Gross--Zagier Revisited
Gross--Zagier Revisited
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格罗斯--扎吉尔再访
DOI：
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发表时间：
2004
期刊：
影响因子：
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通讯作者：
W. Mann
中科院分区：
文献类型：
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作者：
B. Conrad;W. Mann
当环在B中共轭时，它们同构。146 BRIAN CONRAD（W. R.注释A.10.（2）中的阶（A mn A A）是最大阶M2（A）和γ nM 2（A）的交集γ−1 n =（A mn m−n A），其中γn =（0 πn 1 0）。这个例子被称为标准艾希勒顺序。证据对于第一部分，只要证明R上的B积分的元素的集合R是一个阶就足够了。也就是说，我们必须证明R作为A-模是有限的，并且是B的子环。注意，R在对合B 7→ B下是稳定的。子环性质的关键是：如果B ∈ B有N（B）∈ A，则Tr（B）∈ A.事实上，F [B]是一个域，其上的约化范数和迹与通常的范数和迹（相对于F）一致，并且通过A的完备性我们知道F [B]的赋值环的特征在于具有整范数。因此，为了证明R在乘法下是稳定的，我们只需要如果x，y ∈ R，则N（xy）∈ A。但N（xy）= N（x）N（y）。同时，对于加法（一个问题，因为非交换性并不表明整元的和是整数），我们注意到如果x，y ∈ R，则N（x + y）=（x + y）（x + y）= N（x）+ N（y）+ Tr（xy）。但是这个最终的约化迹项位于A中，因为xy ∈ R。因此，R是B的子环。特别地，R是A-子模，因为A <$R。为了证明R是A有限的，我们可以像例A.2那样为B选择一个模型，并假设i = e，j = f ∈ A。因此，i，j ∈ R，所以ij ∈ R。对于任意x ∈ R，我们有x，xi，xj，xij ∈ R。取这四个元素的约化迹，记为x = c + c1 i + c2 j + c3 ij，其中c，c1，c2，c3 ∈ F，我们得到x ∈ 1 2 A + 1
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