# On a Theorem of Ziv Ran concerning Abelian Varieties Which Are Product of Jacobians

Authors: Cristian Anghel; Nicolae Buruiana
Journal: Journal of Mathematics
 (2015)

---

## Abstract

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.


---

## Body

## 1. Introduction

In the classical papers [1] (resp. [2]), Matsusaka and Hoyt gave a necessary and sufficient criterion for an Abelian variety A for being a Jacobian, respectively, a product of Jacobians. In [3], 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 [4] on the ring of the numerical algebraic cycles on A. For the particular case of the Ran-Matsusaka criterion, another proof appeared in [5]. Both proofs are independent of the characteristic of the base field.In the sequel, for an Abelian varietyA, 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. [6] or [4]) 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 onN(A) but also on Z(A). Below, for x,y∈N(A), we will denote the usual product by x·y and the Pontrjagin product by x⋆y.Also, for the two subvarietiesX1 and X2 in A, we denote by X1∔X2 and X1∸X2 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 all1-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

LetA 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=E∔E∔⋯∔Eiterms,   with1≤i≤n,En+1=A.It is clear that this sequence is increasing, dim⁡Ei≤i and dim⁡Ei+1≤dim⁡Ei+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 i≥j and dim⁡Ej=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 [8] 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.
LetE be a curve in A with 0A∈E, 〈E〉 the subvariety of A generated by E and j=dim⁡〈E〉. Then, j is the maximal number i such that E∗i (=E∗E∗⋯∗E with i terms) is nonzero and 〈E〉 is the support of the cycle E∗j.Let us now consider a curveE⊂A which does not necessarily contain the origin. It easily follows that for x∈E the Abelian variety generated by E∸{x} does not depend on x; it is in fact the subgroup of A generated by E∸E. 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 〈E1〉∔〈E2〉∔⋯∔〈Ek〉 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) ForY,E⊂A with Y subvariety and E, a curve, both containing the origin, if Y∗E=0 then E⊂Y.
(b) ForY1,Y2⊂A Abelian subvarieties and m,n nonzero integers, if mY1 and nY2 are numerically equivalent, then Y1=Y2.Proof.
(a) We havedim⁡Y≤dim⁡Y∔E≤1+dim⁡Y and from Y∗E=0 we deduce that dim⁡(Y∔E)<1+dim⁡Y. So Y=Y∔E and because 0A∈Y it follows E⊂Y.
(b) LetE1 be a generating curve for Y1 (it exists cf. Remark 1). Then mY1∗E1=m(Y1∗E1)=0 and so nY2∗E1=n(Y2∗E1)=0. It follows that Y2∗E1=0, and from the first point E1⊂Y2. The last inclusion implies that Y1=〈E1〉⊂Y2. In the same way the reversed inclusion is proved.Remark 5.
The point (b) above, in the casem=n=1, is a result of Matsusaka in [8].Proposition 6.
(a) For two curvesE1,E2⊂A which are numerically equivalent, we have 〈E1〉=〈E2〉.
(b) For a curveE and Z, a 1-cycle which is numerically equivalent with E, we have 〈Z〉=〈E〉.
(c) LetD 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 thatE1 and E2 contain 0A. Let r, s be the dimensions of Yi=〈Ei〉 for i=1,2. Using Lemma 2 and the fact that E1∗a is numerically equivalent with E2∗a for all positive integers a, we find r=s. From Lemma 2 again, E1∗r is a multiple of both Y1 and Y2, and the conclusion follows from Lemma 4(b).
(b) As in (a), denotingZ=m1E1+⋯+mkEk we can suppose that E and all Ei contain 0A. By Lemma 2, for r=dim⁡〈E〉, we have E∗(r+1)=0. Now, Z∗r being numeric equivalent with E∗r, we find Z∗r∗E=0. But again from Lemma 2 we find a nonzero term in the development of Z∗r. With Lemma 4(a), this term which is in fact a subvariety contains E, because all terms in Z∗r are vanished by the Pontrjagin product with E. On the other hand, this term is contained in 〈Z〉 and so 〈E〉⊂〈Z〉.
For the reverse inclusion, we consider the development of the left side of the equalityZ∗E∗r=0. From Lemma 2  E∗r=n〈E〉 with n≥1 is an integer. We find m1nE1∗〈E〉+⋯+mknEk∗〈E〉=0; therefore Ei∗〈E〉=0 for all i and then from Lemma 4(a), Ei⊂〈E〉, so 〈Z〉⊂〈E〉.
(c) Letm 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 Remark1 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 [4] 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 0≤i≤n-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 αE∈Q such that W0(E)=αE·1A, where 1A is the fundamental cycle on A.The result we need from [4] is the following.Proposition 7.
All cyclesWi(E) have integer coefficients and in particular αE∈Z, being evidently positive. Also, W1(E)i=i!αEi-1Wi(E) for 1≤i≤n. In particular, W1(E)n>0 and so W1(E) is ample.Remark 8.
ForE 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) LetD be an effective divisor and Z a generating 1-cycle. Then D·Z>0.
(b) If moreoverD is ample, then D·Z≥n=dim⁡A.Proof.
(a) We can suppose thatD is a prime divisor. Let E1,…,Ek be the components of Z. We have D·Ei≥0 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,y∈Ei 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 whereZ=E is a prime cycle (i.e., E is a curve) and without loss of generality 0A∈E. 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 [7] asserts that all roots of P are real and negative. So the means inequality gives D·E≥n(χ(OA(D))·αE)1/n≥n.
For the general case, letZ=m1E1+⋯+mkEk with all mi>0, Xi=〈Ei〉, and Di be the restriction of D to Xi. The projection formula gives D·Ei=Di·Ei≥dim⁡Xi from the particular case above. So, D·Z=∑imiD·Ei≥∑iD·Ei≥∑idim⁡Xi≥dim⁡A=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.
LetA 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=dim⁡A then mi=nj=1 for all i,j.Proof.
We haven=D·Z=∑i=1rmiDi·Z≥∑i=1rDi·Z≥n 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·Ej≥∑i=1sD·Ej=D·∑i=1sEj≥n 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.
LetD be an ample divisor on the Abelian variety A and let E be a generating curve such that D·E=n=dim⁡A. Then E is smooth, A is its Jacobian, and D is a translation of W1(E).Proof.
In the proof of point (b) from Proposition9 we have the inequality D·E≥n(χ(OA(D))·αE)1/n≥n. 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 normalizationf0:T→E for E, and let f:J→A 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 0A∈E, 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 g≥n.
Let us denote byWi=Wi(T) the canonical cycles on the Jacobian J. Therefore f∗(Wg-i)=Wn-i(E) for 1≤i≤n: 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 A→J compare [7]. 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 [3] and are included only for the sake of completeness. The modifications appear from the replacement of Lemma II.8 from [3] with the result below whose proof is very simple.Lemma 12.
LetD be a prime divisor on an Abelian variety A. Then, there exists an Abelian variety B, a surjective morphism of Abelian varieties f:A→B and an ample divisor F on B such that f-1(F)=D as schemes.Proof.
We consider the closed subgroupK of A defined by K:={x∈A∣{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:A→B the quotient morphism. Finally we denote by F the closed irreducible subset f(D) with the reduced structure. We easily find dim⁡F=dim⁡B-1, so F is a divisor on B and set theoretically f-1(F)=D∔K0=D because K0⊂K. Let x∈A such that {f(x)}∔F=F. Applying f-1 we find {x}∔D∔K0=D∔K0, and because D∔K0=D we find {x}∔D=D and so x∈K. 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.
LetA 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×⋯×Br→A 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 thatmi=nj=1 for all i,j is Corollary 10. For the other points, the proof follows closely the one from [3] 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·Ej≠0. We translate the curves Ej such that they contain the origin and denote the result with the same letter. Let Aj=〈Ej〉 and dj=dim⁡Aj, so that Ej is a generating curve for Aj. Denote by ej the inclusion Aj↪A 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′·Ej≥∑j=1sdj≥n. The first inequality comes from the fact that on Aj one has Dj′·Ej≥dj 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 fixj, 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·Ej≠0, because in such a case Dj′ which is prime would be the sum of two effective divisors. The existence of an i with Di·Ej≠0 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 [3]. 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 n≤D·T1≤D·Z=n implies that s=1 as desired.
For the time being, we replaceD 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×A1→A 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:A→A1 and of a line bundle N on A such that (4)M⊗p∗N≃f×IdA1∗P.Restricting (4) on the fiber {x}×A1, for x∈A, one finds an isomorphism (5)e∗tx∗OAD≃tfx∗OA1D1,where e is the embedding A1↪A. 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 K∩A1={0A}, where K is the kernel of f.
BecauseK cuts A1 only in 0A, the sum morphism K×A1→A is injective and so we will have dim⁡(K∔D1)=n-dim⁡A1+dim⁡A1-1=n-1. Now, for a general x∈A, we have {-f(x)}∔D1=({-x}∔D)|A1⊂{-x}∔D. So, x-fx∈Tran(D1,D)≔{y∈A∣{y}∔D1⊂D}. But Tran(D1,D) is closed and so for any x∈A we have {x-f(x)}∔D1⊂D.
Then forx∈K, {x}∔D1⊂D and therefore K∔D1⊂D. For K0 the connected component of the origin in K, we have K0∔D1⊂D. But K0∔D1 is a divisor and D is prime, so the previous inclusion is an equality. Now, (6)K0∔D=K0∔K0∔D1=K0∔D1=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·Ej≠0. For this, we consider for all i, an Abelian variety Bi, an ample divisor Fi on Bi, and a surjective morphism fi:A→Bi such that fi-1(Fi)=Di. Their existence follows from Lemma 12.
We have(7)n=D·Z=∑i=1rfi-1Fi·Z=∑i=1rFi·fi∗Z≥∑i=1rli,where li=dim⁡Bi 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=codim⁡Ki and so (8)∑i=1rli=∑i=1rcodim⁡Ki≥codim⁡∩Ki=n (by definition of Ki and the ampleness of D, the intersection ∩Ki is finite).
It results in that(9)n≥∑i=1rFi·fi∗Z≥∑i=1rli≥n,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·fi∗Ej. This last number is 0 if j≠ji and nonzero for j=ji because Fi is ample. This conclude the third step.
From the first and third steps we find thati→ji 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·Ej≠0⇔i=j. To conclude the proof, we consider all the requirements supposed above.
In the first place we reviewBi’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=dim⁡Bi 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 j≠i 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 allDi’s cut properly the subvarieties Aj’s, which means that either ej∗(Di) is an effective divisor on Aj or Di∩Aj is empty, in which case ej∗(Di)=0. The former case can happen only for j=i, because in this situation ej∗(Di)·Ej≠0 (more precisely, the projection formula gives ej∗(Di)·Ej=Di·ej∗Ej=Di·Ej). So ej∗(Di)≠0⇔j=i and we have (10)Dj′=ej∗D=ej∗Dj=ej∗fj∗Fj=fj∘ej∗Fj.Let us consider the morphism fj∘ej:Aj→Bj. It sends the generating curve Ej of Aj on the generating curve Tj of Bj, and therefore it is surjective; so dj≥lj. But, from the first and third steps, n=∑j=1rdj=∑j=1rlj; this implies that fj∘ej has a finite nonzero degree. On the other hand fj∘ej 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.
Leth:B1×⋯×Br→A be defined by h(b1,…,br)=∑i=1rgi(bi) and g:A→B1×⋯×Br be defined by g(a)=(f1(a),…,fr(a)). Then h∘g is the identity, being the identity on every Ai. Also, g∘h is the identity, being the identity on every {0B1}×⋯×Bi×⋯×{0Br}.
Soh 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 [2].Corollary 14.
LetA 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 haveDn=(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.



---

*Source: 431098-2015-10-01.xml*