[proofplan]
We first check that every displayed monomial is a modular form of the correct weight. The main spanning argument is an induction on the weight: subtract a scalar multiple of one monomial to kill the constant Fourier coefficient, divide the resulting cusp form by the modular discriminant $\Delta$, and then use
\begin{align*}
\Delta=\frac{E_4^3-E_6^2}{1728}
\end{align*}
to return to monomials in $E_4$ and $E_6$. Finally, the standard dimension formula for $M_k(SL_2(\mathbb{Z}))$ gives that the spanning set has exactly the right cardinality, so it is a basis.
[/proofplan]
[step:Verify that the monomials have weight $k$]
Let $a,b \in \mathbb{Z}_{\geq 0}$ satisfy $4a+6b=k$. Since $E_4 \in M_4(SL_2(\mathbb{Z}))$ and $E_6 \in M_6(SL_2(\mathbb{Z}))$, the product rule for modular forms implies
\begin{align*}
E_4^aE_6^b \in M_{4a+6b}(SL_2(\mathbb{Z}))=M_k(SL_2(\mathbb{Z})).
\end{align*}
Thus every listed monomial lies in $M_k(SL_2(\mathbb{Z}))$.
[/step]
[step:Establish the induction bases]
We use the standard low-weight dimension computation for modular forms on $SL_2(\mathbb{Z})$ (citing a result not yet in the wiki: low-weight dimension formula for modular forms on $SL_2(\mathbb{Z})$):
\begin{align*}
M_0(SL_2(\mathbb{Z})) &= \mathbb{C}\cdot 1,\\
M_2(SL_2(\mathbb{Z})) &= \{0\},\\
M_4(SL_2(\mathbb{Z})) &= \mathbb{C}\cdot E_4,\\
M_6(SL_2(\mathbb{Z})) &= \mathbb{C}\cdot E_6,\\
M_8(SL_2(\mathbb{Z})) &= \mathbb{C}\cdot E_4^2,\\
M_{10}(SL_2(\mathbb{Z})) &= \mathbb{C}\cdot E_4E_6.
\end{align*}
These are exactly the monomials $E_4^aE_6^b$ with $4a+6b=k$ for $k=0,2,4,6,8,10$. Hence the theorem holds in these weights.
[/step]
[step:Subtract a monomial to reduce to a cusp form]
Assume now that $k \geq 12$ is even, and assume inductively that the theorem holds for the even weight $k-12$. Let $f \in M_k(SL_2(\mathbb{Z}))$.
Because $k \geq 12$ is even, there exist $a_0,b_0 \in \mathbb{Z}_{\geq 0}$ such that $4a_0+6b_0=k$. For example, after writing $k=2m$ with $m \geq 6$, choose $b_0 \in \{0,1\}$ with $m-3b_0$ even and nonnegative, and set
\begin{align*}
a_0=\frac{m-3b_0}{2}.
\end{align*}
Let $z \in \mathbb{H}$ be a point in the complex upper half-plane, and define the Fourier parameter $q := e^{2\pi i z}$. The normalized Eisenstein series satisfy
\begin{align*}
E_4(q)&=1+\sum_{n=1}^{\infty} A_n q^n,\\
E_6(q)&=1+\sum_{n=1}^{\infty} B_n q^n,
\end{align*}
for complex coefficients $A_n,B_n \in \mathbb{C}$, so the monomial $E_4^{a_0}E_6^{b_0}$ has constant Fourier coefficient $1$.
Let $c_0 \in \mathbb{C}$ denote the constant Fourier coefficient of $f$. Define
\begin{align*}
h := f-c_0E_4^{a_0}E_6^{b_0}.
\end{align*}
Let $S_k(SL_2(\mathbb{Z}))$ denote the complex vector subspace of $M_k(SL_2(\mathbb{Z}))$ consisting of cusp forms of weight $k$. Both summands lie in $M_k(SL_2(\mathbb{Z}))$, so $h \in M_k(SL_2(\mathbb{Z}))$. Its constant Fourier coefficient is $c_0-c_0=0$, hence
\begin{align*}
h \in S_k(SL_2(\mathbb{Z})).
\end{align*}
[guided]
The induction step begins by separating the constant term from the genuinely cuspidal part. Since $k \geq 12$ is even, write $k=2m$ with $m \geq 6$. We need at least one admissible monomial of weight $k$. Choose $b_0 \in \{0,1\}$ so that $m-3b_0$ is even. This is possible because changing $b_0$ from $0$ to $1$ changes the parity by $3$, hence flips parity. Since $m \geq 6$, the chosen $m-3b_0$ is nonnegative. Define
\begin{align*}
a_0=\frac{m-3b_0}{2}.
\end{align*}
Then $a_0,b_0 \in \mathbb{Z}_{\geq 0}$ and
\begin{align*}
4a_0+6b_0=2(m-3b_0)+6b_0=2m=k.
\end{align*}
Let $z \in \mathbb{H}$ be a point in the complex upper half-plane, and define the Fourier parameter $q := e^{2\pi i z}$. The normalized Eisenstein series have Fourier expansions with constant coefficient $1$:
\begin{align*}
E_4(q)&=1+\sum_{n=1}^{\infty} A_n q^n,\\
E_6(q)&=1+\sum_{n=1}^{\infty} B_n q^n,
\end{align*}
where $A_n,B_n \in \mathbb{C}$. Therefore the product $E_4^{a_0}E_6^{b_0}$ also has constant coefficient $1$.
Let $c_0 \in \mathbb{C}$ be the constant Fourier coefficient of $f$, and define
\begin{align*}
h := f-c_0E_4^{a_0}E_6^{b_0}.
\end{align*}
The space $M_k(SL_2(\mathbb{Z}))$ is a complex [vector space](/page/Vector%20Space), and both $f$ and $E_4^{a_0}E_6^{b_0}$ lie in it, so $h \in M_k(SL_2(\mathbb{Z}))$. Its constant coefficient is $c_0-c_0=0$. By the definition of cusp forms on $SL_2(\mathbb{Z})$, a holomorphic modular form with zero constant Fourier coefficient at the cusp is cuspidal. Hence
\begin{align*}
h \in S_k(SL_2(\mathbb{Z})).
\end{align*}
[/guided]
[/step]
[step:Divide the cusp form by the discriminant]
By the standard discriminant division theorem for cusp forms on $SL_2(\mathbb{Z})$ (citing a result not yet in the wiki: multiplication by the modular discriminant identifies $M_{k-12}(SL_2(\mathbb{Z}))$ with $S_k(SL_2(\mathbb{Z}))$), there exists $g \in M_{k-12}(SL_2(\mathbb{Z}))$ such that
\begin{align*}
h=\Delta g,
\end{align*}
where $\Delta \in S_{12}(SL_2(\mathbb{Z}))$ is the normalized modular discriminant.
By the induction hypothesis applied to the even weight $k-12$, there are complex scalars $r_{a,b} \in \mathbb{C}$, indexed by pairs $(a,b) \in \mathbb{Z}_{\geq 0}^2$ with $4a+6b=k-12$, such that
\begin{align*}
g=\sum_{4a+6b=k-12} r_{a,b}E_4^aE_6^b.
\end{align*}
Using the identity
\begin{align*}
\Delta=\frac{E_4^3-E_6^2}{1728},
\end{align*}
we obtain
\begin{align*}
h
&=\Delta g\\
&=\sum_{4a+6b=k-12} r_{a,b}\Delta E_4^aE_6^b\\
&=\frac{1}{1728}\sum_{4a+6b=k-12} r_{a,b}\left(E_4^{a+3}E_6^b-E_4^aE_6^{b+2}\right).
\end{align*}
For every index pair in the sum,
\begin{align*}
4(a+3)+6b=k
\end{align*}
and
\begin{align*}
4a+6(b+2)=k.
\end{align*}
Thus $h$ is a complex linear combination of monomials $E_4^aE_6^b$ of weight $k$.
[guided]
The point of passing from $h$ to $g$ is that cusp forms of weight $k$ have a common factor $\Delta$, and division by that factor lowers the weight by $12$. The relevant standard result says that multiplication by the modular discriminant gives an isomorphism
\begin{align*}
M_{k-12}(SL_2(\mathbb{Z})) \longrightarrow S_k(SL_2(\mathbb{Z})), \qquad g \mapsto \Delta g
\end{align*}
for $k \geq 12$ (citing a result not yet in the wiki: multiplication by the modular discriminant identifies $M_{k-12}(SL_2(\mathbb{Z}))$ with $S_k(SL_2(\mathbb{Z}))$). We may apply it because the previous step proved $h \in S_k(SL_2(\mathbb{Z}))$. Hence there exists $g \in M_{k-12}(SL_2(\mathbb{Z}))$ satisfying
\begin{align*}
h=\Delta g.
\end{align*}
Now the induction hypothesis applies to $g$, because $k-12$ is a nonnegative even integer. Therefore there are scalars $r_{a,b} \in \mathbb{C}$ such that
\begin{align*}
g=\sum_{4a+6b=k-12} r_{a,b}E_4^aE_6^b.
\end{align*}
Multiplying by $\Delta$ gives
\begin{align*}
h=\sum_{4a+6b=k-12} r_{a,b}\Delta E_4^aE_6^b.
\end{align*}
The key identity is the algebraic formula
\begin{align*}
\Delta=\frac{E_4^3-E_6^2}{1728}.
\end{align*}
Substituting this identity into each summand gives
\begin{align*}
h
&=\frac{1}{1728}\sum_{4a+6b=k-12} r_{a,b}\left(E_4^{a+3}E_6^b-E_4^aE_6^{b+2}\right).
\end{align*}
The two monomials produced from a pair $(a,b)$ both have total weight $k$, since
\begin{align*}
4(a+3)+6b=(4a+6b)+12=k
\end{align*}
and
\begin{align*}
4a+6(b+2)=(4a+6b)+12=k.
\end{align*}
Thus the cusp form $h$ is spanned by the admissible monomials of weight $k$.
[/guided]
[/step]
[step:Conclude spanning in weight $k$]
From the previous steps,
\begin{align*}
f=c_0E_4^{a_0}E_6^{b_0}+h,
\end{align*}
where $E_4^{a_0}E_6^{b_0}$ is an admissible monomial of weight $k$ and $h$ is a complex linear combination of admissible monomials of weight $k$. Therefore every $f \in M_k(SL_2(\mathbb{Z}))$ lies in the span of
\begin{align*}
\{E_4^aE_6^b : a,b \in \mathbb{Z}_{\geq 0},\ 4a+6b=k\}.
\end{align*}
By induction, this spanning statement holds for every even integer $k \geq 0$.
[/step]
[step:Compare the number of monomials with the dimension formula]
It remains to prove [linear independence](/page/Linear%20Independence). Let
\begin{align*}
N_k := \#\{(a,b)\in \mathbb{Z}_{\geq 0}^2 : 4a+6b=k\}.
\end{align*}
Equivalently, writing $k=2m$, $N_k$ is the number of solutions of
\begin{align*}
2a+3b=m.
\end{align*}
The standard dimension formula for modular forms on $SL_2(\mathbb{Z})$ (citing a result not yet in the wiki: dimension formula for $M_k(SL_2(\mathbb{Z}))$) gives
\begin{align*}
\dim_{\mathbb{C}} M_k(SL_2(\mathbb{Z})) =
\begin{cases}
\left\lfloor \frac{k}{12}\right\rfloor+1, & k \not\equiv 2 \pmod{12},\\
\left\lfloor \frac{k}{12}\right\rfloor, & k \equiv 2 \pmod{12}.
\end{cases}
\end{align*}
We check that this equals $N_k$. Since $k=2m$, the equation $4a+6b=k$ is equivalent to $2a+3b=m$. For each $b \in \mathbb{Z}_{\geq 0}$ with $0 \leq b \leq \lfloor m/3\rfloor$, the value $a=(m-3b)/2$ is a nonnegative integer exactly when $b \equiv m \pmod{2}$. Thus $N_k$ is the number of integers in $\{0,\dots,\lfloor m/3\rfloor\}$ having the same parity as $m$. Writing $m=6q+r$ with $q \in \mathbb{Z}_{\geq 0}$ and $0 \leq r \leq 5$, this count is
\begin{align*}
N_k=
\begin{cases}
q+1, & r\neq 1,\\
q, & r=1.
\end{cases}
\end{align*}
Since $k=12q+2r$, the condition $r=1$ is exactly $k \equiv 2 \pmod{12}$, and the displayed count agrees with the dimension formula. Hence
\begin{align*}
\dim_{\mathbb{C}} M_k(SL_2(\mathbb{Z})) = N_k.
\end{align*}
Since the admissible monomials span $M_k(SL_2(\mathbb{Z}))$ and their number equals the dimension of that vector space, they are linearly independent. Hence they form a basis of $M_k(SL_2(\mathbb{Z}))$.
[/step]
