Download e-book for iPad: Hilbert modular surfaces by Gerard Van Der Geer

By Gerard Van Der Geer

ISBN-10: 0387176012

ISBN-13: 9780387176017

Over the past 15 years vital effects were accomplished within the box of Hilbert Modular kinds. although the most emphasis of this publication is at the geometry of Hilbert modular surfaces, either geometric and mathematics features are handled. An abundance of examples - in reality an entire bankruptcy - completes this useful presentation of the topic. This Ergebnisbericht will quickly develop into an indispensible device for graduate scholars and researchers during this box.

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We shall show that M/I n M is flat over A/I n . For 0 ≤ i ≤ n − 1 we have a commutative diagram I i+1 /I n ⊗A M −−−−→ I i /I n ⊗A M −−−−→ I i /I i+1 ⊗A M −−−−→ 0       αi+1 γi α i 0 −−−−→ I i+1 M/I n M −−−−→ I i M/I n M −−−−→ I i M/I i+1 M −−−−→ 0. 8 Local criteria of flatness 3 with exact rows. It follows from the assumptions that γi is an isomorphism for i = 0, 1, . . By descending induction on i, starting with αn = 0, it follows that αi is an isomorphism for i = 0, . . , n. In particular we have that α1 : I/I n ⊗A M = I/I n ⊗A/I n M/I n M → IM/I n M is an isomorphism.

Given a map ϕ: F → G of finite complexes of A-modules of finite length. Then we have that A (ϕ) → = A (G) − A (F ). Proof. ). 6) Proposition. Given a commutative diagram 0 −−−−→ F −−−−→ F −−−−→ F −−−−→ 0       ϕ ϕ ϕ 0 −−−−→ G −−−−→ G −−−−→ G of complexes of A-modules. If two of the lengths defined, then the third is, and we have that A (ϕ) → = A (ϕ )+ A (ϕ −−−−→ 0 A (ϕ ), A (ϕ) and A (ϕ ) are ). Proof. That the third length is defined when the two others are is an immediate consequence of the properties of length of modules.

9 Generic flatness 2 Proof. Let K be the quotient field of A. Then B ⊗A K is a K-algebra of finite type and N ⊗A K is a B ⊗A K-module of finite type. Let s = dim supp(N ⊗A K) be the Krull dimension of the support of N ⊗A K in Spec(B ⊗A K). We shall prove the Lemma by induction on s. When s < 0 we have that N ⊗A K = 0. Since K is flat over A we have that N ⊗A K = 0 implies that each element in N has A torsion, and since N is a finitely generated B-module there is an element f ∈ A such that f N = 0.

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Hilbert modular surfaces by Gerard Van Der Geer

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Categories: Algebraic Geometry