Gross-zagier Revisited

ly isomorphic as rings are conjugate in B. 146 BRIAN CONRAD (APPENDIX BY W. R. MANN) Remark A.10. The order ( A mn A A ) in (2) is the intersection of the maximal orders M2(A) and γnM2(A)γ−1 n = ( A mn m−n A ) , where γn = ( 0 πn 1 0 ) . This example is called a standard Eichler order. Proof. For the first part, it suffices to show that the set R of elements of B integral over R is an order. That is, we must show R is finite as an A-module and is a subring of B. Note that R is stable under the involution b 7→ b. The key to the subring property is that if b ∈ B has N(b) ∈ A, then Tr(b) ∈ A. Indeed, F [b] is a field on which the reduced norm and trace agree with the usual norm and trace (relative to F ), and by completeness of A we know that the valuation ring of F [b] is characterized by having integral norm. Thus, to show that R is stable under multiplication we just need that if x, y ∈ R then N(xy) ∈ A. But N(xy) = N(x)N(y). Meanwhile, for addition (an issue because noncommutativity does not make it evident that a sum of integral elements is integral), we note that if x, y ∈ R then N(x + y) = (x + y)(x + y) = N(x) + N(y) + Tr(xy). But this final reduced trace term lies in A because xy ∈ R. Hence, R is a subring of B. In particular, R is an A-submodule since A ⊆ R. To show that R is A-finite, we may pick a model for B as in Example A.2, and may assume i = e, j = f ∈ A. Thus, i, j ∈ R, so ij ∈ R. For any x ∈ R, we have x, xi, xj, xij ∈ R. Taking reduced traces of all four of these elements and writing x = c + c1i + c2j + c3ij for c, c1, c2, c3 ∈ F , we get x ∈ 1 2 A + 1