Corollary 53 If A ≃ E 1 × . . . × E n , where E 1 , . . . , E n ∈ Isog(E/K), thenf A = lcm(f E1 , . . . , f En ).Pro<strong>of</strong>. By Proposition 36 there is an E 0 ∈ Isog(E/K) such that f E0 = lcm(f E1 , . . . , f En ).Thus, if π i : E 0 → E i is an isogeny, then H i = Ker(π i ) is an ideal subgroup <strong>of</strong> E 0 by(46) (because f Ei |f E0 by construction). Thus, H i = H(I i ), for some R 0 -ideal I i , whereR 0 = End(E 0 ). Note that by (45) we have that End(E i ) ≃ E(H i ) = (I i : I i ) = R(I i ),and so f Ei = f R(Ii ).Put H = H 1 × . . . × H n . Then A H ≃ E H1 × . . . × E Hn ≃ E 1 × . . . × E n ≃ A.By Proposition 41 we have H = H(I), where I = (I 1 | . . . |I n ). Thus, by (95) wehave End(A) ≃ End(A H ) ≃ E(H) ≃ End R0 (M) op , where M = I 1 ⊕ . . . ⊕ I n ⊂ R0 n ,and so Z(End(A)) ≃ Z(End R0 (M)) op = (M : M) F = R F (M) by (94). Now clearly(M : M) F = R(I 1 ) ∩ . . . ∩ R(I n ). Since the latter has conductor lcm(f R(I1 ), . . . , f R(In ))by (37), the assertion follows.We next want to characterize the ideal subgroups <strong>of</strong> E n . The following characterizationshows that the necessary condition <strong>of</strong> Corollary 11 is also sufficient. Moreover,this result can also be viewed as a generalization <strong>of</strong> the criterion (46) <strong>of</strong> Theorem 18.Theorem 54 Let E/K is a <strong>CM</strong> <strong>elliptic</strong> curve, and let H be a finite subgroup scheme<strong>of</strong> E n . Then:(97)H is an ideal subgroup <strong>of</strong> E n ⇔ Z(End(E n )) ⊂ E(H) ⇔ f E(H) | f E .Before proving this, we note the following immediate application.Corollary 55 If A ∼ E n is an abelian variety which is isogenous to E n with f A |f E ,then there exist <strong>elliptic</strong> <strong>curves</strong> E 1 , . . . , E n ∈ Isog + (E/K) such that A ≃ E 1 × . . . × E n .Pro<strong>of</strong>. Let π : E n → A be an isogeny, and let H = Ker(π). Then E n H ≃ A andEnd(A) ≃ E(H), so f A = f E(H) . It thus follows from the hypothesis and (97) that His an ideal subgroup. Thus, H = H(I), for some regular left ideal I <strong>of</strong> End(E n ), andhence by (95) we have A ≃ A H ≃ E 1 × . . . × E n , where E i := E H(Ii ), for suitable idealsI i <strong>of</strong> End(E). Since H(I i ) is an ideal subgroup <strong>of</strong> E, we have by (46) that f Ei |f E , andso E i ∈ Isog + (E/K), as claimed.Another application <strong>of</strong> Theorem 54 is the following result which can be viewedas a solution <strong>of</strong> the isomorphism problem which was posed in the introduction <strong>of</strong>this paper. Note that it applies to an arbitrary tuple E 1 , . . . , E n /K <strong>of</strong> isogenous <strong>CM</strong><strong>curves</strong> because by Proposition 36 we can always find an <strong>elliptic</strong> curve E/K such thatE 1 , . . . , E n ∈ Isog + (E/K).42
Corollary 56 Let E/K be a <strong>CM</strong> <strong>elliptic</strong> curve and E 1 , . . . , E n ∈ Isog + (E/K). IfA ∼ E n is an abelian variety isogenous to E n , then the following conditions areequivalent:(98)(99)(100)A ≃ E 1 × . . . × E n ;I E n(A) ≃ (I E (E 1 ) : . . . : I E (E n )) and f A |f E ;End(E) n ⊗ End(E n ) I E n(A) ≃ I E (E 1 ) ⊕ . . . ⊕ I E (E n ) and f A |f E .Pro<strong>of</strong>. Fix isogenies π : E n → A and π i : E → E i , for 1 ≤ i ≤ n, and put H = Ker(π)and H i = Ker(π i ). Also, put A ′ = E 1 × . . . × E n . Since f Ei |f E by hypothesis, weknow by (46) that H i = H(I(H i )) is an ideal subgroup <strong>of</strong> E, so H ′ := H 1 × . . . × H nis an ideal subgroup <strong>of</strong> E n by Proposition 42. Furthermore, we have that I E n(A ′ ) ≃I(H ′ ) = (I(H 1 ) : . . . : I(H n )) ≃ (I E (E 1 ) : . . . : I E (E n )) by (87).Now suppose that (98) holds. Then by Corollary 53 we have that f A = f E1 ×...×E n=lcm(f E1 , . . . , f En )|f E . Moreover, I E n(A) ≃ I E n(A ′ ) ≃ (I E (E 1 ) : . . . : I E (E n )), and so(99) holds. Conversely, if (99) holds, then H is an ideal subgroup <strong>of</strong> E n by Theorem54, and so A ≃ A ′ by (21). Thus, conditions (98) and (99) are equivalent.To prove the equivalence <strong>of</strong> conditions (99) and (100), consider the submodulesM 1 := End(E) n ⊗ End(E n ) I E n(A) and M 2 = I E (E 1 ) ⊕ . . . ⊕ I E (E n ). Then by definitionwe have T −1E,n (I n(M 2 )) = (I E (E 1 ) : . . . : I E (E n )) and by Morita equivalence (cf. thepro<strong>of</strong> <strong>of</strong> Proposition 40) we have T −1E,n (I n(M 1 )) ≃ I E n(A). Thus, since M 1 ≃ M 2 (asEnd(E)-modules) ⇔ T −1E,n I n(M 1 ) ≃ T −1E,n I n(M 2 ) (as End(E n )-modules) by Proposition40, the equivalence <strong>of</strong> conditions (99) and (100) follows.As first step towards establishing Theorem 54, we prove the following result.Proposition 57 Let H be a finite subgroup scheme <strong>of</strong> E n , where E/K is a <strong>CM</strong> <strong>elliptic</strong>curve. Then H is an ideal subgroup <strong>of</strong> E n if and only if(101)A H ≃ E 1 × . . . × E n , where E i ∈ Isog + (E/K), for 1 ≤ i ≤ n.Pro<strong>of</strong>. Suppose first that H = H(I) is an ideal subgroup. Then by (95) we haveEH n ≃ E 1 × . . . × E n where E i = E H(Ii ). Since H(I i ) is an ideal subgroup <strong>of</strong> E, wehave by (46) that f Ei |f E , i.e. that E i ∈ Isog + (E/K).Conversely, suppose that (101) holds, and fix an isogeny π i : E → E i for 1 ≤i ≤ n. Then by (46) we know that H i := Ker(π i ) is an ideal subgroup <strong>of</strong> E, soH ′ := H 1 × . . . × H n is an ideal subgroup <strong>of</strong> E n by Proposition 42. Since E i ≃ E Hi ,we have by the hypothesis (101) that EH n ≃ E H 1× . . . × E Hn ≃ EH n ′, and so it followsfrom (22) that H is also an ideal subgroup <strong>of</strong> E n .As we shall see presently, Theorem 54 follows easily from the following result.43
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