Let $N+1=\dim M_k(SL_2(\mathbb Z))$. For each $0\leq i\leq N$, the dimension formula, equivalently the <valence formula for the modular group>, ensures that
$$
k-12i=4a_i+6b_i
$$
for some nonnegative integers $a_i,b_i$. Define
$$
h_i=\Delta^iE_4^{a_i}E_6^{b_i}.
$$
The <modular discriminant>, $E_4$, and $E_6$ have integral Fourier coefficients and leading terms $q$, $1$, and $1$, respectively. Hence
$$
h_i=q^i+\sum_{n>i}c_{i,n}q^n,
\qquad c_{i,n}\in\mathbb Z.
$$
Their distinct orders of vanishing make the $h_i$ linearly independent, so they form a basis.

Starting with $f_N=h_N$, define $f_i$ downwards by subtracting from $h_i$ the integral multiples of $f_{i+1},\ldots,f_N$ needed to kill the coefficients of $q^{i+1},\ldots,q^N$. This integer Gaussian elimination preserves all integral coefficients and gives
$$
f_i=q^i+\sum_{n\geq N+1}a_n(f_i)q^n,
\qquad a_n(f_i)\in\mathbb Z.
$$
This is the <integral echelon basis of level-one modular forms>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 137 / 1 / b / Solution

If $N+1=\dim M_k(SL_2(\mathbb Z))$, there is a basis $f_0,\ldots,f_N$ with integral Fourier coefficients and
$$
f_i=q^i+\sum_{n\geq N+1}a_n(f_i)q^n.
$$
Start from the integral forms $\Delta^iE_4^aE_6^b$ and use integer elimination on their unitriangular leading coefficients.


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