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

In the classical papers [1] (resp. [2]),Matsusaka andHoyt 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 probablymore natural criterion for this.However, hismethod 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 onA. 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 byZ(A) theQ-vector space of the algebraic cycles withQ-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 on N(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 subvarieties X 1 and X 2 in A, we denote by X 1 cX 2 and X 1 aX 2 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 1cycles will be considered effective. Finally, a prime cycle is an irreducible subvariety of A of a corresponding codimension.


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 Ran'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 Q-vector space of the algebraic cycles with Q-coefficients on  and by () the quotient by numerical equivalence.It is well known (cf.[6] or [4]) that on () 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 () but also on ().Below, for ,  ∈ (), we will denote the usual product by  ⋅  and the Pontrjagin product by  ⋆ .
Also, for the two subvarieties  1 and  2 in , we denote by  1 c 2 and  1 a 2 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 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 1cycles will be considered effective.Finally, a prime cycle is an irreducible subvariety of  of a corresponding codimension.

Generating Curves on an Abelian Variety
Let  be an Abelian variety of dimension  and  be a curve on it containing the origin 0  of .We consider a sequence of closed subsets in , defined as follows: It is clear that this sequence is increasing, dim   ≤  and dim  +1 ≤ dim   + 1 for every .As long as  is a curve,   is irreducible and there is a first index  such that   =  +1 .Also, we have   =   for all  ≥  and dim   = .It follows that   is stable for the group law on  and the induced operation has 0  as unity.Using a result of Ramanujan in ( [7, chapter II, Section 4]),   is an Abelian subvariety of  and the points of  generate the group   .We denote by ⟨⟩ 2 Journal of Mathematics 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 0  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 0  ∈ , ⟨⟩ the subvariety of  generated by  and  = dim⟨⟩.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 a{} does not depend on ; it is in fact the subgroup of  generated by a.This Abelian variety will also be denoted by ⟨⟩.If  =  1  1 +⋅ ⋅ ⋅+    is an effective 1-cycle, we denote by ⟨⟩ the Abelian subvariety given by ⟨ 1 ⟩c⟨ 2 ⟩c ⋅ ⋅ ⋅ c⟨  ⟩ and we will say that  is a generating 1-cycle for ⟨⟩.
Remark 3. From the definition above, we see that the construction of ⟨⟩ is independent of the numbers   .In particular,  and  red 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  *  = 0 then  ⊂ .
(c) Let  be an ample divisor and  a 1-cycle which is numerically equivalent with  −1 .Then  is a generating 1cycle for .
Proof.(a) Using convenient translations we can suppose that  1 and  2 contain 0  .Let ,  be the dimensions of   = ⟨  ⟩ for  = 1, 2. Using Lemma 2 and the fact that  1 *  is numerically equivalent with  2 *  for all positive integers , we find  = .From Lemma 2 again,  1 *  is a multiple of both  1 and  2 , and the conclusion follows from Lemma 4(b).
(b) As in (a), denoting  =  1  1 + ⋅ ⋅ ⋅ +     we can suppose that  and all   contain 0  .By Lemma 2, for  = dim⟨⟩, we have  * (+1) = 0. Now,  *  being numeric equivalent with  *  , we find  *  *  = 0.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 ⟨⟩ ⊂ ⟨⟩.
(c) Let  be a positive integer with || very ample.The cycles  −1  −1 and  −1  are therefore numerical equivalents and there will exist an integer curve  in the same numeric class with  −1  −1 and so with 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.

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  be a generating curve of the -dimensional Abelian variety .We consider on  the following cycles:   () = {0  } and   () = (1/( − )!) * (−) for 0 ≤  ≤  − 1.From the definition of the Pontrjagin product,   () is a cycle with irreducible support of codimension  on .In particular  1 () is a divisor and there exists   ∈ Q such that  0 () =   ⋅ 1  , where 1  is the fundamental cycle on .
The result we need from [4] is the following.
Proposition 7. All cycles   () have integer coefficients and in particular   ∈ Z, being evidently positive.Also,  1 ()  = !  −1   () for 1 ≤  ≤ .In particular,  1 ()  > 0 and so  1 () is ample.Proof.(a) We can suppose that  is a prime divisor.Let  1 , . . .,   be the components of .We have  ⋅   ≥ 0 for all , because the general translation of   cuts properly .It is therefore sufficient to find an  such that ⋅  > 0. Suppose there is no such .Then using a result from [7, chapter2, Section 6] translations with elements of the form {}a{} with ,  ∈   leave  invariant.But  is a generating 1-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 0, 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 0  ∈ .Let  be a variable and consider the polynomial 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  ⋅  ≥ ((O  ()) ⋅   ) 1/ ≥ .
For the general case, let  =  1  1 + ⋅ ⋅ ⋅ +     with all   > 0,   = ⟨  ⟩, and   be the restriction of  to   .The projection formula gives  ⋅   =   ⋅   ≥ dim   from the particular case above.So, The following consequence of the above proposition will be useful in the last part of the paper.In the same way,  =  ⋅  = ∑  =1    ⋅   ≥ ∑  =1  ⋅   =  ⋅ ∑  =1   ≥  because ∑  =1   remains a generator 1-cycle by Remark 3.So    ⋅   =  ⋅   and  being ample, the last term is positive.It results that   = 1 for all .
We can now prove the following result, which is nothing else but Ran'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  ⋅  =  = dim .Then  is smooth,  is its Jacobian, and  is a translation of  1 ().
Consider the normalization  0 :  →  for , and let  :  →  be a prolongation of  0 , where  is a Jacobian of .If we choose a base point in the construction of , one on  which sits above 0  ∈ ,  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 1 ≤  ≤ : for  = 1 this is clear because  −1 =  and for  > 1 it is a consequence of the definitions for  − and  − () and also from the fact that  * commute with the Pontrjagin product.In particular  * ( − ) =  0 () =   ⋅ 1  and so   = 1 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  −1 =  which generates  and so  − = .In this case we have  =  and  is birational from  to  hence an isomorphism.

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. 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  −1 () =  as schemes.
Proof.We consider the closed subgroup  of  defined by  := { ∈  | {}c = } and the Abelian subvariety  0 of  which is the connected component of 0  in .We denote by  the quotient / 0 and by  :  →  the quotient morphism.Finally we denote by  the closed irreducible subset () with the reduced structure.We easily find dim  = dim  − 1, so  is a divisor on  and set theoretically  −1 () = c 0 =  because  0 ⊂ .Let  ∈  such that {()}c = .Applying  −1 we find {}cc 0 = c 0 , and because c 0 =  we find {}c =  and so  ∈ .Therefore, the elements in  which leave  invariant by translations are from ().They are then in a finite number, because the index [ :  0 ] is finite.So  is an ample divisor on .Finally the equality  −1 () =  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 ,  = ∑  =1     an ample effective divisor, and  = ∑  =1     a generating 1-cycle such that  ⋅  = .Then   =   = 1 for all , ,  =  and there are  smooth curves  1 , . . .,   with Jacobians  1 , . . .,   and an isomorphism of Abelian varieties ℎ : Proof.The fact that   =   = 1 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   ⋅   ̸ = 0. We translate the curves   such that they contain the origin and denote the result with the same letter.Let   = ⟨  ⟩ and   = dim   , 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 1-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   ⋅   = 0.But the sum of these restrictions is numerically equivalent with    and so there cannot be two indexes  with   ⋅  ̸ = 0, because in such a case    which is prime would be the sum of two effective divisors.The existence of an  with   ⋅   ̸ = 0 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 1-cycle  =  1 + ⋅ ⋅ ⋅ +   with  ⋅  =  one has  = 1 (i.e.,  is ireducible and reduced).
The proof is due to Ran compare Lemma III.2 from [3].Denote by  1 = ⟨ 1 ⟩.From the first step, we know that  1 is in fact the Jacobian of the smooth curve  1 ; in particular it is principally polarized and isomorphic with its dual.It will suffice to prove that  1 = , because in this case  1 will be a generating curve, and the fact that  is ample together with the inequalities  ≤  ⋅  1 ≤  ⋅  =  implies that  = 1 as desired.
For the time being, we replace  with a translation whose restriction  | 1 :=  1 is well defined as divisor on  1 .As in the proof of Step 1,  1 is numerically equivalent with  1 ( 1 ).Let  :  ×  1 →  the morphism given by (, ) =  +  and ,  1 be the projections.Consider on  ×  1 the line bundle ), where  1 ,  2 are the projections on the factors of  1 ×  1 .Using the fact that  1 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  :  →  1 and of a line bundle N on  such that Restricting ( 4) on the fiber {} ×  1 , for  ∈ , one finds an isomorphism where  is the embedding  1 → .Because  1 is a principal polarisation, the point () is uniquelly defined by the above property, which can be written in divisorial terms as ({−}c) | 1 = {−()}c 1 , at least for  general such that the divisor ({−} + ) | 1 is well defined.From this one deduces that points in  1 are fixed by  and so  is surjective with  ∩  1 = {0  }, where  is the kernel of .
Because  cuts  1 only in 0  , the sum morphism  ×  1 →  is injective and so we will have dim(c 1 ) =  − dim  1 + dim  Then for  ∈ , {}c 1 ⊂  and therefore c 1 ⊂ .For  0 the connected component of the origin in , we have  0 c 1 ⊂ .But  0 c 1 is a divisor and  is prime, so the previous inclusion is an equality.Now, But ample  implies that  0 is finite and prime  implies that  0 = {0  } which is equivalent with  =  1 .
Step 3. Within this step we prove that for any  there is a unique  such that   ⋅   ̸ = 0.For this, we consider for all , an Abelian variety   , an ample divisor   on   , and a surjective morphism   :  →   such that   −1 (  ) =   .Their existence follows from Lemma 12.
We have where   = dim   and the last inequality is from Proposition 9(b).We examine the last sum using the effective

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 1-cycle.Then  ⋅  > 0.(b) If moreover  is ample, then  ⋅  ≥  = dim .