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.
The result we are interested in is the following theorem of Ran.
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 [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=dimAj, 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 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·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 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×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.
Because K cuts A1 only in 0A, the sum morphism K×A1→A is injective and so we will have dim(K∔D1)=n-dimA1+dimA1-1=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 for x∈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=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=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 that i→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 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 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 all Di’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.
Let h: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}.
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.