By Fred Diamond

This ebook introduces the idea of modular types, from which all rational elliptic curves come up, with an eye fixed towards the Modularity Theorem. dialogue covers elliptic curves as complicated tori and as algebraic curves; modular curves as Riemann surfaces and as algebraic curves; Hecke operators and Atkin-Lehner thought; Hecke eigenforms and their mathematics homes; the Jacobians of modular curves and the Abelian kinds linked to Hecke eigenforms. because it offers those principles, the ebook states the Modularity Theorem in a number of varieties, concerning them to one another and relating their purposes to quantity idea. The authors imagine no historical past in algebraic quantity conception and algebraic geometry. workouts are included.

**Quick preview of A First Course in Modular Forms (Graduate Texts in Mathematics, Vol. 228) PDF**

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**Extra info for A First Course in Modular Forms (Graduate Texts in Mathematics, Vol. 228)**

This is often the union of 2 conjugacy sessions below Γ0 (N ) (Exercise three. 7. 1(c)), so the variety of interval 2 (or interval three) elliptic issues of Γ0 (N ) is the variety of prolonged conjugacy periods of order four (or order 6) components of Γ0 (N ). The prolonged conjugacy category of γ lower than Γ0± (N ) isn't really more often than not the union of the conjugacy periods of γ and γ −1 below Γ0 (N ) (see workout three. 7. 1(d)). Counting the elliptic issues of Γ0 (N ) by way of counting those conjugacy sessions is completed within the comparable atmosphere because the facts of Proposition 2. three. three. reflect on elliptic issues τ of interval three with Γ0 (N )τ = γ , γ 6 = I. enable µ6 = e2πi/6 be the complicated 6th root of team spirit and permit A = Z[µ6 ], a vital excellent area. each one prolonged conjugacy classification will correspond to an excellent J of A such that A/J ∼ = Z/N Z as an Abelian workforce, {αγα−1 : α ∈ Γ0± (N )} ←→ Jγ . to build the perfect Jγ from γ, word that the lattice L = Z2 is an A-module with the multiplication γ : A × L −→ L given by means of 94 three measurement formulation (a + bµ6 ) γ l = (aI + bγ)l for a, b ∈ Z and l ∈ L. Deﬁne L0 (N ) = {[ xy ] ∈ L : y ≡ zero (mod N )}, a subgroup of L such that the quotient L/L0 (N ) is isomorphic to Z/N Z. Then L0 (N ) is usually an Asubmodule of L because γ ∈ Γ0 (N ) (Exercise three. 7. 2), making L/L0 (N ) an Amodule. permit Jγ be its annihilator, Jγ = Ann(L/L0 (N )) = {a + bµ6 ∈ A : (a + bµ6 ) γ L/L0 (N ) = 0}. ∼ Z/N Z as wanted. ∼ A γ L/L0 (N ) = L/L0 (N ) = therefore A/Jγ = to work out that the proper Jγ relies purely at the prolonged conjugacy type of γ, allow γ = αγα−1 with α ∈ Γ0± (N ). Then γ is usually an order-6 section of Γ0 (N ), and L is additionally an A-module linked to γ with multiplication (a + bµ6 ) γ l = (aI + bγ )l. permit L denote L because the A-module linked to γ during this type, and equally for L0 (N ). For any a, b ∈ Z and l ∈ L compute that α((a + bµ6 ) γ l) = α(aI + bγ)l = (aI + bγ )αl = (aI + bµ6 ) γ αl. in view that αγ = γ α ∼ This exhibits that multiplication by way of α supplies an A-module isomorphism L −→ L , and because αL0 (N ) ⊂ L0 (N ) simply because α ∈ Γ0± (N ) this induces an isomorphism ∼ of quotients L/L0 (N ) −→ L /L0 (N ). Compute that for any j = a + bµ6 ∈ A, j ∈ Jγ ⇐⇒ j γ L/L0 (N ) = 0L/L0 (N ) ⇐⇒ α(j γ L/L0 (N )) = 0L /L0 (N ) ⇐⇒ j γ α(L/L0 (N )) = 0L /L0 (N ) ⇐⇒ j γ L /L0 (N ) = 0L /L0 (N ) ⇐⇒ j ∈ Jγ , exhibiting that certainly Jγ relies in simple terms at the prolonged conjugacy category. Conversely to the map from conjugacy sessions to beliefs, commence now from an incredible J of A with A/J ∼ = Z/N Z. through the constitution theorem for modules over a imperative perfect area back, there exists a foundation (u, v) of A over Z such that (u, N v) is a foundation of J over Z. on account that (u, v) is a foundation of A over Z, (µ6 u, µ6 v) = (ua + vc, ub + vd) for a few integers a, b, c, d, or µ6 (u, v) = (u, v)γJ , for a few γJ = a b c d ∈ M2 (Z). So the matrix γJ has order 6 and has an analogous minimum polynomial X 2 −X +1 as µ6 , displaying that γJ ∈ SL2 (Z) (Exercise three. 7. 3(a)). and because u ∈ J and J is an A-submodule of A, the formulation µ6 u = ua + vc exhibits that c ≡ zero (mod N ), i. e. , γJ ∈ Γ0 (N ).

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