Products of CM elliptic curves - Universität Duisburg-Essen

ecause by part (a) we know that r E Hkis surjective if and only if p ∤ f EH .Now suppose that H is an ideal subgroup scheme. Since f EH |f E by (46), it followsfrom (62) that we have equality in the second part of (58). Moreover, since H = H(I)for some ideal I, we see from (59) that H k = H(r k (I)) is also an ideal subgroup. PutR = End(E). Since r k (R) = End(E k ), we obtain from Corollary 23 that [R : I(H)] =|H| = |H k | = [r k (R) : I(H k )], and so we must have equality in the first part of (58) aswell.Finally, to prove that equality holds in (58) when (|H|, p) = 1, we first observethat although the map H ↦→ H k is not injective in general, we do have(63)H 1 ≤ H 2 ⇔ (H 1 ) k ≤ (H 2 ) k , if (|H 1 |, p) = 1.To see this, recall first that we also have a reduction map r E(K)k: E(K) → E k (k)on the group of rational points and that by base-change this extends to a map r k :E(K) → E(k) whose kernel has no non-trivial points of order prime to p = char(k).We thus have an isomorphismr (p)k: E(K) (p)tor∼→ E k (k) (p)tor,where E(K) (p)tor denotes the group of torsion points in E(K) of order prime to p, andE k (k) (p)tor is defined similarly. Now if H is a finite group scheme of E with (|H|, p) = 1,then ˜H is an etale group scheme, and so we can identify H k ⊗ k and H ⊗ K withsubgroups of E k (k) and of E(K) of the same order and we have r (p)k(H ⊗K) = H k ⊗k.From this, the assertion (63) follows immediately, provided that also (p, |H 2 |) = 1.To prove it without this hypothesis, note that we can reduce the assertion to this caseas follows. Put n 1 = |H 1 | = |(H 1 ) k |. Since H 1 ≤ Ker([n 1 ]), we see that H 1 ≤ H 2 ⇔H 1 ≤ H 2 [n 1 ] := H 2 ∩ Ker([n 1 ])), and similarly, (H 1 ) k ≤ (H 2 ) k ⇔ (H 1 ) k ≤ (H 2 ) k [n 1 ] =(H 2 [n 1 ]) k . Thus, since p ∤ |H 2 [n 1 ]|, it follows that (63) is true.We are now ready to prove that equality holds in (58) when p ∤ |H|. Indeed,since f E(H) |n 2 f E by (27), it follows from (62) that equality holds in the second partof (58). To prove this also for the first part, let g ∈ I(H k ), so H k ≤ Ker(g). Sincer k : End(E) → End(E k ) is an isomorphism, there exists a unique g 1 ∈ End(E) suchthat r k (g 1 ) = g. By (60) we know that Ker(g 1 ) k = Ker(g), and so it follows from (63)that H ≤ Ker(g 1 ). Thus g 1 ∈ I(H) and so g = r k (g 1 ) ∈ r k (I(H)). This shows thatequality holds in the first part of (58) as well.3.2 The case K = CIn the case that K = C, every elliptic curve E/C has an analytic description, i.e. thereexists a lattice L ⊂ C and an isomorphism of compact Riemann surfaces E C ≃ C/L,where E C denotes the compact Riemann surface associated to the (algebraic) curve22