= Solution
The normalized <Eisenstein series> satisfy $E_4=1+240\sum_{m\ge1}\sigma_3(m)q^m$ and $E_6=1-504\sum_{m\ge1}\sigma_5(m)q^m$. Their weights are four and six. The <valence formula for the modular group> makes the normalized weight-twelve <cusp form> $\Delta$ have one simple cusp zero and no zero in $\mathbb H$. Comparing its first coefficient gives $\Delta=(E_4^3-E_6^2)/1728$.
Odd weights vanish because $-I$ acts by $(-1)^k$. Weight two vanishes by the <valence formula for the modular group>: the forced zero at $i$ alone contributes $1/2>2/12$. For every remaining nonnegative even weight, except two, choose $a,b\ge0$ with $4a+6b=k$. Subtract the constant coefficient of $f$ times $E_4^aE_6^b$. The remainder is a <cusp form>; division by $\Delta$ produces a holomorphic <modular form> of weight $k-12$. The cusp and interior orders justify this division. Induction on the weight proves
$$
\boxed{M_*(SL_2(\mathbb Z))=\mathbb C[E_4,E_6].}
$$
The <integral triangular basis of level-one modular forms> is $F_j=\Delta^jE_4^{a_j}E_6^{b_j}$, with $4a_j+6b_j=k-12j\ge0$ and $k-12j\ne2$. The induction just given shows these span, and their distinct leading terms $q^j$ show linear independence. Every coefficient is integral by the hypothesis on $\Delta$ and the displayed Eisenstein expansions. In $f=\sum_jc_jF_j$, compare coefficients successively: $c_0=a_0$, then $c_1$ is $a_1$ minus an integer combination of already integral coefficients, and similarly for every $j$. The largest permitted $j$ is at most $\lfloor(k+1)/12\rfloor$, so the stipulated initial coefficients make every $c_j$ integral. Therefore \b[all Fourier coefficients of $f$ are integers].
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