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The graded ring of level-one modular forms

Statement

The graded C-algebra M∗=⨁k evenMk is freely generated by E4 and E6: the substitution X↦E4, Y↦E6 defines an isomorphism of graded C-algebras C[X,Y]→M∗, where deg⁡X=4 and deg⁡Y=6. Equivalently, for every even k≥0 the forms E4aE6b with 4a+6b=k, a,b≥0, form a basis of Mk, and the cusp forms form the principal ideal S∗=ΔM∗; explicitly Sk=ΔMk−12 for all k≥0 (with Mk−12=0 for k<12).

Facts & Assumptions

Given: The graded spaces Mk,Sk and the Eisenstein forms E4,E6 with E4(∞)=E6(∞)=1, together with Δ=(E43−E62)/1728∈S12 of order one at the cusp and without zeros on H (Level-one modular forms and cusp forms, Eisenstein series are modular forms; their Fourier coefficients, The discriminant is a nonvanishing cusp form of weight 12, The modular discriminant and the j-invariant).

[F1]

For even k the monomials E4aE6b with 4a+6b=k are linearly independent in Mk, and their number equals dim⁡Mk from the dimension formula; every such monomial has value 1 at the cusp (The dimension of the space of level-one modular forms, Eisenstein series are modular forms; their Fourier coefficients).

[F2]

A nonzero weight-k form has a well-defined constant term f(∞) at the cusp, and Sk is exactly the kernel of f↦f(∞) (Level-one modular forms and cusp forms).

[F3]

Δ has vanishing order 1 at the cusp and no zeros on H; therefore f↦f/Δ maps Sk bijectively onto Mk−12 for every even k: it is holomorphic on H, has the correct weight, and at the cusp the order drops by exactly 1; conversely Δh∈Sk for h∈Mk−12 (The discriminant is a nonvanishing cusp form of weight 12, The modular discriminant and the j-invariant, The level-one valence formula, The zeros of E4 and E6 at the elliptic points).

[F4]

For every even k≥4 there is a monomial of weight k (and for k=0 the empty monomial 1); M2=0 and Mk is one-dimensional for k=0,4,6,8,10 (The dimension of the space of level-one modular forms).

Proof

technique · induction on the weight $k$ in steps of $12$
1.1baseF1F2F4givenalgebra

Base cases: for k=0, M0=C consists of constants and the empty monomial 1 has weight 0; for k=2, M2=0; for k=4,6,8,10 the space Mk is one-dimensional and contains the monomial E4,E6,E42,E4E6 respectively, whose value at the cusp is 1, so every f∈Mk equals f(∞) times that monomial. Hence in these degrees the monomials span Mk.

1.2ihgiven

Assume, as induction hypothesis, that for some even k≥12 the weight-(k−12) monomials span Mk−12; equivalently every element of Mk−12 is a polynomial in E4,E6.

2.1F2F3F4step 1.1step 1.2givenalgebra

Let f∈Mk and choose a monomial P=E4aE6b of weight k [F4]; then P(∞)=1 and f−f(∞)P∈Mk has zero constant term at the cusp, so f−f(∞)P∈Sk [F2]. By [F3] there is h∈Mk−12 with f−f(∞)P=Δh; by the induction hypothesis 1.2 the form h is a polynomial in E4,E6, so f=f(∞)P+Δh is a polynomial in E4,E6 (using Δ=11728(E43−E62)). Hence the monomials span Mk for every even k≥0.

3.1F1F3step 2.1discharge-induction∎

The graded algebra map φ:C[X,Y]→M∗ with φ(X)=E4, φ(Y)=E6 is well defined because E4aE6b∈M4a+6b and multiplication of modular forms adds weights. In degree k the source is spanned by the monomials of weight k, whose images are linearly independent and as numerous as dim⁡Mk [F1]; since they also span Mk by the induction result, φ is an isomorphism in each degree and hence an isomorphism of graded algebras. Therefore the monomials form a basis and E4,E6 are algebraically independent. Finally [F3] gives Sk=ΔMk−12 for every k, that is S∗=ΔM∗.

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