.e., a must be bigger than 4, so a � 5. Again, thedesired result follows.

To handle C � 11L � 4E � �� 4E � 3E , note that we know1 7 8Ž . Ž . Ž .that a, b is either 4, 7 or 5, 6 and, as before, the dimension of the

kernel of � tells which. Let D � 6L � 3E � 2 E � �� 2 E4, C 1 2 8and E � 6L � 2 E � 3E � 2 E � �� 2 E as before and let F �1 2 3 82 L � E � �� E . Following a process similar to before, we find that3 7

Ž Ž .. Ž Ž ..dim ker � � dim ker � , but �2 K � C � D ��2 K �C�D�E�F 4, C XXŽ Ž .. Ž .E � F � �K and we know dim ker � � 0 see Example 8 . ThusX �K X

a � 4 and hence a � 5, as needed.

PROPOSITION 13. Let X be a good surface with exceptional configurationL, E , . . . , E . Let F be a monotone di�isor on X such that F � E � for all1 8 Eexceptional cur�es E but F � C � for some exceptional cur�e C. ThenCeither

Ž . Ž .i F � L � E � E � 0, in which case1 2

dim cok � � h1 X , F � L � E � h1 X , F � L � E ,Ž . Ž . Ž .Ž . Ž . Ž .F 1 2

Ž .ii � has maximal rank, orF

Ž . Ž .iii F � E � �� F � E i.e., F is nearly uniform , F is numerically1 7� � � �effecti�e, and F � C � 2, where C � 5L � 2 E � �� 2 E � E � E .1 6 7 8

Proof. By Lemma 4, we see that F is effective and numerically effec-Ž Ž .. 1Ž .tive, and hence by A1 h X, F vanishes. From Lemma 3 and mono-

� � � � �tonicity of F, we can assume that C is one of L � E � E , 3L � 2 E1 2 1� � �� E � �� E , or 5L � 2 E � �� 2 E � E � E , since for all other2 7 1 6 7 8

cases � . We also may assume that � fails to have maximal rank.C C F� � � � Ž .If C is L � E � E , then F � L � E � E � 0, and our result1 2 1 2

Ž .follows by Proposition 9. Thus we may assume that F � L � E � E � 0.1 2� � � �If C is 3L � 2 E � E �� E , then F � C � 1 and, by Lemma 10,1 2 7

Ž . 1Ž .either q* F � h X, F � E � 0 or F � rH � K � E for some r � 2,1 X 1� �where H is primitive and H is smooth, rational, and numerically effec-

tive with C � H � 0 and H 2 � 0.Ž . � �In the latter case, since F � L � E � E � 0, F is monotone and H1 2

is, up to permutation of the indices, one of the classes listed in LemmaŽ . � � �3 b , we see that H can only be 8 L � 4E � 3E � �� 3E � 2 E �1 2 5 6

� � � � � Ž2 E � 2 E . Thus F � 5L � 2 E � �� 2 E � E � E � E � r �7 8 1 5 6 7 8.� �1 H , which for the purposes of induction we denote by F . By ar

Ž . Ž .calculation, q F � 0 and l F � 0, so � is injective by Lemma 2. Now1 1 F1Ž . Ž . Ž .consider the exact sequence 0 � OO F � OO F � OO 2 � 0. ByX r�1 X r H

Ž .Lemma 12, � has maximal rank and hence here must be injective , so2, Happlying Proposition 7 and inducting, we see that � is injective for allFr

r � 1, contradicting our assumption that � fails to have maximal rank.F


Ž .On the other hand, suppose that q* F � 0. We still have F � C � 1 andŽF � E � for all exceptional curves E. If F � E � F � E , then F � L �E 1 2

. Ž Ž ..E is monotone and F � L � E � E � 0 for all exceptional curves E1 1Ž . Žunless E � 5L � 2 E � �� E � E � E or one obtained from this1 6 7 8

.class by permuting the indices and F � E � 2, in which case we at leastŽ Ž ..have F � L � E � E � �1. As before, by applying Lemma 10, either1

Ž . Ž . � �l* F � 0 or F � L � E � rH � K for some r � 2 with H primitive1 Xand H smooth, rational, and numerically effective with H 2 � 0. If C � H

Ž .� 0, then F � C � 0 contradicting F � C � 1 , while if C � H � 1, thenŽ . Ž .F � C � r � 2 contradicting F � C � 1 . Thus we see that l* F � 0 inŽ .addition to q* F � 0, so � is surjective by Lemma 2, contradicting ourF

assumption that � fails to have maximal rank.FIf, however, F � E � F � E , then by checking each possible exceptional1 2

curve E, we see that either

Ž . Ž Ž ..1 F � L � E � E � �1 for all exceptional curves E;1

Ž . Ž Ž .. � � � �2 F � L � E � E � �2 for E � 3L � 2 E � E � �� E1 2 3 8and hence F � E � F � C � 1 and so F � E � �� F � E ; or1 8

Ž . Ž Ž .. � � �3 F � L � E � E � �2 for E � 5L � E � 2 E �� 2 E1 1 2 7� � � �� E and hence F � E � 2 and, since, for E� � 5L � 2 E � �� 2 E8 1 6

�� E � E , we have 2 � F � E � F � E� � � 2 by monotonicity, we7 8 E �

see that F � E � �� F � E .1 7

Ž . Ž .In case 1 , by applying Lemma 10, we see that as before l* F � 0,Ž .since having F � L � E � rH � K again contradicts F � C � 1. Thus1 X

Ž .� is surjective by Lemma 2, contrary to assumption. In case 2 , F � tLFŽ .� m E � �� E for some t and m. Since F is numerically effective, we1 8

Ž Ž ..must have m � 0 and F � 6L � 3E � 2 E � �� E � 0, so t �1 2 817m�6. Thus m�2 � 17m�2 � 8m � F � C � 1, so m � 2, in which caseby Example 8, we know that � for the particular F that we have here willF

Ž . Ž .have maximal rank, contrary to assumption. Case 3 is just case iii ofProposition 13.

Ž .Now we may assume that F � 3L � 2 E � E � �� E � 1, and we1 2 7� � � �consider the case where C � 5L � 2 E � �� 2 E � E � E with1 6 7 8

Ž .F � C � 2. Note now that F � E � E � 0 for all exceptional curves E, so1F � E is, by Lemma 4, numerically effective and effective, and hence1

Ž .q* F � 0.Ž Ž ..If F � L � E � E � �1 for some exceptional curve E, then E can1

be taken to be 5L � E � 2 E � �� 2 E � E and F � E � �� F �1 2 7 8 1E . Otherwise, we may assume that F � E � �1 for all exceptional curves7

Ž . ŽE; hence, by Lemma 10, either l* F � 0 and � is surjective, contrary toF. � �assumption , or F � rH � K � L � E , where r � 2, H is primitive,X 1

and H is smooth, rational, and numerically effective with C � H � 0 andH 2 � 0.


ŽIn the latter case, keeping in mind that F is monotone which, since. Žr � 2, means that H must also be monotone , F � C � 2, and F � 3L �

. � �2 E � E � �� E � 1, from Lemma 3 we see that H must be either1 2 7� � �10L � 4E � �� 4E � 3E � �� 3E or 11L � 4E � �� 4E �1 4 5 8 1 7

�3E . In each case, � ends up having maximal rank. The argument in8 Feach case is similar; here are the details for the latter case.

� � � � � � �Let H be 11L � 4E � �� 4E � 3E and F � 9L � 4E �1 7 8 1� Ž .� � Ž3E � �� 3E � 2 E � r � 1 H ; we denote 9L � 4E � 3E2 7 8 1 2

.� �� 3E � 2 E � tH by F , for t � 0. Consider the exact sequence7 8 tŽ . Ž . Ž .0 � OO F � OO F � OO 5 � 0. By Lemma 2, � is surjective, andX t�1 X t H F0

by Lemma 12, � also is surjective. Applying Proposition 7 and induct-5, Hing, we see that � is surjective for all t � 0, contradicting our assumptionFt

that � fails to have maximal rank.F

To complete the proof of Theorem 1, it still remains to analyze numeri-cally effective monotone divisors F which are nearly uniform, have F � E

� � �� for all exceptional curves E, and have F � C � 2, where C � 5L �E�2 E � �� 2 E � E � E . It is useful to determine all nearly uniform1 6 7 8

monotone numerically effective classes which is what we do now. To� �simplify notation, we denote the class dL � aE � �� aE � bE by1 7 8

Ž .the triple d, a, b .

PROPOSITION 14. If X is the blow up of P 2 at eight general points withŽ . Ž .exceptional configuration L, E , . . . , E , then the classes 1, 0, 0 , 3, 1, 0 ,1 8

Ž . Ž . Ž . Ž . Ž .3, 1, 1 , 8, 3, 0 , 8, 3, 1 , 11, 4, 3 , and 17, 6, 6 generate the cone of mono-tone, numerically effecti�e, nearly uniform classes.

Ž . Ž . Ž . Ž . Ž . Ž .Proof. Since 1, 0, 0 , 3, 1, 0 , 3, 1, 1 , 8, 3, 0 , 8, 3, 1 , 11, 4, 3 , andŽ .17, 6, 6 are all numerically effective, monotone, and nearly uniform, anynonnegative Z-linear combination is also numerically effective, monotone,and nearly uniform.

Ž .Conversely, let F � d, a, b be a nearly uniform class which is mono-tone and numerically effective. Since F is monotone, we have a � b, and

Žsince F is numerically effective, we have F � E � 0, F � 3L � 2 E � E8 1 2. Ž . Ž� �� E � 0, F � 5L � 2 E � �� 2 E � E � E � 0, and F � 6L7 1 6 7 8

.� 3E � 2 E � �� 2 E � 0; i.e., we have a � b, b � 0, 3d � 8a � 0,1 2 85d � 13a � b � 0, and 6d � 15a � 2b � 0.

It is not hard to check that the rational solution set to these inequalitiesŽ .is the cone � Q given by all nonnegative rational linear combinations of

Ž . Ž . Ž . Ž . Ž .1, 0, 0 , 8, 3, 0 , 8, 3, 1 , 11, 4, 3 , and 17, 6, 6 , since each of these classesŽ . Ž .satisfies all of the inequalities, but a � b for 17, 6, 6 and 1, 0, 0 , b � 0

Ž . Ž . Ž . Ž .for 1, 0, 0 and 8, 3, 0 , 3d � 8a � 0 for 8, 3, 0 and 8, 3, 1 , 5d � 13a �Ž . Ž . Ž .b � 0 for 8, 3, 1 and 11, 4, 3 , and 6d � 15a � 2b � 0 for 11, 4, 3 and

Ž .17, 6, 6 . Thus we see that the cone of monotone, numerically effective,


Ž .nearly uniform classes is just the cone � � � Z of integer lattice points inŽ .� Q .We now show that � is in fact the set of nonnegative Z-linear combina-

Ž . Ž . Ž . Ž . Ž . Ž . Ž .tions of 1, 0, 0 , 8, 3, 0 , 8, 3, 1 , 11, 4, 3 , 17, 6, 6 , 3, 1, 0 , and 3, 1, 1 .² : ² :Let �� denote the cone generated over Z, and let �� denote theQ

Ž .cone generated over Q. It is easy to see that � Q is the union of therational cones

² :� � 11, 4, 3 , 8, 3, 1 , 8, 3, 0 ,Ž . Ž . Ž . Q1

² :� � 11, 4, 3 , 17, 6, 6 , 8, 3, 0Ž . Ž . Ž . Q2

and

² :� � 1, 0, 0 , 17, 6, 6 , 8, 3, 0 .Ž . Ž . Ž . Q3

First, consider � . Since1

11 8 8� 1,4 3 3

3 1 0

every integer lattice point which is a rational linear combination ofŽ . Ž . Ž .11, 4, 3 , 8, 3, 1 , and 8, 3, 0 is in fact a Z-linear combination; i.e.,

²Ž . Ž . Ž .:� � � 11, 4, 3 , 8, 3, 1 , 8, 3, 0 .1Ž . Ž .Suppose that d, a, b � � . Then the integer triple d, a, b is2

Ž . Ž . Ž .� 17, 6, 6 � � 8, 3, 0 � 11, 4, 3 for some nonnegative rational numbers� , � , and , and we have

�1 3 �8 �1�3� 17 8 11 d d� � � .6 3 4 a 2 �5 �2�3 a 6 0 3 b b�6 16 1

Thus 3d � 8a � b�3 � � , 2 d � 5a � 2b�3 � � , and �6d � 16a � b �Ž . , so is an integer and � and � in lowest terms, of course have

denominators of 1 or 3. Thus any element of � � is of the form2²Ž . Ž . Ž .: Ž .A � B, where A 11, 4, 3 , 17, 6, 6 , 8, 3, 0 and B � � � 17, 6, 6 �

Ž . � 4�� 8, 3, 0 with � �, �� 0, 1�3, 2�3 . But the only such B which areŽ .Ž . Ž .Ž . Ž . Ž .integer triples are 1�3 17, 6, 6 � 2�3 8, 3, 0 � 11, 4, 2 � 8, 3, 1 �

Ž . Ž .Ž . Ž .Ž . Ž . Ž .3, 1, 1 and 2�3 17, 6, 6 � 1�3 8, 3, 0 � 14, 5, 4 � 11, 4, 3 �Ž .3, 1, 1 . This argument shows that all elements of � � are contained2

²Ž . Ž . Ž . Ž . Ž .:in 11, 4, 3 , 17, 6, 6 , 8, 3, 0 , 8, 3, 1 , 3, 1, 1 .Ž . Ž . Ž . Ž .Finally, suppose that d, a, b � � 1, 0, 0 � � 8, 3, 0 � 17, 6, 6

� � ; then 6 Z, 3� � 6 Z, and � � 8� � 17 Z. Thus we3may assume that � and are multiples of 1�6 and � is a multiple of 1�3.


²Ž .Thus any element of � � is of the form A � B, where A 1, 0, 0 ,3Ž . Ž .: Ž . Ž . Ž .17, 6, 6 , 8, 3, 0 and B � � � 1, 0, 0 � �� 8, 3, 0 � � 17, 6, 6 with

� 4 � 4� �, � 0, 1�6, . . . , 5�6 and �� 0, 1�3, 2�3 . By direct check, for²Ž . Ž . Ž . Ž .every integer triple B we have B 1, 0, 0 , 17, 6, 6 , 8, 3, 0 , 3, 1, 1 ,

Ž .:3, 1, 0 .Ž . Ž . Ž . Ž . Ž . Ž . Ž .Thus 1, 0, 0 , 3, 1, 0 , 3, 1, 1 , 8, 3, 0 , 8, 3, 1 , 11, 4, 3 , and 17, 6, 6

generate the cone of monotone, numerically effective, nearly uniform2classes on a blow up of P at eight general points.

Ž .We now analyze those classes falling into case iii of Proposition 13.

PROPOSITION 15. Let F be a monotone, numerically effecti�e, nearlyuniform di�isor class such that F � E � for all exceptional cur�es E andE

Ž .F � 5L � 2 E � �� 2 E � E � E � 2. Then either1 6 7 8

Ž . Ž . Ž .a F is 3, 1, 0 � r 8, 3, 1 for some r � 0, in which caseŽ Ž .. Ž Ž ..dim cok � � r and dim ker � � r � 1, orF F

Ž .b � has maximal rank.F

Proof. By Proposition 14, we know that

² :F 1, 0, 0 , 3, 1, 0 , 3, 1, 1 , 8, 3, 0 , 8, 3, 1 , 11, 4, 3 , 17, 6, 6 .Ž . Ž . Ž . Ž . Ž . Ž . Ž .

Ž .Since F � 5L � 2 E � �� 2 E � E � E � 2, we deduce that F must1 6 7 8Ž . Ž .be of the form H � r 8, 3, 1 � s 11, 4, 3 for some nonnegative integers r

Ž . Ž . Ž . Ž . Ž .and s, where H is one of 3, 1, 0 , 6, 2, 2 , 16, 6, 0 , 20, 7, 7 , 25, 9, 6 , orŽ . Ž Ž .34, 12, 12 . Note that we need not consider the possibility H � 11, 4, 1 ,

Ž . Ž .since 11, 4, 1 has been accounted for by taking H � 3, 1, 0 with r � 1. Ž . Ž .and s � 0. If F � H � r 8, 3, 1 � s 11, 4, 3 , for F � E � for all excep-E

Ž .tional curves E see Lemma 3 , it is easy to check that the followingadditional restrictions are necessary:

� Ž . Ž .If H � 6, 2, 2 , then r � 0; hence we can replace H � 6, 2, 2 byŽ . Ž . Ž .H � 6, 2, 2 � 8, 3, 1 � 14, 5, 3 and remove the requirement that r � 0.

� Ž . Ž .If H � 16, 6, 0 , then s � 0, so we replace H � 16, 6, 0 by H �Ž .27, 10, 3 .

� Ž . Ž .If H � 20, 7, 7 , then r � 1, so we replace H � 20, 7, 7 by H �Ž .36, 13, 9 .

� Ž . Ž .If H � 34, 12, 12 , then r � 2, so we replace H � 34, 12, 12 byŽ .H � 58, 21, 15 .

Ž . Ž .Thus F must be of the form H � r 8, 3, 1 � s 11, 4, 3 for some non-Ž . Ž . Ž .negative integers r and s, where H is one of 3, 1, 0 , 14, 5, 3 , 27, 10, 3 ,

Ž . Ž . Ž .36, 13, 9 , 25, 9, 6 , or 58, 21, 15 . We consider each possibility for H inŽ .turn, beginning with H � 3, 1, 0 .


Ž . Ž . Ž .So F � H � r 8, 3, 1 � s 11, 4, 3 , with H � 3, 1, 0 . We first considerŽ .the case where s � 0. Note that cok � � 0 by Theorem 11. NowH

Ž . Ž .F � 8, 3, 1 � 3, so for r � 1 we have the exact sequence 0 � OO F � DXŽ . Ž . Ž .� OO F � OO 3 � 0 with D � 8, 3, 1 . By Lemma 12 and an easyX D

calculation, � has a one-dimensional kernel and cokernel. Applying3, DŽ .Proposition 7 to the foregoing exact sequence with r � 1 and s � 0 , we

Ž . Ž .see that the induced map ker � � ker � is surjective. Since theF 3, DŽ . 0Ž .restriction of OO D to D is trivial, the image of the map H X, H � rDX

0Ž Ž .. 0Ž . 0Ž .� H D, OO 3 , and hence of H X , H � rD H X , L �D0Ž Ž .. 0Ž . Ž .H D, OO 3 H X, L is the same for all r � 1. Thus ker � �D FŽ .ker � is surjective for all r � 1. We therefore see that the exact3, D

sequence0 � ker � � ker � � ker �F� D F 3, D

� cok � � cok � � cok � � 0F� D F 3, D

coming from Proposition 7 is exact separately on kernels and cokernels. ItŽ Ž ..now follows for s � 0 and all r � 0 that dim ker � � r � 1 andF

Ž Ž .. Ž .dim cok � � r, as claimed in part a .FŽ . Ž .Now assume that F � H � s 11, 4, 3 ; we find that F � 11, 4, 3 � 5, so

Ž . Ž . Ž .we have the exact sequence 0 � OO F � D � OO F � OO 5 � 0,X X DŽ .where this time we take D � 11, 4, 3 . By Lemma 12, � has maximal5, D

rank, and one easily checks that � is thus surjective. Applying Proposi-5, DŽ .tion 7 and inducting on s, we see that cok � � 0.F

Ž . Ž .Finally, consider F � H � s 11, 4, 3 � r 8, 3, 1 with s � 0. We find thatŽ . Ž .F � 8, 3, 1 � 3 � s, so we have the exact sequence 0 � OO F � D �XŽ . Ž . Ž .OO F � OO 3 � s � 0, where this time we take D � 8, 3, 1 . By LemmaX D

12, � is surjective. Applying Proposition 7 and inducting on r, we see3�s, DŽ .that cok � � 0 for all r � 0 and s � 0.F

Ž . Ž . Ž . Ž .Now let H � 14, 5, 3 . Note that 14, 5, 3 � 3, 1, 0 � 11, 4, 3 . ThusŽ . Ž . Ž . Ž . Ž .Ž .F � H � r 8, 3, 1 � s 11, 4, 3 is just 3, 1, 0 � r 8, 3, 1 � s � 1 11, 4, 3 ,

Ž . Ž .and our preceding analysis shows that cok � � 0 for F � 3, 1, 0 �FŽ . Ž .Ž .r 8, 3, 1 � s � 1 11, 4, 3 .

The remaining possibilities for H reduce in a similar way to the caseŽ . Ž . Ž . Ž . Ž .H � 3, 1, 0 treated earlier; 27, 10, 3 is 3, 1, 0 � 3 8, 3, 1 , 36, 13, 9 is

Ž . Ž . Ž . Ž . Ž . Ž .3, 1, 0 � 3 11, 4, 3 , 25, 9, 6 is 3, 1, 0 � 2 11, 4, 3 , and 58, 21, 15 isŽ . Ž .3, 1, 0 � 5 11, 4, 3 . In each instance, the reader will easily verify that

Ž . Ž .either a or b of Proposition 15 is obtained.

4. EXAMPLES

In this section we show by example how our results give minimal freeresolutions for fat point subschemes Z � m p � �� m p of P 2 with1 1 n n


n � 8, where the points p are assumed to be general. We also give twoiexamples for n � 8 general points on a smooth plane cubic curve, oneshowing that our results sometimes determine resolutions even thoughn � 8 and one showing that sometimes they do not.

If we are interested in a resolution of I for Z � m p � �� m pZ 1 1 n nwith n � 8, then we might as well assume n � 8 and simply set somemultiplicities m equal to 0; i.e., having n � 8 is no different from havingi

Ž .n � 8. Now, for our first example, consider Z � 54 p � �� p . Recall1 8Ž . Ž . ŽŽ . .that h t is the Hilbert function of I in degree t; thus h t � dim I .Z Z Z Z t

Let X be the surface obtained by blowing up the points p , with theicorresponding exceptional configuration being L, E , . . . , E . Note that1 8

Ž .D � 17L � 6 E � �� E is numerically effective by Lemma 4. Denote1 8Ž .tL � 54 E � �� E by H . Since H � D � 17t � 54 � 6 � 8 is negative1 8 t t

0Ž .for t � 153, we see that h X, H � 0 for t � 153. Since H � E � 0 for allt texceptional curves E when t � 153, we know that H is numericallyt

1Ž . 0Ž .effective for all t � 153, and hence that h X, H � 0 and h X, H �t t2 t � 2 54 � 1ŽŽ . . Ž . Ž .H � K � H �2 � 1 � � 8 for t � 153. In other words,t X t 2 2

t � 2 54 � 1Ž . Ž . Ž . Ž .h t � 0 for t � 153, and h t � � 8 for t � 153.Z Z 2 2

Ž .By vanishing of h t for t � 153, we see that � � 0 for t � 153, and byZ t1Ž .vanishing of h X, H for t � 153 and by the general fact at the end oft

Example 8 we see that � � 0 for all t � 153. Since it is obvious thatt�2Ž .� � h 153 � 55, we are left with finding � . But H � E � 0 � 153 Z 154 153 E

for E � 6L � 3E � 2 E � �� 2 E , and likewise for E � 6L � 2 E �1 2 8 1Ž .3E � 2 E � �� 2 E , . . . , so by Theorem 1 b , � has the same2 4 8 H153

Žkernel as the maps corresponding to H � 6 L � 3E � 2 E153 1 2. Ž . Ž� �� 2 E , H � 6L � 3E � 2 E � �� 2 E � 6L � 2 E � 3E8 153 1 2 8 1 2

.� 2 E � �� 2 E , and so on, and we find eventually that � has the4 8 H153Ž .same kernel as � with H � 9L � 3E � �� 3E . By Theorem 1 a , �H 1 8 H

0Ž . 0Ž .has maximal rank, and since h X, H � 7 and h X, H � L � 18 byLemma 4, we see that � must be surjective with a three-dimensionalH

0Ž .kernel. Thus � has a three-dimensional kernel, and from h X, HH 1531530Ž . Ž� 55 and h X, H � 210, we see that � equals 48 rather than the154 154

.maximal rank value of 45 .3 Ž .Using the relation � � s � � h t , we find that s is 0 for t � 154 ort t Z t

t � 155, s � 3 and s � 99. Thus the minimal free resolution of I is154 155 Z55� � 48� �0 � F � F � I � 0, where F � R �153 � R �154 and F �1 0 Z 0 1

3� � 99� �R �154 � R �155 .Consider now another example. Suppose that Z � mp � �� mp for1 n

n � 9 general points of a smooth cubic curve in P 2. We show that ourŽresults recover the resolution of I , which is known in this case seeZ

� �. Ž .Section 3.2.1 of H7 . So let H � tL � m E � �� E with m � 0 andt 1 nlet D as usual be a smooth section of �K , where X is obtained fromX X


2 Ž .P by blowing up the points p . If t � 3m, then H � �K � 0, soi t X0Ž . 0Ž . Ž . Ž .h X, H � h X, H � K , but we still have H � K � �K � 0,t t X t X X

0Ž . 0Ž .and iterating, we eventually find that h X, H � h X, H � mK � 0.t t XŽ Ž . . Ž .The last equality follows since L � H � mK � 0. Thus h t � 0 fort X Z

Ž .t � 3m, and hence � Z has maximal rank for t � 3m. For t � 3m,t� � � 4 Ž . Ž .H � mD , so � Z has maximal rank. For t � 3m, H � t � 3m Lt X t t

Ž .� mK , so E � H � t � 3m E � L � mE � K � E � L � for every ex-X t X EŽ . Ž Ž .ceptional curve E and either H � �K � 2 � in which case � Zt X D tXŽ .. Ž . Žhas maximal rank by Theorem 1 a or H � �K � 2 � in whicht X D X

case � and � have kernels of the same dimension by TheoremH H �Kt t XŽ .1 b , and iterating, for some l we eventually obtain H � lK falling undert X

Ž . .case a of Theorem 1 . Thus in any case we can compute the rank of �H t

for every t, which makes it easy to work out the resolution for anyparticular m and n.

Ž .Finally, consider Z � 156 p � 121 p � �� p � 104 p � 78 p .1 2 7 8 9Ž Ž . .Again, let H � tL � 156E � 108 E � �� E � 104E � 78 E . Itt 1 2 7 8 9

� Ž . �turns out that 27L � 12 E � 9 E � �� E � 8 E � 6E is the class1 2 7 8 9� � � � �Ž . �E of an exceptional curve E, and that H � t � 351 L � 13E . Fromt

Ž 0Ž .this it is easy to check that � is injective for t � 351 since h X, H � 0H tt0Ž . . Ž 1Ž .for t � 351, while h X, H � 1 and by Example 8, since h X, H � 0351 t

.for t � 351 is surjective for t � 352. However, for t � 352, we haveH � C � for all C � except C � E, for which we have � 12 �t C X EH � E � 14 � 15 � , and hence Theorem 1 does not apply and indeedt Ethe rank of � is not known.H t

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Resolutions of Fat Point Ideals Involving Eight General Points of P2