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.

In the classical papers [

The aim of this paper is to reprove Ran’s criterion, using results from [

In the sequel, for an Abelian variety

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

Also, for the two subvarieties

The term curve, is reserved for integral ones, and all

Let

In [

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

Let

Let us now consider a curve

From the definition above, we see that the construction of

The next lemma will be useful in the sequel.

(a) For

(b) For

(a) We have

(b) Let

The point (b) above, in the case

(a) For two curves

(b) For a curve

(c) Let

(a) Using convenient translations we can suppose that

(b) As in (a), denoting

For the reverse inclusion, we consider the development of the left side of the equality

(c) Let

The point (c) above is a slight generalization of the result from Remark

We recall a result from [

The result we need from [

All cycles

For

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

(a) Let

(b) If moreover

(a) We can suppose that

(b) Consider a first case where

For the general case, let

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

Let

We have

In the same way,

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

Let

In the proof of point (b) from Proposition

Consider the normalization

Let us denote by

The purpose of this section is to give a proof for Ran’s full theorem. Some points are as in [

Let

We consider the closed subgroup

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

Let

The fact that

Let us fix

The proof is due to Ran compare Lemma III.2 from [

For the time being, we replace

Because

Then for

We have

It results in that

From the first and third steps we find that

In the first place we review

Let us recall that in the first step we supposed (using appropriate translations) that all

Let

So

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

Let

We have

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