MODULAR FORMS-page77 - 73 Proof. Let Γ be a normal...

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Unformatted text preview: 73 Proof. Let Γ be a normal subgroup of finite index in Γ(1) which is contained in Γ. It always can be found by taking the intersection of conjugate subgroups g- 1 · Γ · g,g ∈ Γ(1). We first apply Lemma 7.3 to the case when B = M (Γ ) ,A = M (Γ(1)). Since A ∼ = C [ T 1 ,T 2 ] is finitely generated, B is finitely generated. It follows easily from (7.15) that the field of fractions of B is a finite extension of the field of fractions of A of degree equal to the order of the group Γ / Γ . Next we apply the same lemma to the case when B = M (Γ ) ,A = M (Γ) . Then B is finitely generated, hence A is finitely generated. Corollary 7.3. The linear spaces M k (Γ) are finite-dimensional. Proof. Let f 1 ,...,f k be a set of generators of the algebra M k (Γ). Writing each f i as a linear combination of modular forms of different weights, and then adding to the set of generators all the summands, we may assume that M k (Γ) is generated by finitely many modular forms f i ∈ M k i...
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This note was uploaded on 01/08/2012 for the course MATH 300 taught by Professor Ontonkong during the Fall '09 term at SUNY Stony Brook.

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