JMATH Journal of Mathematics 2314-4785 2314-4629 Hindawi Publishing Corporation 10.1155/2015/431098 431098 Research Article On a Theorem of Ziv Ran concerning Abelian Varieties Which Are Product of Jacobians Anghel Cristian Buruiana Nicolae Kazmi Kaleem R. Department of Mathematics Institute of Mathematics of the Romanian Academy Calea Grivitei No. 21, 010702 Bucuresti Romania imar.ro 2015 1102015 2015 21 07 2015 21 09 2015 1102015 2015 Copyright © 2015 Cristian Anghel and Nicolae Buruiana. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We give a new proof for a theorem of Ziv Ran which generalizes some results of Matsusaka and Hoyt. These results provide criteria for an Abelian variety to be a Jacobian or a product of Jacobians. The advantage of our method is that it works in arbitrary characteristic.

1. Introduction

In the classical papers  (resp. ), Matsusaka and Hoyt gave a necessary and sufficient criterion for an Abelian variety A for being a Jacobian, respectively, a product of Jacobians. In , Ran reconsiders the subject and gives a more general and probably more natural criterion for this. However, his method seems unsatisfactory in positive characteristic.

The aim of this paper is to reprove Ran’s criterion, using results from  on the ring of the numerical algebraic cycles on A. For the particular case of the Ran-Matsusaka criterion, another proof appeared in . Both proofs are independent of the characteristic of the base field.

In the sequel, for an Abelian variety A, we denote by Z(A) the Q-vector space of the algebraic cycles with Q-coefficients on A and by N(A) the quotient by numerical equivalence. It is well known (cf.  or ) that on N(A) there are two Q-algebra structures given by the usual product and by the Pontrjagin product.

Throughout this paper, the latter will be very useful thanks to its geometric definition and to the fact that it gives a ring structure not only on N(A) but also on Z(A). Below, for x,yN(A), we will denote the usual product by x·y and the Pontrjagin product by xy.

Also, for the two subvarieties X1 and X2 in A, we denote by X1X2 and X1X2 their sum and difference in the group law of A, to avoid confusion with the corresponding operations on the cycles. In the current paper, the algebraic cycles will often be divisors and 1-cycles (the latter ones being formal sums of curves) and they will always have integer coefficients.

The term curve, is reserved for integral ones, and all 1-cycles will be considered effective. Finally, a prime cycle is an irreducible subvariety of A of a corresponding codimension.

2. Generating Curves on an Abelian Variety

Let A be an Abelian variety of dimension n and E be a curve on it containing the origin 0A of A. We consider a sequence of closed subsets in A, defined as follows: (1)E0=0AEi=EEEi  terms,   with  1in,En+1=A.It is clear that this sequence is increasing, dimEii and dimEi+1dimEi+1 for every i. As long as E is a curve, Ei is irreducible and there is a first index j such that Ej=Ej+1. Also, we have Ei=Ej for all ij and dimEj=j. It follows that Ej is stable for the group law on A and the induced operation has 0A as unity. Using a result of Ramanujan in ([7, chapter II, Section 4]), Ej is an Abelian subvariety of A and the points of E generate the group Ej. We denote by E the subvariety Ej. If j=n, then E=A and we say that E is a generating curve on A.

Remark 1.

In  Matsusaka proves that every Abelian variety has a generating curve. Moreover, from his proof, for a projective embedding of A, every linear section with a convenient linear subspace of appropriate dimension which contains 0A is a generating curve for A.

Using the Pontrjagin product (for cycles, not for numerical classes) it is easy to deduce the following useful fact.

Lemma 2.

Let E be a curve in A with 0AE, E the subvariety of A generated by E and j=dimE. Then, j is the maximal number i such that Ei (=EEE with i terms) is nonzero and E is the support of the cycle Ej.

Let us now consider a curve EA which does not necessarily contain the origin. It easily follows that for xE the Abelian variety generated by E{x} does not depend on x; it is in fact the subgroup of A generated by EE. This Abelian variety will also be denoted by E. If Z=m1E1++mkEk is an effective 1-cycle, we denote by Z the Abelian subvariety given by E1E2Ek and we will say that Z is a generating 1-cycle for Z.

Remark 3.

From the definition above, we see that the construction of Z is independent of the numbers mi. In particular, Z and Zred generate the same subvariety and also for Z and mZ.

The next lemma will be useful in the sequel.

Lemma 4.

(a) For Y,EA with Y subvariety and E, a curve, both containing the origin, if YE=0 then EY.

(b) For Y1,Y2A Abelian subvarieties and m,n nonzero integers, if mY1 and nY2 are numerically equivalent, then Y1=Y2.

Proof.

(a) We have dimYdimYE1+dimY and from YE=0 we deduce that dim(YE)<1+dimY. So Y=YE and because 0AY it follows EY.

(b) Let E1 be a generating curve for Y1 (it exists cf. Remark 1). Then mY1E1=m(Y1E1)=0 and so nY2E1=n(Y2E1)=0. It follows that Y2E1=0, and from the first point E1Y2. The last inclusion implies that Y1=E1Y2. In the same way the reversed inclusion is proved.

Remark 5.

The point (b) above, in the case m=n=1, is a result of Matsusaka in .

Proposition 6.

(a) For two curves E1,E2A which are numerically equivalent, we have E1=E2.

(b) For a curve E and Z, a 1-cycle which is numerically equivalent with E, we have Z=E.

(c) Let D be an ample divisor and Z a 1-cycle which is numerically equivalent with Dn-1. Then Z is a generating 1-cycle for A.

Proof.

(a) Using convenient translations we can suppose that E1 and E2 contain 0A. Let r, s be the dimensions of Yi=Ei for i=1,2. Using Lemma 2 and the fact that E1a is numerically equivalent with E2a for all positive integers a, we find r=s. From Lemma 2 again, E1r is a multiple of both Y1 and Y2, and the conclusion follows from Lemma 4(b).

(b) As in (a), denoting Z=m1E1++mkEk we can suppose that E and all Ei contain 0A. By Lemma 2, for r=dimE, we have E(r+1)=0. Now, Zr being numeric equivalent with Er, we find ZrE=0. But again from Lemma 2 we find a nonzero term in the development of Zr. With Lemma 4(a), this term which is in fact a subvariety contains E, because all terms in Zr are vanished by the Pontrjagin product with E. On the other hand, this term is contained in Z and so EZ.

For the reverse inclusion, we consider the development of the left side of the equality ZEr=0. From Lemma 2Er=nE with n1 is an integer. We find m1nE1E++mknEkE=0; therefore EiE=0 for all i and then from Lemma 4(a), EiE, so ZE.

(c) Let m be a positive integer with mD very ample. The cycles mn-1Dn-1 and mn-1Z are therefore numerical equivalents and there will exist an integer curve E in the same numeric class with mn-1Dn-1 and so with mn-1Z. From (b) we have E=mn-1Z=Z. But from Remark 1  E=A, so Z is a generating 1-cycle.

The point (c) above is a slight generalization of the result from Remark 1 and will be used to deduce the Matsusaka-Hoyt criterion from that of Ran.

3. Algebraic Cycles Constructed from Generating Curves

We recall a result from  which will be the main tool in the proof of Ran’s theorem. Let E be a generating curve of the n-dimensional Abelian variety A. We consider on A the following cycles: Wn(E)={0A} and Wi(E)=1/(n-i)!E(n-i) for 0in-1. From the definition of the Pontrjagin product, Wi(E) is a cycle with irreducible support of codimension i on A. In particular W1(E) is a divisor and there exists αEQ such that W0(E)=αE·1A, where 1A is the fundamental cycle on A.

The result we need from  is the following.

Proposition 7.

All cycles Wi(E) have integer coefficients and in particular αEZ, being evidently positive. Also, W1(E)i=i!αEi-1Wi(E) for 1in. In particular, W1(E)n>0 and so W1(E) is ample.

Remark 8.

For E a smooth curve and A its Jacobian, these divisors are well-known.

A first application of the proposition above is point (b) in the following.

Proposition 9.

(a) Let D be an effective divisor and Z a generating 1-cycle. Then D·Z>0.

(b) If moreover D is ample, then D·Zn=dimA.

Proof.

(a) We can suppose that D is a prime divisor. Let E1,,Ek be the components of Z. We have D·Ei0 for all i, because the general translation of Ei cuts properly D. It is therefore sufficient to find an i such that D·Ei>0. Suppose there is no such i. Then using a result from [7, chapter 2, Section 6], translations with elements of the form {x}{y} with x,yEi leave D invariant. But Z is a generating 1-cycle, and therefore every element in A is a sum of elements of this form. So D is invariant with respect to any translation and then numerically equivalent with 0, in contradiction with its effectiveness.

(b) Consider a first case where Z=E is a prime cycle (i.e., E is a curve) and without loss of generality 0AE. Let t be a variable and consider the polynomial P(t)=(t·W1(E)+D)n=n!αEn-1tn+n!αEn-2(D·E)tn-1++Dn. Because W1(D) is ample and D is nondegenerate, the index theorem for Abelian varieties compare  asserts that all roots of P are real and negative. So the means inequality gives D·En(χ(OA(D))·αE)1/nn.

For the general case, let Z=m1E1++mkEk with all mi>0, Xi=Ei, and Di be the restriction of D to Xi. The projection formula gives D·Ei=Di·EidimXi from the particular case above. So, D·Z=imiD·EiiD·EiidimXidimA=n, because Z is a generating 1-cycle.

The following consequence of the above proposition will be useful in the last part of the paper.

Corollary 10.

Let A be an Abelian variety, D=i=1rmiDi an ample effective divisor, and Z=i=1snjEj a generating 1-cycle of A (the coefficients are supposed to be nonzero). If D·Z=n=dimA then mi=nj=1 for all i,j.

Proof.

We have n=D·Z=i=1rmiDi·Zi=1rDi·Zn because i=1rDi is ample and one can apply Proposition 9(b). So miDi·Z=Di·Z and because the last term is nonzero by Proposition 9(b), we find mi=1 for all i.

In the same way, n=D·Z=i=1snjD·Eji=1sD·Ej=D·i=1sEjn because i=1sEj remains a generator 1-cycle by Remark 3. So njD·Ej=D·Ej and D being ample, the last term is positive. It results that nj=1 for all j.

We can now prove the following result, which is nothing else but Ran’s version of the Matsusaka theorem.

Theorem 11.

Let D be an ample divisor on the Abelian variety A and let E be a generating curve such that D·E=n=dimA. Then E is smooth, A is its Jacobian, and D is a translation of W1(E).

Proof.

In the proof of point (b) from Proposition 9 we have the inequality D·En(χ(OA(D))·αE)1/nn. If D·E=n we will have χ(OA(D))=αE=1 and so Dn=n!. In this case the polynomial P(t) from the same proposition becomes P(t)=n!tn+n!·n·tn-1++n!. It follows that the arithmetic and geometric means of the roots coincide and so all the roots have the form -λ for a positive value of λ. So P(t)=n!(t+λ)n and by identification, λn=1. It follows that λ=1 and then W1(E)n-1·D=W1(E)n-2·D2=n!. These relations imply that (D-W1(E))·W1(E)n-1=(D-W1(E))2·W1(E)n-2=0. The Hodge index theorem asserts that D is numerically equivalent with W1(E), and because W1(E) is a principal polarization (from Proposition 7 and equality αE=1), one may deduce that D is a translation of W1(E).

Consider the normalization f0:TE for E, and let f:JA be a prolongation of f0, where J is a Jacobian of T. If we choose a base point in the construction of J, one on T which sits above 0AE, f will be a morphism of Abelian varieties, sending origin to origin. Also, f is surjective because E is generating for A and for g= genus of T we have gn.

Let us denote by Wi=Wi(T) the canonical cycles on the Jacobian J. Therefore f(Wg-i)=Wn-i(E) for 1in: for i=1 this is clear because Wg-1=T and for i>1 it is a consequence of the definitions for Wg-i and Wg-i(E) and also from the fact that f commute with the Pontrjagin product. In particular f(Wg-n)=W0(E)=αE·1A and so αE=1 is the degree of the restriction of f to Wg-n. Therefore this restriction is a birational morphism and has an inverse: A---Wg-n. This inverse, considered as a rational map from A to J can be extended over all the A giving a morphism AJ compare . As a consequence, the restriction g of f to Wg-n will be an isomorphism and Wg-n will be an Abelian subvariety of J. But Wg-n contains Wg-1=T which generates J and so Wg-n=J. In this case we have g=n and f is birational from J to A hence an isomorphism.

4. Proof of Ran Theorem

The purpose of this section is to give a proof for Ran’s full theorem. Some points are as in  and are included only for the sake of completeness. The modifications appear from the replacement of Lemma II.8 from  with the result below whose proof is very simple.

Lemma 12.

Let D be a prime divisor on an Abelian variety A. Then, there exists an Abelian variety B, a surjective morphism of Abelian varieties f:AB and an ample divisor F on B such that f-1(F)=D as schemes.

Proof.

We consider the closed subgroup K of A defined by K:={xA{x}D=D} and the Abelian subvariety K0 of A which is the connected component of 0A in K. We denote by B the quotient A/K0 and by f:AB the quotient morphism. Finally we denote by F the closed irreducible subset f(D) with the reduced structure. We easily find dimF=dimB-1, so F is a divisor on B and set theoretically f-1(F)=DK0=D because K0K. Let xA such that {f(x)}F=F. Applying f-1 we find {x}DK0=DK0, and because DK0=D we find {x}D=D and so xK. Therefore, the elements in B which leave F invariant by translations are from f(K). They are then in a finite number, because the index [K:K0] is finite. So F is an ample divisor on B. Finally the equality f-1(F)=D also holds at the schemes level, because f is smooth from its construction.

The result we are interested in is the following theorem of Ran.

Theorem 13.

Let A be an Abelian variety of dimension n, D=i=1rmiDi an ample effective divisor, and Z=j=1snjEj a generating 1-cycle such that D·Z=n. Then mi=nj=1 for all i,j, r=s and there are r smooth curves T1,,Tr with Jacobians B1,,Br and an isomorphism of Abelian varieties h:B1××BrA such that, for every i, Ei is a translation of h({0}××{0}×Ti×{0}××{0}) (Ti on the ith place) and Di is a translation of h(B1××Bi-1×Wi×Bi+1××Br), where Wi is the canonical divisor W1(Ti) on Bi.

Proof.

The fact that mi=nj=1 for all i,j is Corollary 10. For the other points, the proof follows closely the one from  with some modifications of the arguments. We began with three preliminary steps.

Step 1. We prove that for every j there is a unique i such that Di·Ej0. We translate the curves Ej such that they contain the origin and denote the result with the same letter. Let Aj=Ej and dj=dimAj, so that Ej is a generating curve for Aj. Denote by ej the inclusion AjA and by the same letter D a translation of the divisor D which has a proper intersection with every Aj. Therefore, ej(D):=Dj is defined as a cycle and is an ample divisor on Aj. The projection formula gives (2)D·Ej=Dj·Ejand so (3)n=D·E=j=1sD·Ej=j=1sDj·Ejj=1sdjn. The first inequality comes from the fact that on Aj one has Dj·Ejdj according to Proposition 9(b), and the last one is due to the fact that Z is a generating 1-cycle. So Dj·Ej=dj, and Ej being a generating curve for Aj, from Theorem 11 one finds that Ej is smooth, Aj is its Jacobian, and Dj is a translation of the canonical divisor on Aj; so Dj is prime as any divisor numeric equivalent with it (it is a principal polarization).

Let us fix j, and consider for any i a translation of Di which cuts Aj properly. Every such translation, also denoted by Di, restricted to Aj either is an effective divisor or has an empty intersection with Aj, in which case Di·Ej=0. But the sum of these restrictions is numerically equivalent with Dj and so there cannot be two indexes i with Di·Ej0, because in such a case Dj which is prime would be the sum of two effective divisors. The existence of an i with Di·Ej0 comes from the fact that D is ample.

Step 2. This part consists in the proof of the following fact: for an n-dimensional Abelian variety A, a prime ample divisor D, and a generating 1-cycle Z=T1++Ts with D·Z=n one has s=1 (i.e., Z is ireducible and reduced).

The proof is due to Ran compare Lemma III.2 from . Denote by A1=T1. From the first step, we know that A1 is in fact the Jacobian of the smooth curve T1; in particular it is principally polarized and isomorphic with its dual. It will suffice to prove that A1=A, because in this case T1 will be a generating curve, and the fact that D is ample together with the inequalities nD·T1D·Z=n implies that s=1 as desired.

For the time being, we replace D with a translation whose restriction D|A1:=D1 is well defined as divisor on A1. As in the proof of Step 1, D1 is numerically equivalent with W1(T1). Let s:A×A1A the morphism given by s(r,y)=r+y and p,p1 be the projections. Consider on A×A1 the line bundle M=s(OA(D))p1(OA1(-D1)) and on A1×A1 the line bundle P=(s|A1×A1)(OA1(D1))q1(OA1(-D1))q2(OA1(-D1)), where q1,q2 are the projections on the factors of A1×A1. Using the fact that A1 is a Jacobian (and therefore it is its own Picard variety whith the Poincare bundle equal to P), we deduce the existence of a morphism f:AA1 and of a line bundle N on A such that (4)MpNf×IdA1P.Restricting (4) on the fiber {x}×A1, for xA, one finds an isomorphism (5)etxOADtfxOA1D1,where e is the embedding A1A. Because D1 is a principal polarisation, the point f(x) is uniquelly defined by the above property, which can be written in divisorial terms as ({-x}D)|A1={-f(x)}D1, at least for x general such that the divisor ({-x}+D)|A1 is well defined. From this one deduces that points in A1 are fixed by f and so f is surjective with KA1={0A}, where K is the kernel of f.

Because K cuts A1 only in 0A, the sum morphism K×A1A is injective and so we will have dim(KD1)=n-dimA1+dimA1-1=n-1. Now, for a general xA, we have {-f(x)}D1=({-x}D)|A1{-x}D. So, x-fxTran(D1,D){yA{y}D1D}. But Tran(D1,D) is closed and so for any xA we have {x-f(x)}D1D.

Then for xK, {x}D1D and therefore KD1D. For K0 the connected component of the origin in K, we have K0D1D. But K0D1 is a divisor and D is prime, so the previous inclusion is an equality. Now, (6)K0D=K0K0D1=K0D1=D.But ample D implies that K0 is finite and prime D implies that K0={0A} which is equivalent with A=A1.

Step 3. Within this step we prove that for any i there is a unique j such that Di·Ej0. For this, we consider for all i, an Abelian variety Bi, an ample divisor Fi on Bi, and a surjective morphism fi:ABi such that fi-1(Fi)=Di. Their existence follows from Lemma 12.

We have (7)n=D·Z=i=1rfi-1Fi·Z=i=1rFi·fiZi=1rli,where li=dimBi and the last inequality is from Proposition 9(b). We examine the last sum using the effective construction of Bi’s from Lemma 12. There, Bi is of the form A/Ki where Ki is an Abelian subvariety of A. As consequence, li=codimKi and so (8)i=1rli=i=1rcodimKicodimKi=n (by definition of Ki and the ampleness of D, the intersection Ki is finite).

It results in that (9)ni=1rFi·fiZi=1rlin,and so Fi·(fi)Z=li. But Fi is a prime divisor and from Step 2 there is a unique ji with (fi)Eji a curve. All the other curves from the support of Z will be therefore contracted. We now fix i and compute Di·Ej=fi-1(Fi)·Ej=Fi·fiEj. This last number is 0 if jji and nonzero for j=ji because Fi is ample. This conclude the third step.

From the first and third steps we find that iji is a bijection and so r=s. Also one can reorder the curves Ej (such that Eji will be numbered by Ei) and so we can suppose that for all i,j we have Di·Ej0i=j. To conclude the proof, we consider all the requirements supposed above.

In the first place we review Bi’s. Let Ti be the cycle (fi)Z. From the third step, Ti is in fact a curve, namely, fi(Ei). Also we have seen that Fi·Ti=li=dimBi and therefore Theorem 11 implies that Bi is the Jacobian of Ti. To see this, we need only to prove that Ti is a generating curve of Bi and this is implied by the fact that, as we have seen, fi contracts all the curves Ej for ji and as far as these contain 0A, the contraction will be to 0Bi. So fi(A)=fi(Ai)=Bi and because Ei generates Ai, Ti generates Bi. So, by Theorem 11, Fi is a translation of the canonical divisor on Bi.

Let us recall that in the first step we supposed (using appropriate translations) that all Di’s cut properly the subvarieties Aj’s, which means that either ej(Di) is an effective divisor on Aj or DiAj is empty, in which case ej(Di)=0. The former case can happen only for j=i, because in this situation ej(Di)·Ej0 (more precisely, the projection formula gives ej(Di)·Ej=Di·ejEj=Di·Ej). So ej(Di)0j=i and we have (10)Dj=ejD=ejDj=ejfjFj=fjejFj.Let us consider the morphism fjej:AjBj. It sends the generating curve Ej of Aj on the generating curve Tj of Bj, and therefore it is surjective; so djlj. But, from the first and third steps, n=j=1rdj=j=1rlj; this implies that fjej has a finite nonzero degree. On the other hand fjej pull back the principal polarization Fj from Bj to the principal polarization Dj on Aj. So its degree is 1 and it is an isomorphism with inverse denoted by gj.

Let h:B1××BrA be defined by h(b1,,br)=i=1rgi(bi) and g:AB1××Br be defined by g(a)=(f1(a),,fr(a)). Then hg is the identity, being the identity on every Ai. Also, gh is the identity, being the identity on every {0B1}××Bi××{0Br}.

So h is an isomorphism, Bi is the Jacobian of Ti, and the last part of the theorem concerning the form of the divisors Di and curves Ti is obvious due to the fact that the transformations of Di and Ti were translations.

Finally, we formulate the following corollary which is the result of Hoyt from .

Corollary 14.

Let A be an Abelian variety, D an ample divisor with Dn=n!, and Z a 1-cycle such that Dn-1 is numerically equivalent with (n-1)!Z. Then the conclusion of Theorem 13 holds true.

Proof.

We have Dn=(n-1)!D·Z·, so D·Z=n. On the other hand, from Proposition 6(c), (n-1)!Z is a generating 1-cycle and therefore Z is a generating 1-cycle. Now everything is a consequence of Theorem 13.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

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