Products of CM elliptic curves - Universität Duisburg-Essen

Corollary 11 If H is an ideal subgroup of A, then Z(R) ⊂ E(H).Proof. By hypothesis, H = H(I), for some regular left R-ideal I, and so (I : I) ⊂ E(H)by (30). Now Z(R) ⊂ (I : I) because I is a left R-ideal (for z ∈ Z(R) ⇒ Iz = zI ⊂ I),and so the assertion follows.The following result, which is a variant of a result of [26], p. 534, shows that thesubgroup scheme associated to a product of ideals has a natural interpretation in termsof composition of maps.Proposition 12 Let I be a regular left R-ideal and let J be a regular left R ′ -ideal,where R ′ = E(H(I)). Then IJ is a regular left R-ideal and(34)H(IJ) = Ker(π H(Φ−1I (J)) ◦ π H(I)),where Φ I = Φ H(I) : End(A H(I) ) → ∼ R ′ and π H(Φ−1I (J)) : A H(I) → (A H(I) ) H(Φ−1Icanonical quotient map.(J))is theProof. Put H = H(I). Since J ⊂ E(H) ⊂ (I(H) : I(H)) by (29), we have IJ ⊂I(H)J ⊂ I(H) ⊂ R, and hence IJ is an R-ideal. Moreover, IJ is regular because byhypothesis ∃α ∈ I ∩ ˜R × and β ∈ J ∩ ˜R × , and so αβ ∈ IJ ∩ ˜R × , which means that IJis regular.To prove (34), note first that it follows from (25) that nΦ −1In = n H , π = π H , and π ′ = π H ′ , and so Φ−1I(35)(J) = πJπ ′ , where(J)π = πJ. From this it follows thatKer(πf) = ∩ g∈I Ker(gf), ∀f ∈ J,where we view πf ∈ Hom(A, A H ) since πf = f 1 π for some f 1 ∈ Φ −1I(J) ⊂ End(A H ).To verify (35), write f = f 2 /n with f 2 ∈ R and n ∈ N. Then[n] −1 (Ker(πf)) = Ker(πf 2 ) = f −12 (Ker(π)) = f −12 (∩ g∈I Ker(g))= ∩ g∈I f −12 (Ker(g)) = ∩ g∈I Ker(gf 2 ) = [n] −1 (∩ g∈I Ker(gf)),the latter because gf 2 = gfn and gf ∈ R, for all g ∈ I. From this, equation (35)follows because [n] is faithfully flat.Using (35), we therefore obtain thatKer(π H(Φ−1Iwhich proves (34).(J)) π) = π−1 (Ker(π H(Φ−1(J)) )) = π−1 (∩ f∈Φ−1(J) Ker(f))I= ∩ f∈Φ−1I (J) Ker(fπ)) = ∩ f ′ ∈Φ −1I (J)π Ker(f ′ )) = ∩ f ′ ∈πJKer(f ′ ))= ∩ f∈J Ker(πf)) = ∩ f∈J f −1 (Ker(π)) = ∩ f∈J f −1 (∩ g∈I Ker(g))= ∩ f∈J ∩ g∈I Ker(gf) = H(IJ),12I