Products of CM elliptic curves - Universität Duisburg-Essen

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Lemma 13 If L 1 , L 2 ∈ Lat F are any two lattices of F , then(40)(L 1 : L 2 )L 2 = (R(L 1 ) : R(L 2 ))L 1 .Thus, if f i = f R(Li ) = [O F : R(L i )], for i = 1, 2, and if f = (f 1 , f 2 ), then we have(41)(L 1 : L 2 ) = [R(L 1 )R(L 1 ) : R(L 1 )]L 1 L −12 = f 1f L 1L −12 .Proof. Put R i = R(L i ) and R 0 = R 1 R 2 = R(L 1 L 2 ). Note that (L 1 : L 2 ) is anR 0 -module, for if c ∈ (L 1 : L 2 ) and r i ∈ R(L i ), then cr 1 r 2 ∈ (L 1 : L 2 ) becauser 1 cr 2 L 2 ⊂ r 1 cL 2 ⊂ r 1 L 1 ⊂ L 1 . In particular, (R 1 : R 2 ) is also an R 0 -module (becauseR(R i ) = R i ).Now let c ∈ (L 1 : L 2 ). Then (cL −11 L 2 )R 2 = cL −11 L 2 = cL 2 L −11 ⊂ L 1 L −11 = R 1 , socL 2 L −11 ⊂ (R 1 : R 2 ), and hence (L 1 : L 2 )L 2 L −11 ⊂ (R 1 : R 2 ). Thus (L 1 : L 2 )L 2 R 1 =(L 1 : L 2 )L −12 L 1 ⊂ (R 1 : R 2 )L 1 . But since (L 1 : L 2 ) is an R 1 -module (as R 1 ⊂ R 0 ), wehave that (L 1 : L 2 )L 2 R 1 = (L 1 : L 2 )L 2 , and so the left hand side of (40) is containedin the right hand side.To prove the other inclusion, let r ∈ (R 1 : R 2 ). Then (rL 1 L −12 )L 2 = rL 1 R 2 =rR 2 L 2 ⊂ R 1 L 1 = L 1 , so rl 1 L −12 ⊂ (L 1 : L 2 ), and hence (R 1 : R 2 )L 1 L −12 ⊂ (L 1 : L 2 ).Thus (R 1 : R 2 )L 1 R 2 = (R 1 : R 2 )L 1 L −12 L 2 ⊂ (L 1 : L 2 )L 2 , and so the other inclusion of(40) holds because (R 1 : R 2 )L 1 R 2 = (R 1 : R 2 )L 1 (since (R 1 : R 2 ) is an R 2 -module).This proves (40).The formula (41) follows immediately from (40) once we have shown that(42)(R(L 1 ) : R(L 1 )) = f 1f R(L 1)R(L 2 ) and f 1f = [R(L 1)R(L 2 ) : R(L 1 )].Indeed, multiplying (40) by L −12 , we obtain with (42) that (L 1 : L 2 ) = f 1R f 1R 2 L 1 L −12 =f 1L f 1L −12 , where the last equality follows from the obvious fact that L ±1i is an R i -module, for i = 1, 2.It thus remains to verify (42). For this, we first note that since f = f R0 by (39), wehave that [R 0 : R 1 ] = f 1f , which is the second equality of (42). Thus, f 1f R 0 is largestR 0 -module which is contained in R 1 , and hence (R 1 : R 2 ) ⊂ f 1f R 0 because (R 1 : R 2 ) isan R 0 -module which is contained in R 1 . On the other hand, since f 1f R 2 ⊂ f 1f R 0 ⊂ R 1 ,we have the opposite inclusion f 1f R 0 ⊂ (R 1 : R 2 ), which proves (42).Corollary 14 Every non-zero ideal of an order in F is a divisorial ideal.Proof. Let R be an order of F and let I be a nonzero R-ideal. Then I ∈ Lat FR ⊂ R(I). Applying (41) to L 1 = R and L 2 = I yieldsand(43)(R : I) = [R(I) : R]I −114
ecause here f 2 |f 1 , so f = f 2 and f 1= [R(I) : R]. Next, apply (41) to L f1 = R andL 2 = (R : I). Since R(L 2 ) = R(I −1 ) = R(I) by (43), we see that in this case (41)gives(R : (R : I)) = [R(I) : R]((R : I)) −1 = [R(I) : R][R(I) : R] −1 (I −1 ) −1 = ISince I ∗ = (R : (R : I)) by [5], p. 476, we thus have that I = I ∗ , and so I is a divisorialideal in the sense of Remark 7(d).We now apply the preceding results to abelian varieties.Proposition 15 Let A/K be an abelian variety such that ˜R = End 0 (A) is a quadraticfield. Then every non-zero ideal of R = End(A) is a kernel ideal, and hence we havefor any two non-zero ideals I, J of R that Φ I,J := Φ H(I),H(J) defines an isomorphism(44)Φ I,J := Φ H(I),H(J) : Hom(A H(I) , A H(J) )∼→(I : J).Moreover, we have that A H(I) ≃ A H(J) ⇔ I ≃ J (as R-modules).Proof. Since R is an order of ˜R = F , we know by Corollary 14 that every non-zeroideal of R is divisorial and hence a kernel ideal by Remark 7(d). This proves thefirst assertion, and hence the other assertions follow from Proposition 10 and from thediscussion after (20).Corollary 16 In the above situation, let R ′ be an order of ˜R with R ⊂ R ′ . Then thereexists an abelian variety A ′ /K which is isogenous to A/K such that End(A ′ ) ≃ R ′ .Proof. Take I = [R ′ : R]R ′ ⊂ R. Then I is a non-zero R-ideal with R(I) = R ′ . ThenA ′ := A H(I) is isogenous to A and End(A ′ ) ≃ (I : I) = R(I) = R ′ by (44).Remark 17 (a) Note that if we drop the hypothesis R ⊂ R ′ in Corollary 16, thenthe corresponding statement is in general no longer true; cf. §3.3 below.(b) We can use the above Corollary 16 to construct an abelian variety A ′ with afinite subgroup scheme H ′ such that E(H ′ ) ⊄ End(A ′ ). In particular, H ′ is not anideal subgroup by Corollary 11.Indeed, suppose there exists an abelian variety A/K such that R := End(A) ⊂ Fbut R ≠ O F . Then by Corollary 16 there is an R-ideal I such that A ′ := A H(I) satisfiesEnd(A ′ ) ≃ O F . Consider H ′ = Ker(π ′ H(I) ). Since π′ H(I) : A′ → A is an isogeny, wehave (A ′ ) H ′ ≃ A, so E(H ′ ) ≃ End(A) = R. Thus O F = End(A ′ ) ⊄ E(H ′ ), and so H ′is not an ideal subgroup of A ′ .15
Page 1: Products of CM elliptic curves1 Int
Page 7 and 8: Indeed, if If is a kernel ideal, th
Page 9 and 10: Indeed, by (18) we have I 1 f = I 2
Page 11 and 12: Proof. To prove (29), let h ∈ Hom
Page 13: 2.4 The quadratic caseWe now specia
Page 17 and 18: For this, put f = [R ′ : R], I =
Page 19 and 20: From (54) we can conclude that(55)d
Page 21 and 22: If H is any finite subgroup scheme
Page 23 and 24: E. As we shall see, it is very illu
Page 25 and 26: From this we see on the one hand th
Page 27 and 28: Proposition 30 Let L ∈ Lat F , wh
Page 29 and 30: (b) Deuring[10], p. 263. Note that
Page 31 and 32: integral quadratic formq L,α,β (x
Page 33 and 34: Corollary 39 Let E 1 /K and E 2 /K
Page 35 and 36: Proof. Write π i = π Hi : A → A
Page 37 and 38: (a) The order R is cyclic if and on
Page 39 and 40: for k = 1, . . . , n − 2. Moreove
Page 41 and 42: 4.3 Products of CM elliptic curvesW
Page 43 and 44: Corollary 56 Let E/K be a CM ellipt
Page 45 and 46: Case 2. char(K) = 0.As was mentione
Page 47 and 48: Let h A/K (X) ∈ Z[X] (respectivel
Page 49 and 50: Proof. To prove this, we shall use
Page 51 and 52: where xy denotes the quadratic form
Page 53 and 54: denote the set of isomorphism class

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