Almost isomorphic abelian varieties

Abstract

We study abelian varieties over finitely generated fields K of characteristic zero, whose $\ell$-adic Tate modules are isomorphic as Galois modules for all primes $\ell$.

1 Introduction

Let K be a field, $\overline{K}$ its separable algebraic closure, the absolute Galois group of K. If A is an abelian variety over K then we write for its ring of all K-endomorphisms and $\mathrm{End}^0(A)$ for the corresponding (finite-dimensional semisimple) ${\mathbb {Q}}$-algebra .

If $\ell$ is a prime different from then we write $T_{\ell }(A)$ for the ${\mathbb {Z}}_{\ell }$-Tate module of A [7, 9] which is a free ${\mathbb {Z}}_{\ell }$-module of rank provided with the natural continuous group homomorphism

$$\begin{aligned} \rho _{\ell ,A}{:} G_K \rightarrow \mathrm{Aut}_{{\mathbb {Z}}_{\ell }}(T_{\ell }(A)) \end{aligned}$$

and the ${\mathbb {Z}}_{\ell }$-ring embedding

The image of commutes with $\rho _{\ell ,A}(G_K)$. Tensoring by ${\mathbb {Q}}_{\ell }$ (over ${\mathbb {Z}}_{\ell }$), we obtain the ${\mathbb {Q}}_{\ell }$-Tate module of A

which is a -dimensional ${\mathbb {Q}}_{\ell }$-vector space containing as a ${\mathbb {Z}}_{\ell }$-lattice of maximal rank. We may view $\rho _{\ell ,A}$ as an $\ell$-adic representation [11]

$$\begin{aligned} \rho _{\ell ,A}{:} G_K\rightarrow \mathrm{Aut}_{{\mathbb {Z}}_{\ell }}(T_{\ell }(A))\subset \mathrm{Aut}_{{\mathbb {Q}}_{\ell }}(V_{\ell }(A)) \end{aligned}$$

and extend $e_{\ell }$ by ${\mathbb {Q}}_{\ell }$-linearity to the embedding of ${\mathbb {Q}}_{\ell }$-algebras

which we still denote by $e_{\ell }$. Further we will identify with its image in $\mathrm{End}_{{\mathbb {Q}}_{\ell }}(V_{\ell }(A))$. This provides $V_{\ell }(A)$ with the natural structure of $G_K$-module; in addition, is a ${\mathbb {Q}}_{\ell }$-(sub)algebra of endomorphisms of the Galois module $V_{\ell }(A)$. In other words,

Let K be a field of characteristic zero that is finitely generated over ${\mathbb {Q}}$. Suppose we are given an abelian variety A of positive dimension over K. Let B be an abelian variety over K such that the ${\mathbb {Z}}_{\ell }$-Tate modules of A and B are isomorphic as Galois modules for all $\ell$ (we call such A and B almost isomorphic). In this paper we discuss the structure of the corresponding right -module . Using a theorem of Faltings [4, 5] (conjectured by Tate [12]), we prove that is a locally free module of rank 1. In addition, using a special case of Serre’s tensor construction ([3, Section 7], [2, Section 1.7.4]), we prove that there is a natural bijection between isomorphism classes of locally free modules of rank 1 over and isomorphism classes of abelian varieties B over K, whose Tate modules are isomorphic to ones of A.

The paper is organized as follows. Section 2 deals with isogenies of abelian varieties and corresponding homomorphisms of their Tate modules. In Sect. 3 we discuss locally free modules of rank 1 over orders in semisimple ${\mathbb {Q}}$-algebras. In Sect. 4 we apply results of Sect. 3 to a construction of almost isomorphic abelian varieties.

2 Isogenies

If $\ell$ is a prime then we write ${\mathbb {Z}}_{(\ell )}$ for the subring in ${\mathbb {Q}}$ that consists of all the rational numbers, whose denominators are prime to $\ell$. We have

$$\begin{aligned} {\mathbb {Z}}\subset {\mathbb {Z}}_{(\ell )}={\mathbb {Z}}_{\ell }\cap {\mathbb {Q}}\subset {\mathbb {Z}}_{\ell }. \end{aligned}$$

(Here the intersection is taken in ${\mathbb {Q}}_{\ell }$.) In addition, if m is a positive integer that is prime to $\ell$ then

The intersection of all ${\mathbb {Z}}_{(\ell )}$ (in ${\mathbb {Q}}$) coincides with ${\mathbb {Z}}$.

Let K be an arbitrary field. If and X is an abelian variety over K then we write for the kernel of multiplication by $\ell$ in $X(\overline{K})$. It is well known that is a finite $G_K$-submodule in $X(\overline{K})$ of order and there is a natural isomomorphism of $G_K$-modules .

Lemma 2.1

Let A and B be abelian varieties of positive dimension over K.

• (a) If A and B are isogenous over K then the right -module is free of rank 1. In addition, one may choose as a generator of  any isogeny $\phi {:} A \rightarrow B$.

• (b) The following conditions are equivalent:

  • (i) The right -module is free of rank 1.

  • (ii) and there exists a -dimensional abelian K-subvariety $B_0\subset B$ such that A and $B_0$ are isogenous over K and

    

    In particular, the image of every K-homomorphism of abelian varieties $A \rightarrow B$ lies in $B_0$.

• (c) If the equivalent conditions (i) and (ii) hold and then , $B=B_0$, and A and B are isogenous over K.

Proof

(a) is obvious.

Suppose (bii) is true. Let us pick an isogeny $\phi {:} A \rightarrow B_0$. It follows that is a free right -module of rank 1 generated by $\phi$. Now (bi) follows from the equality

Suppose that (bi) is true. We may choose a homomorphism of abelian varieties $\phi {:} A \rightarrow B$ as a generator (basis) of the free right -module . In other words, for every homomorphism of abelian varieties $\psi {:} A \rightarrow B$ there are and a nonzero integer n such that $n\psi =\phi u$. In addition, for each nonzero the composition $\phi u$ is a nonzero element of . Clearly, $B_0=\phi (A)\subset B$ is an abelian K-subvariety of B with . We have

$$\begin{aligned} n\psi (A) =\phi u(A) \subset \psi (A)\subset B_0. \end{aligned}$$

It follows that the identity component of $\psi (A)$ lies in $B_0$. Since $\psi (A)$ is a (connected) abelian K-subvariety of B, we have $\psi (A)\subset B_0$. This proves that . On the other hand, if then $\phi {:} A \rightarrow B_0$ is an isogeny and we get (bii) under our additional assumption. If then has positive dimension that is strictly less than . By the Poincaré complete reducibility theorem [7], there is an endomorphism such that the image $u_0(A)$ coincides with the identity component of ; in particular, $u_0 \ne 0$, . This implies that $\phi u_0=0$ in and we get a contradiction, which proves (bii).

(c) follows readily from (bii).$\square$

Lemma 2.2

Suppose that A, B, C are abelian varieties over K of positive dimension that are mutually isogenous over K. We view and as right -modules. Then the natural map

that associates to $\tau {:} B \rightarrow C$ a homomorphism of right -modules

is a group isomorphism.

Proof

Clearly, $m_{B,C}$ is injective. In order to check the surjectiveness, notice that the statement is clearly invariant by isogeny, so we can assume that $B=A$ and $C=A$, in which case it is obvious.$\square$

Now till the end of this paper we assume that K is a field of characteristic zero that is finitely generated over ${\mathbb {Q}}$, and A and B are abelian varieties of positive dimension over K. By a theorem of Faltings [4, 5],

Lemma 2.3

Let $\ell$ be a prime. Then the following conditions are equivalent:

• (i) There is an isogeny $\phi _{\ell }{:} A \rightarrow B$, whose degree is prime to $\ell$.

• (ii) The Tate modules $T_{\ell }(A)$ and $T_{\ell }(B)$ are isomorphic as -modules.

If the equivalent conditions (i) and (ii) hold then the right -module is free of rank 1 and the right -module is free of rank 1.

Proof

(i) implies (ii). Indeed, let $\phi _{\ell }{:} A \rightarrow B$ be an isogeny such that its degree is prime to $\ell$. Then there exists an isogeny $\varphi _{\ell }{:} B \rightarrow A$ such that is multiplication by d in B and is multiplication by d in A. This implies that $\phi _{\ell }$ induces a $G_K$-equivariant isomorphism of the ${\mathbb {Z}}_{\ell }$-Tate modules of A and B.

Suppose that (ii) holds. Since the rank of the free ${\mathbb {Z}}_{\ell }$-module $T_{\ell }(A)$ (resp. $T_{\ell }(B)$) is (resp. ), we conclude that , i.e., . By the theorem of Faltings ($\star$), there is an isomorphism of the ${\mathbb {Z}}_{\ell }$-Tate modules of A and B that lies in . Since is dense in in the $\ell$-adic topology, and the set of isomorphisms $T_{\ell }(A)\cong T_{\ell }(B)$ is open in , there is $\phi _{\ell }\in \mathrm{Hom}(A,B)$ that induces an isomorphism $T_{\ell }(A)\cong T_{\ell }(B)$. Clearly, does not contain points of order $\ell$ and therefore is finite. This implies that $\phi _{\ell }$ is an isogeny, whose degree is prime to $\ell$. This proves (i).

In order to prove the last assertion of Lemma 2.3, one has only to observe that is a generator of the (obviously) free right ${\mathbb {Z}}_{(\ell )}$-module and of the free right ${\mathbb {Z}}_{\ell }$-module . $\square$

We say that A and B are almost isomorphic if for all primes $\ell$ the equivalent conditions (i) and (ii) of Lemma 2.3 hold. Clearly, if A and B are isomorphic over K then they are almost isomorphic. It is also clear that if A and B are almost isomorphic then they are isogenous over K. Obviously, the property of being almost isomorphic is an equivalence relation on the set of (nonzero) abelian varieties over K.

Corollary 2.4

Suppose that A and B are almost isomorphic. Then A and B are isomorphic over K if and only if is a free -modules of rank 1. In particular, if is a principal ideal domain (for example, $)$ then every abelian variety over K, which is almost isomorphic to A, is actually isomorphic to A.

Proof

Suppose is a free -module, i.e., there is a homomorphism of abelian varieties $\phi {{:}} A \rightarrow B$ such that . We know that for any prime $\ell$ there is an isogeny $\phi _{\ell }{{:}} A \rightarrow B$ of degree prime to $\ell$. (In particular, .) Therefore there is with $\phi _{\ell }=\phi u_{\ell }$. In particular, $\phi _{\ell }(A)\subset \phi (A)$ and is divisible by . Since $\phi _{\ell }(A)=B$ and is prime to $\ell$, we conclude that $\phi (A)=B$ (i.e., $\phi$ is an isogeny) and $\deg (\phi )$ is prime to $\ell$. Since the latter is true for all primes $\ell$, we conclude that , i.e., $\phi$ is an isomorphism.

Conversely, if $A\cong B$ then is obviously a free -module generated by an isomorphism between A and B.

The last assertion of corollary follows from the well-known fact that every finitely generated module without torsion over a principal ideal domain is free.$\square$

Remark 2.5

The special case of Corollary 2.4 when was actually done in [10, second paragraph of p. 1205].

The next statement is a generalization of Corollary 2.4.

Corollary 2.6

Suppose that A, B, C are abelian varieties of positive dimension over K that are almost isomorphic to each other. Then B and C are isomorphic over K if and only if the right -modules and are isomorphic.

Proof

We know that all A, B, C are mutually isogenous over K. Let us choose an isogeny $\phi {{:}} B \rightarrow C$. We are given an isomorphism of right -modules that obviously extends by ${\mathbb {Q}}$-linearity to the isomorphism of right -modules, which we continue to denote by $\delta$. By Lemma 2.2, there exists such that $\delta =m_{B,C}(\tau _0)$, i.e.,

There exists a positive integer n such that and $\tau$ is not divisible in . This implies that

Since B and C are almost isomorphic, for each $\ell$ there is an isogeny $\phi _{\ell }{:} B \rightarrow C$ of degree prime to $\ell$. Since , we conclude that $\tau$ is an isogeny and is prime to $\ell$ if $\ell$ does not divide n. We need to prove that $\tau$ is an isomorphism. Suppose it is not, then there is a prime $\ell$ that divides and therefore divides n. We need to arrive to a contradiction. Since A and B are almost isomorphic, there is an isogeny $\psi _{\ell }{:} A \rightarrow B$ of degree prime to $\ell$. We have . This implies that $\tau$ kills all points of order $\ell$ on B and therefore is divisible by $\ell$ in , which is not the case. This gives us the desired contradiction.$\square$

Remark 2.7

Let ${\mathcal {Z}}(A)$ $(\hbox {resp.}\;{\mathcal {Z}}(B))$ be the center of (resp. ). Then (resp. ) is the center of (resp. ) and for all primes $\ell$ the ${\mathbb {Z}}_{(\ell )}$-subalgebra

(resp. the ${\mathbb {Z}}_{(\ell )}$-subalgebra ) is the center of (resp. of ). Every K-isogeny $\phi {:} A \rightarrow B$ gives rise to an isomorphism of ${\mathbb {Q}}$-algebras

such that $i_{\phi }({\mathcal {Z}}(A)_{{\mathbb {Q}}})={\mathcal {Z}}(B)_{{\mathbb {Q}}}$ and the restriction $i_{\mathcal {Z}}{:} {\mathcal {Z}}(A)_{{\mathbb {Q}}} \cong {\mathcal {Z}}(B)_{{\mathbb {Q}}}$ of $i_{\phi }$ to the center(s) does not depend on a choice of $\phi$ [14]. If $\phi _{\ell }{:} A \rightarrow B$ is a K-isogeny of degree prime to $\ell$ then and therefore $i_{\mathcal {Z}}({\mathcal {Z}}(A)_{(\ell )}) = {\mathcal {Z}}(B)_{(\ell )}$. This implies that if A and B are almost isomorphic then $i_{\mathcal {Z}}({\mathcal {Z}}(A))$ coincides with ${\mathcal {Z}}(B)$ and therefore $i_{\mathcal {Z}}$ defines a canonical isomorphism of commutative rings ${\mathcal {Z}}(A)\cong {\mathcal {Z}}(B)$. In particular, if is commutative then is also commutative (because and are isomorphic) and there is a canonical ring isomorphism .

3 Locally free modules of rank 1

Throughout this section, ${\Lambda }$ is a ring with 1 that, viewed as an additive group, is a free ${\mathbb {Z}}$-module of finite positive rank. In addition, we assume that the finite-dimensional ${\mathbb {Q}}$-algebra is semisimple. We write ${\Lambda }_{\ell }$ (resp. ${\Lambda }_{(\ell )}$) for the ${\mathbb {Z}}_{\ell }$-algebra (resp. for the ${\mathbb {Z}}_{(\ell )}$-algebra ). We have

In addition, the intersection of ${\Lambda }_{\ell }$ and ${\Lambda }_{{\mathbb {Q}}}$ (in ) coincides with ${\Lambda }_{(\ell )}$.

Let M be an arbitrary free commutative group of finite positive rank that is provided with the structure of a right ${\Lambda }$-module. We write $M_{{\mathbb {Q}}}$ for the right ${\Lambda }_{{\mathbb {Q}}}$-module , $M_{\ell }$ for the right ${\Lambda }_{\ell }$-module and $M_{(\ell )}$ for the right ${\Lambda }_{(\ell )}$-module . We have

In addition, the intersection of $M_{\ell }$ and $M_{{\mathbb {Q}}}$ (in ) coincides with $M_{(\ell )}$.

Definition 3.1

We say that M is a locally free right ${\Lambda }$-module of rank 1 if for all primes $\ell$ the right ${\Lambda }_{\ell }$-module $M_{\ell }$ is free of rank 1, see [6].

Theorem 3.2

Let M be a locally free right ${\Lambda }$-module of rank 1. Then it enjoys the following properties:

• (i) M is a projective ${\Lambda }$-module. More precisely, M is isomorphic to a direct summand of a free right ${\Lambda }$-module of rank 2.

• (ii) The right ${\Lambda }_{\mathbb {Q}}$-module $M_{{\mathbb {Q}}}$ is free of rank 1.

• (iiii) The right ${\Lambda }_{(\ell )}$-module $M_{(\ell )}$ is free of rank 1 for all primes $\ell$.

Proof

Let $J({\Lambda }_{{\mathbb {Q}}})$ be the (multiplicative) idele group of ${\Lambda }_{{\mathbb {Q}}}$, i.e., the group of invertible elements of the adele ring of ${\Lambda }_{{\mathbb {Q}}}$ [6, p. 114] (in the notation of [6, Section 2], $\mathfrak {o}={\mathbb {Z}}$, $K={\mathbb {Q}}$, $A={\Lambda }_{{\mathbb {Q}}}$, $\mathfrak {U}={\Lambda }$). To each $\alpha \in J({\Lambda }_{{\mathbb {Q}}})$ corresponds a certain right ${\Lambda }$-submodule $\alpha {\Lambda }\subset {\Lambda }_{{\mathbb {Q}}}$ that is a locally free ${\Lambda }$-module of rank 1 and a ${\mathbb {Z}}$-lattice of maximal rank in the ${\mathbb {Q}}$-vector space ${\Lambda }_{{\mathbb {Q}}}$, i.e., the natural homomorphism of ${\mathbb {Q}}$-vector spaces is an isomorphism [6, p. 114]. This implies that $(\alpha {\Lambda })_{{\mathbb {Q}}}$ is a free ${\Lambda }_{{\mathbb {Q}}}$-module of rank 1. In addition, the direct sum is a free right ${\Lambda }$-module of rank 2 [6, Theorem 1, pp. 114–115]. This implies that $\alpha {\Lambda }$ is isomorphic to a direct summand of a rank 2 free module; in particular, it is projective. By the same [6, Theorem 1], every right locally free ${\Lambda }$-module M of rank 1 is isomorphic to $\alpha {\Lambda }$ for a suitable $\alpha$. This proves (i) and (ii).

Let $f_0$ be a generator of the free ${\Lambda }_{\mathbb {Q}}$-module $M_{{\mathbb {Q}}}$ of rank 1. Multiplying $f_0$ by a sufficiently divisible positive integer, we may and will assume that . Clearly, the right -module

is free of rank 1 for all primes $\ell$ and $f_0$ is also a generator of . It is also clear that every generator $f_{\ell }$ of the ${\Lambda }_{\ell }$-module $M_{\ell }$ is a generator of the -module . We claim that there is a generator $f_{\ell }$ that lies in M. Indeed, with respect to the $\ell$-adic topology, the subset

is dense in $M_{\ell }$ while the set of generators of the free ${\Lambda }_{\ell }$-module $M_{\ell }$ is open, because the group of units $({\Lambda }_{\ell })^{*}$ is open in ${\Lambda }_{\ell }$. This implies that there exists a (nonzero) generator $f_{\ell } \in M\subset M_{\ell }$ of the ${\Lambda }_{\ell }$-module $M_{\ell }$. Recall that $f_{\ell }$ is also a generator of the free -module . This implies that there exists such that . On the other hand, since $f_{\ell }$ lies in the free rank 1 ${\Lambda }_{{\mathbb {Q}}}$-module $M_{{\mathbb {Q}}}=f_0{\Lambda }_{{\mathbb {Q}}}$, we have $\mu _0\in {\Lambda }_{{\mathbb {Q}}}$. This implies that $\mu _{0}$ is not a zero divisor in the finite-dimensional ${\mathbb {Q}}$-algebra ${\Lambda }_{{\mathbb {Q}}}$ (because it is invertible in ) and therefore lies in ${\Lambda }_{{\mathbb {Q}}}^{*}$. It follows that $f_{\ell }$ is also a generator of the free ${\Lambda }_{{\mathbb {Q}}}$-module $M_{{\mathbb {Q}}}$ of rank 1.

We want to prove that (this would prove that $M_{(\ell )}$ is a free right ${\Lambda }_{(\ell )}$-module of rank 1 with the generator $f_{\ell }$). For each $x \in M_{(\ell )}$ there exists a unique $\lambda \in {\Lambda }_{\ell }$ with $x= f\lambda$. We need to prove that $\lambda \in {\Lambda }_{(\ell )}$. Notice that $x \in M_{(\ell )}\subset M_{{\mathbb {Q}}}$. Since $f_{\ell }$ is a generator of the free ${\Lambda }_{{\mathbb {Q}}}$-module $M_{{\mathbb {Q}}}$, there exists exactly one $\mu _0 \in {\Lambda }_{{\mathbb {Q}}}$ such that $x=f \mu _0$. We get the equalities $f \mu _0=x=f \mu$ in .

Since $f_{\ell }$ is a generator of the free -module , we get $\mu =\mu _0$. Since ${\Lambda }_{(\ell )}$ coincides with intersection of ${\Lambda }_{\ell }$ and ${\Lambda }_{{\mathbb {Q}}}$ in , we conclude that $\mu =\mu _0 \in {\Lambda }_{(\ell )}$ and therefore . This implies that $M_{(\ell )}$ is a free right ${\Lambda }_{(\ell )}$-module of rank 1, which proves (iii).$\square$

Corollary 3.3

Let M be a free commutative group of finite positive rank that is provided with a structure of a right ${\Lambda }$-module. Then M is a locally free ${\Lambda }$-module of rank 1 if and only if the right ${\Lambda }_{(\ell )}$-module $M_{(\ell )}$ is free of rank 1 for all primes $\ell$.

Proof

Clearly, if $M_{(\ell )}$ is a free right ${\Lambda }_{(\ell )}$-module of rank 1 then the right ${\Lambda }_{\ell }$-module $M_{\ell }$ is free of rank 1. The converse follows from Theorem 3.2 (iii).$\square$

Remark 3.4

Suppose that ${\Lambda }$ is an order in a number field E, i.e., ${\Lambda }$ is a finitely generated over ${\mathbb {Z}}$ a subring (with 1) of E such that ${\Lambda }_{{\mathbb {Q}}}=E$. Let M be a finitely generated ${\Lambda }$-submodule in E, i.e., a free commutative additive (sub)group of finite rank in E such that . In particular, $M_{{\mathbb {Q}}}=E$ is a free $E={\Lambda }_{{\mathbb {Q}}}$-module of rank 1.

• (i) If ${\Lambda }$ is the ring of all integers in E then it is a Dedekind ring and each of its localizations ${\Lambda }_{(\ell )}$ is a Dedekind ring with finitely many maximal ideals and therefore is a principal ideal domain [8, Chapter III, Proposition 2.12, p. 93]. This implies that $M_{(\ell )}$ is a free ${\Lambda }_{(\ell )}$-module, whose rank is obviously 1. By Corollary 3.3, M is locally free of rank 1.

• (ii) Suppose that E is a quadratic field. We do not impose any restrictions on ${\Lambda }$ but instead assume that . Then it is known [1, Lemma 2, p. 55] that for each prime $\ell$ there is a nonzero ideal $\mathfrak {J}\subset {\Lambda }$ such that the order of the finite quotient ${\Lambda }/\mathfrak {J}$ is prime to $\ell$ and the ${\Lambda }$-modules M and $\mathfrak {J}$ are isomorphic. This implies that the ${\Lambda }_{(\ell )}$-module $J_{(\ell )}={\Lambda }_{(\ell )}$ is free and therefore the ${\Lambda }_{(\ell )}$-module $M_{(\ell )}$ is also free and its rank is obviously 1. By Corollary 3.3, M is locally free of rank 1.

4 Tensor products

Now we are going to use Theorem 3.2, in order to construct abelian varieties over K that are almost isomorphic to a given A. Notice that our are a rather special naive case of powerful Serre’s tensor construction [3, Section 7], [2, Section 1.7.4].

Suppose we are given a free commutative group M of finite (positive) rank that is provided with the structure of a right locally free -module of rank 1. Let $F_2$ be a free right ${\Lambda }$-module of rank 2. It follows from Theorem 3.2 (i) that there is an endomorphism $\gamma {:} F_2 \rightarrow F_2$ of the right ${\Lambda }$-module $F_2$ such that $\gamma ^2=\gamma$ and whose image is isomorphic to M. Notice that is the matrix algebra ${\mathbb {M}}_2({\Lambda })$ of size 2 over ${\Lambda }$. So, the idempotent

where . Let us define the K-abelian (sub)variety

Clearly, B is a direct factor of $A^2$. More precisely, if we consider the K-abelian (sub)variety then the natural homomorphism , $(x,y) \mapsto x+y$ of abelian varieties over K is an isomorphism, i.e., . This implies that the right -module coincides with

and therefore the right -module is canonically isomorphic to $\gamma (F_2)=M' \cong M$. It also follows that for every prime $\ell$

Theorem 4.1

Let M be a free commutative group of finite rank which is a right locally free $\mathrm{End}(A)$-module of rank 1. Let us consider the abelian variety over K. Then:

• (i) A and B are isogenous over K.

• (ii) The right -module is isomorphic to M.

• (ii) A and B are almost isomorphic.

Proof

We have already seen that , which proves (ii). Since the right -module is free of rank 1, the same is true for the right -module . By Lemma 2.1, and there exists a -dimensional abelian K-subvariety $B_0\subset B$ such that A and $B_0$ are isogenous over K and

We claim that $B=B_0$. Indeed, if $B_0 \ne B$ then, by the Poincaré Complete Reducibility theorem [7, Theorem 6, p. 28], there is an “almost complimentary” abelian K-subvariety $B_1\subset B$ of positive dimension such that the intersection $B_0\cap B_1$ is finite and $B_0+B_1=B$. It follows from ($\star \star \star$) that . However, $B_1\subset B \subset A^2$ is an abelian K-subvariety of $A^2$ and therefore there is a surjective homomorphism and therefore there exists a nonzero homomorphism $A\rightarrow B_1$. This is a contradiction, which proves that $B=B_0$, the right -module is isomorphic to M, and A and B are isogenous over K. In particular, . This proves (i).

Let $\ell$ be a prime. Since is a free right -module of rank 1, is a free right -module of rank 1. Let us choose a generator of the module . Thesurjection is defined by a certain pair of homomorphisms $\phi _1, \phi _2{:} A\rightarrow B$, i.e.,

$$\begin{aligned} \gamma (x_1,x_2)=\phi _1(x_1)+\phi _2(x_2) \qquad \text {for all}\quad (x_1,x_2) \in A^2. \end{aligned}$$

Since $\phi$ is a generator, there are such that

$$\begin{aligned} \phi _1=\phi u_1, \qquad \phi _2=\phi u_2 \end{aligned}$$

in . It follows that

By ($\star \star$), . This implies that $\phi$ induces a surjective homomorphism . Since finite groups and have the same order, $\phi$ induces an isomorphism . This implies that does not contain points of order $\ell$ and therefore is an isogeny of degree prime to $\ell$. This proves (iii).$\square$

Corollary 4.2

Suppose that for each $i=1,2$ we are given a commutative free group $M_i$ of finite positive rank provided with the structure of a right locally free -module of rank 1. Then abelian varieties and are isomorphic over K if and only if the -modules $M_1$ and $M_2$ are isomorphic.

Proof

By Theorem 4.1 (ii), the right -module is isomorphic to $M_i$. Now the result follows from Theorem 4.1 (iii) combined with Corollary 2.6.$\square$

Corollary 4.3

Let A and B be abelian varieties over K of positive dimension. Suppose that the Galois modules $T_{\ell }(A)$ and $T_{\ell }(B)$ are isomorphic for all primes $\ell$. Then abelian varieties B and are isomorphic over K.

Proof

By Theorem 4.1 (ii), the right -module is isomorphic to . Now the result follows from Theorem 4.1 (iii) combined with Corollary 2.6.$\square$

Remark 4.4

Let $g \geqslant 2$ be an integer and a g-dimensional abelian variety A is a product where $A_1$ and $A_2$ are abelian varieties of positive dimension over K with . Then . Suppose that for each $i=1,2$ we are given a commutative free group $M_i$ of finite positive rank provided with the structure of a right locally free -module of rank 1.

Then the direct sum becomes a right locally free module of rank 1 over the ring .

There is an obvious canonical isomorphism between abelian varieties and over K.

For example, we may take as $A_2$ (for a suitable number field K) an elliptic curve such that is the ring of integers in an imaginary quadratic field with class number $>1$ while $A_1$ is a -dimensional principally polarized abelian variety with

(If $g>2$ then one may take as $A_1$ the $(g-1)$-dimensional jacobian of the hyperelliptic curve , see [13].) Clearly, all $\overline{K}$-endomorphisms of A are defined over K; in particular, $A_1$ is absolutely simple. Let us take $M_1={\mathbb {Z}}$. Clearly, . Actually, every $\overline{K}$-homomorphism between $A_1$ and $A_2$ is 0. Let $M_2$ be a non-principal ideal in . Then the elliptic curves $A_2$ and are almost isomorphic but are not isomorphic over K and even over $\overline{K}$. This implies that is almost isomorphic over K but is not isomorphic to over $\overline{K}$. Notice that both A and are principally polarized, since $A_1$ is principally polarized while both $A_2$ and are elliptic curves.

Remark 4.5

See last section of [15] for examples of almost isomorphic but not isomorphic elliptic curves over finite fields.