#### 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.

#### 1. Introduction

In the classical papers [1] (resp. [2]), Matsusaka and Hoyt gave a necessary and sufficient criterion for an Abelian variety 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 Ranâ€™s criterion, using results from [4] on the ring of the numerical algebraic cycles on . 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 variety , we denote by the -vector space of the algebraic cycles with -coefficients on and by the quotient by numerical equivalence. It is well known (cf. [6] or [4]) that on there are two -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 but also on . Below, for , we will denote the usual product by and the Pontrjagin product by .

Also, for the two subvarieties and in , we denote by and their sum and difference in the group law of , to avoid confusion with the corresponding operations on the cycles. In the current paper, the algebraic cycles will often be divisors and -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 -cycles will be considered effective. Finally, a prime cycle is an irreducible subvariety of of a corresponding codimension.

#### 2. Generating Curves on an Abelian Variety

Let be an Abelian variety of dimension and be a curve on it containing the origin of . We consider a sequence of closed subsets in , defined as follows: It is clear that this sequence is increasing, and for every . As long as is a curve, is irreducible and there is a first index such that . Also, we have for all and . It follows that is stable for the group law on and the induced operation has as unity. Using a result of Ramanujan in ([7, chapter II, Section ]), is an Abelian subvariety of and the points of generate the group . We denote by the subvariety . If , then and we say that is a generating curve on .

*Remark 1. *In [8] Matsusaka proves that every Abelian variety has a generating curve. Moreover, from his proof, for a projective embedding of , every linear section with a convenient linear subspace of appropriate dimension which contains is a generating curve for .

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

Lemma 2. *Let be a curve in with , the subvariety of generated by and . Then, is the maximal number such that ( with terms) is nonzero and is the support of the cycle .*

Let us now consider a curve which does not necessarily contain the origin. It easily follows that for the Abelian variety generated by does not depend on ; it is in fact the subgroup of generated by . This Abelian variety will also be denoted by . If is an effective -cycle, we denote by the Abelian subvariety given by and we will say that is a generating -cycle for .

*Remark 3. *From the definition above, we see that the construction of is independent of the numbers . In particular, and generate the same subvariety and also for and .

The next lemma will be useful in the sequel.

Lemma 4. *(a) For with subvariety and , a curve, both containing the origin, if then .**(b) For Abelian subvarieties and nonzero integers, if and are numerically equivalent, then .*

* Proof. *(a) We have and from we deduce that . So and because it follows .

(b) Let be a generating curve for (it exists cf. Remark 1). Then and so . It follows that , and from the first point . The last inclusion implies that . In the same way the reversed inclusion is proved.

*Remark 5. *The point (b) above, in the case , is a result of Matsusaka in [8].

Proposition 6. *(a) For two curves which are numerically equivalent, we have .**(b) For a curve and , a 1-cycle which is numerically equivalent with , we have .**(c) Let be an ample divisor and a 1-cycle which is numerically equivalent with . Then is a generating 1-cycle for .*

* Proof. *(a) Using convenient translations we can suppose that and contain . Let , be the dimensions of for . Using Lemma 2 and the fact that is numerically equivalent with for all positive integers , we find . From Lemma 2 again, is a multiple of both and , and the conclusion follows from Lemma 4(b).

(b) As in (a), denoting we can suppose that and all contain . By Lemma 2, for , we have . Now, being numeric equivalent with , we find . But again from Lemma 2 we find a nonzero term in the development of . With Lemma 4(a), this term which is in fact a subvariety contains , because all terms in are vanished by the Pontrjagin product with . On the other hand, this term is contained in and so .

For the reverse inclusion, we consider the development of the left side of the equality . From Lemma 2â€‰ with is an integer. We find ; therefore for all and then from Lemma 4(a), , so .

(c) Let be a positive integer with very ample. The cycles and are therefore numerical equivalents and there will exist an integer curve in the same numeric class with and so with . From (b) we have . But from Remark 1â€‰â€‰, so is a generating -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 [4] which will be the main tool in the proof of Ranâ€™s theorem. Let be a generating curve of the -dimensional Abelian variety . We consider on the following cycles: and for . From the definition of the Pontrjagin product, is a cycle with irreducible support of codimension on . In particular is a divisor and there exists such that , where is the fundamental cycle on .

The result we need from [4] is the following.

Proposition 7. *All cycles have integer coefficients and in particular , being evidently positive. Also, for . In particular, and so is ample.*

*Remark 8. *For a smooth curve and its Jacobian, these divisors are well-known.

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

Proposition 9. *(a) Let be an effective divisor and a generating -cycle. Then .**(b) If moreover is ample, then .*

* Proof. *(a) We can suppose that is a prime divisor. Let be the components of . We have for all , because the general translation of cuts properly . It is therefore sufficient to find an such that . Suppose there is no such . Then using a result from [7, chapter , Section ], translations with elements of the form with leave invariant. But is a generating -cycle, and therefore every element in is a sum of elements of this form. So is invariant with respect to any translation and then numerically equivalent with , in contradiction with its effectiveness.

(b) Consider a first case where is a prime cycle (i.e., is a curve) and without loss of generality . Let be a variable and consider the polynomial . Because is ample and is nondegenerate, the index theorem for Abelian varieties compare [7] asserts that all roots of are real and negative. So the means inequality gives .

For the general case, let with all , , and be the restriction of to . The projection formula gives from the particular case above. So, , because is a generating -cycle.

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

Corollary 10. *Let be an Abelian variety, an ample effective divisor, and a generating -cycle of (the coefficients are supposed to be nonzero). If then for all .*

* Proof. *We have because is ample and one can apply Proposition 9(b). So and because the last term is nonzero by Proposition 9(b), we find for all .

In the same way, because remains a generator -cycle by Remark 3. So and being ample, the last term is positive. It results that for all .

We can now prove the following result, which is nothing else but Ranâ€™s version of the Matsusaka theorem.

Theorem 11. *Let be an ample divisor on the Abelian variety and let be a generating curve such that . Then is smooth, is its Jacobian, and is a translation of .*

* Proof. *In the proof of point (b) from Proposition 9 we have the inequality . If we will have and so . In this case the polynomial from the same proposition becomes . It follows that the arithmetic and geometric means of the roots coincide and so all the roots have the form for a positive value of . So and by identification, . It follows that and then . These relations imply that . The Hodge index theorem asserts that is numerically equivalent with , and because is a principal polarization (from Proposition 7 and equality ), one may deduce that is a translation of .

Consider the normalization for , and let be a prolongation of , where is a Jacobian of . If we choose a base point in the construction of , one on which sits above , will be a morphism of Abelian varieties, sending origin to origin. Also, is surjective because is generating for and for genus of we have .

Let us denote by the canonical cycles on the Jacobian . Therefore for : for this is clear because and for it is a consequence of the definitions for and and also from the fact that commute with the Pontrjagin product. In particular and so is the degree of the restriction of to . Therefore this restriction is a birational morphism and has an inverse: . This inverse, considered as a rational map from to can be extended over all the giving a morphism compare [7]. As a consequence, the restriction of to will be an isomorphism and will be an Abelian subvariety of . But contains which generates and so . In this case we have and is birational from to hence an isomorphism.

#### 4. Proof of Ran Theorem

The purpose of this section is to give a proof for Ranâ€™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 from [3] with the result below whose proof is very simple.

Lemma 12. *Let be a prime divisor on an Abelian variety . Then, there exists an Abelian variety , a surjective morphism of Abelian varieties and an ample divisor on such that as schemes.*

* Proof. *We consider the closed subgroup of defined by and the Abelian subvariety of which is the connected component of in . We denote by the quotient and by the quotient morphism. Finally we denote by the closed irreducible subset with the reduced structure. We easily find , so is a divisor on and set theoretically because . Let such that . Applying we find , and because we find and so . Therefore, the elements in which leave invariant by translations are from . They are then in a finite number, because the index is finite. So is an ample divisor on . Finally the equality also holds at the schemes level, because is smooth from its construction.

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

Theorem 13. *Let be an Abelian variety of dimension , an ample effective divisor, and a generating 1-cycle such that . Then for all , and there are smooth curves with Jacobians and an isomorphism of Abelian varieties such that, for every , is a translation of ( on the th place) and is a translation of (), where is the canonical divisor on .*

* Proof. *The fact that for all 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 there is a unique such that . We translate the curves such that they contain the origin and denote the result with the same letter. Let and , so that is a generating curve for . Denote by the inclusion and by the same letter a translation of the divisor which has a proper intersection with every . Therefore, is defined as a cycle and is an ample divisor on . The projection formula gives and so The first inequality comes from the fact that on one has according to Proposition 9(b), and the last one is due to the fact that is a generating -cycle. So , and being a generating curve for , from Theorem 11 one finds that is smooth, is its Jacobian, and is a translation of the canonical divisor on ; so is prime as any divisor numeric equivalent with it (it is a principal polarization).

Let us fix , and consider for any a translation of which cuts properly. Every such translation, also denoted by , restricted to either is an effective divisor or has an empty intersection with , in which case . But the sum of these restrictions is numerically equivalent with and so there cannot be two indexes with , because in such a case which is prime would be the sum of two effective divisors. The existence of an with comes from the fact that is ample.*Step **2*. This part consists in the proof of the following fact: for an -dimensional Abelian variety , a prime ample divisor , and a generating -cycle with one has (i.e., is ireducible and reduced).

The proof is due to Ran compare Lemma III.2 from [3]. Denote by . From the first step, we know that is in fact the Jacobian of the smooth curve ; in particular it is principally polarized and isomorphic with its dual. It will suffice to prove that , because in this case will be a generating curve, and the fact that is ample together with the inequalities implies that as desired.

For the time being, we replace with a translation whose restriction is well defined as divisor on . As in the proof of Step , is numerically equivalent with . Let the morphism given by and be the projections. Consider on the line bundle and on the line bundle , where are the projections on the factors of . Using the fact that is a Jacobian (and therefore it is its own Picard variety whith the Poincare bundle equal to ), we deduce the existence of a morphism and of a line bundle on such that Restricting (4) on the fiber , for , one finds an isomorphism where is the embedding . Because is a principal polarisation, the point is uniquelly defined by the above property, which can be written in divisorial terms as , at least for general such that the divisor is well defined. From this one deduces that points in are fixed by and so is surjective with , where is the kernel of .

Because cuts only in , the sum morphism is injective and so we will have . Now, for a general , we have . So, . But is closed and so for any we have .

Then for , and therefore . For the connected component of the origin in , we have . But is a divisor and is prime, so the previous inclusion is an equality. Now, But ample implies that is finite and prime implies that which is equivalent with .*Step **3*. Within this step we prove that for any there is a unique such that . For this, we consider for all , an Abelian variety , an ample divisor on , and a surjective morphism such that . Their existence follows from Lemma 12.

We have where and the last inequality is from Proposition 9(b). We examine the last sum using the effective construction of â€™s from Lemma 12. There, is of the form where is an Abelian subvariety of . As consequence, and so (by definition of and the ampleness of , the intersection is finite).

It results in that and so . But is a prime divisor and from Step there is a unique with a curve. All the other curves from the support of will be therefore contracted. We now fix and compute . This last number is if and nonzero for because is ample. This conclude the third step.

From the first and third steps we find that is a bijection and so . Also one can reorder the curves (such that will be numbered by ) and so we can suppose that for all we have . To conclude the proof, we consider all the requirements supposed above.

In the first place we review â€™s. Let be the cycle . From the third step, is in fact a curve, namely, . Also we have seen that and therefore Theorem 11 implies that is the Jacobian of . To see this, we need only to prove that is a generating curve of and this is implied by the fact that, as we have seen, contracts all the curves for and as far as these contain , the contraction will be to . So and because generates , generates . So, by Theorem 11, is a translation of the canonical divisor on .

Let us recall that in the first step we supposed (using appropriate translations) that all â€™s cut properly the subvarieties â€™s, which means that either is an effective divisor on or is empty, in which case . The former case can happen only for , because in this situation (more precisely, the projection formula gives ). So and we have Let us consider the morphism . It sends the generating curve of on the generating curve of , and therefore it is surjective; so . But, from the first and third steps, ; this implies that has a finite nonzero degree. On the other hand pull back the principal polarization from to the principal polarization on . So its degree is and it is an isomorphism with inverse denoted by .

Let be defined by and be defined by . Then is the identity, being the identity on every . Also, is the identity, being the identity on every .

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

Finally, we formulate the following corollary which is the result of Hoyt from [2].

Corollary 14. *Let be an Abelian variety, an ample divisor with , and a 1-cycle such that is numerically equivalent with . Then the conclusion of Theorem 13 holds true.*

* Proof. *We have , so . On the other hand, from Proposition 6(c), is a generating 1-cycle and therefore 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.