By Wiles A.J.

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Example text

Lemma (Ribet). Suppose that (M, N ) = 1. If M is odd then TM = T. If M is even then TM has finite index in T equal to a power of 2. As the rings are finitely generated free Z-modules, it suffices to prove that TM ⊗ Fl → T ⊗ Fl is surjective unless l and M are both even. The claim follows from 1. TM ⊗ Fl → TM/p ⊗ Fl is surjective if p|M and p lN . 2. Tl ⊗ Fl → T ⊗ Fl is surjective if l 2N. Proof of 1. Let A denote the Tate module Tal (J1 (N )). Then R = TM/p ⊗ Zl acts faithfully on A. Let R = (R ⊗ Ql ) ∩ EndZl A and choose d so that Zl ¯ ld R ⊂ lR.

We define L∗ to be the orthogonal complement of L under the perfect pairing (local Tate duality) H 1 (Qq , X ∗ ) → Qp /Zp H 1 (Qq , X) × q∈Σ q∈Σ where X ∗ = Hom(X, µp∞ ). Let λX : H 1 (QΣ /Q, X) → H 1 (Qq , X) q∈Σ be the localization map and similarly λX ∗ for X ∗ . Then we set −1 1 ∗ ∗ HL1 (QΣ /Q, X) = λ−1 X (L), HL∗ (QΣ /Q, X ) = λX ∗ (L ). The following result was suggested by a result of Greenberg (cf. [Gre1]) and is a simple consequence of the theorems of Poitou and Tate. Recall that p is always assumed odd and that p ∈ Σ.

Also J1 (N p)´em/O m-divisible group over O. This makes sense because J1 (N p)m does extend to 484 ANDREW JOHN WILES a p-divisible group over O (again by a theorem of Deligne and Rapoport [DR] and because ∆(p) is nontrivial mod m). 2) when we use the main theorem of Tate ([Ta]) since D0 and DE clearly correspond to ordinary p-divisible groups. Now the q-expansion principle implies that dimFp X[m ] ≤ 1 where X = {H 0 (Σµ1 , Ω1 ) ⊕ H 0 (Σ´e1t , Ω1 )} and m is defined by embedding T/m → Fp and setting m = ker : T⊗Fp → Fp under the map t ⊗ a → at mod m.

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