Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 26 / 1 / i / Solution

Write $\Gamma =S{L}_2(\mathbb{Z})$, $\mathbb{H}=\{\tau :\mathrm{Im}\tau >0\}$ and $q=e^{2\pi i\tau }$. The space $M_k$ consists of holomorphic functions on the complex upper half-plane satisfyingand holomorphic at the cusp: their Fourier expansion of a modular form is $f={\sum }_{n\ge 0}{a}_n{q}^n$. The subspace $S_k$ of cusp forms has $a_0=0$. Since all cusps are equivalent at level one, this condition treats every cusp. The element $-I$ forces odd-weight forms to vanish.

Use the permitted properties of the normalized Eisenstein series and modular discriminant:They have integer Fourier coefficients and weights $4,6,12$. For even $k\ge 0$, the assumed dimension formula gives $d=\lfloor k/12\rfloor +1$ except when $k\equiv 2(\mathrm{mod}12)$, when $d=\lfloor k/12\rfloor$. For $0\le j<d$, the remaining weight $k-12j$ is nonnegative and is not two. Choose $b_j\in \{0,1\}$ and $a_j\ge 0$ with $4a_j+6b_j=k-12j$. Every allowable remaining weight admits such a choice. The formsgive an integral triangular basis of level-one modular forms: their distinct first exponents prove independence, and their number $d$ proves that they span.

Now take $f$ with integral coefficients. Subtract its constant coefficient times $F_0$. The remainder still has integral coefficients and has no constant term. Subtract its coefficient at $q$ times $F_1$, then its current coefficient at $q^2$ times $F_2$, and continue. Every subtraction uses an integer. After $d$ steps the remainder's first $d$ coefficients vanish; triangularity and the basis property force that remainder to be zero. ConsequentlyThis proves integral polynomial generation without dividing by $1728$. The relation $E_4^3-E_6^2=1728\Delta$ explains why the additional integral generator is useful, although over $\mathbb{C}$ two generators suffice. For $k=0$ the basis is the constant one; when $M_k=0$, including weight two and odd weights, the claim is vacuous.