[proofplan] We construct the graded algebra map sending the formal variables $X$ and $Y$ to $E_4$ and $E_6$. Surjectivity is proved by induction on the weight: subtract a suitable monomial in $E_4$ and $E_6$ to kill the constant term, then divide the resulting cusp form by the modular discriminant defined below. Injectivity follows by comparing the number of monomials of weighted degree $k$ with the standard dimension formula for $M_k(SL_2(\mathbb{Z}))$, which is a consequence of the valence formula. The only external input is the standard discriminant and dimension theory for modular forms on $SL_2(\mathbb{Z})$; these are currently cited as results not yet in the wiki. [/proofplan]

[step:Define the graded algebra map generated by $E_4$ and $E_6$] Let \begin{align*} R := \bigoplus_{k \geq 0} M_k(SL_2(\mathbb{Z})) \end{align*} denote the graded $\mathbb{C}$-algebra whose multiplication is pointwise multiplication of modular forms. Since the product of a weight-$k$ modular form and a weight-$\ell$ modular form is a weight-$(k+\ell)$ modular form, $R$ is a graded algebra. Define the weighted polynomial algebra $\mathbb{C}[X,Y]$ by assigning $\deg X = 4$ and $\deg Y = 6$. Define the graded $\mathbb{C}$-algebra homomorphism \begin{align*} \Phi: \mathbb{C}[X,Y] &\to R \\ P(X,Y) &\mapsto P(E_4,E_6). \end{align*} Because $E_4 \in M_4(SL_2(\mathbb{Z}))$ and $E_6 \in M_6(SL_2(\mathbb{Z}))$, the image of a weighted homogeneous polynomial of degree $k$ lies in $M_k(SL_2(\mathbb{Z}))$. [/step]

[step:Record the discriminant divisor step for cusp forms] For every integer $m < 0$, use the convention \begin{align*} M_m(SL_2(\mathbb{Z})) := \{0\}. \end{align*} For every integer $k \geq 0$, let $S_k(SL_2(\mathbb{Z}))$ denote the $\mathbb{C}$-vector subspace of $M_k(SL_2(\mathbb{Z}))$ consisting of modular forms whose $q$-expansion at the cusp has constant term zero. Define the modular discriminant \begin{align*} \Delta: \mathbb{H} &\to \mathbb{C} \\ \tau &\mapsto \frac{E_4(\tau)^3 - E_6(\tau)^2}{1728}. \end{align*} Using the displayed $q$-expansions of $E_4$ and $E_6$, we obtain \begin{align*} E_4(\tau)^3 &= 1 + 720q + O(q^2), \\ E_6(\tau)^2 &= 1 - 1008q + O(q^2), \end{align*} and therefore \begin{align*} \Delta(\tau) = q + O(q^2). \end{align*} Thus $\Delta \in M_{12}(SL_2(\mathbb{Z}))$ has zero constant term, so $\Delta \in S_{12}(SL_2(\mathbb{Z}))$. We use the standard discriminant fact for $SL_2(\mathbb{Z})$: $\Delta$ has no zeros on $\mathbb{H}$ and multiplication by $\Delta$ gives an isomorphism \begin{align*} M_{k-12}(SL_2(\mathbb{Z})) &\to S_k(SL_2(\mathbb{Z})) \\ F &\mapsto \Delta F \end{align*} for every integer $k$. This follows from the product formula \begin{align*} \Delta(\tau) = q\prod_{n=1}^{\infty}(1-q^n)^{24}, \end{align*} together with the transformation law of $\Delta$ and holomorphy at the cusp. Here we are citing a result not yet in the wiki: the nonvanishing and cusp-divisibility theorem for the modular discriminant. Indeed, if $F \in M_{k-12}(SL_2(\mathbb{Z}))$, then $\Delta F$ is a cusp form of weight $k$. Conversely, if $f \in S_k(SL_2(\mathbb{Z}))$, then the quotient $f/\Delta$ is holomorphic on $\mathbb{H}$ because $\Delta$ has no zeros there, transforms with weight $k-12$, and is holomorphic at the cusp because both $f$ and $\Delta$ vanish to order at least one at $q=0$ while $\Delta$ has first nonzero term $q$. Hence $f/\Delta \in M_{k-12}(SL_2(\mathbb{Z}))$. [/step]

[step:Prove that every modular form is a polynomial in $E_4$ and $E_6$]We prove by induction on $k \geq 0$ that every element of $M_k(SL_2(\mathbb{Z}))$ lies in the image of $\Phi$. If $k$ is odd, then $M_k(SL_2(\mathbb{Z})) = \{0\}$ because the matrix $-I \in SL_2(\mathbb{Z})$ acts on a weight-$k$ modular form $f$ by \begin{align*} f(\tau) = (-1)^k f(\tau), \end{align*} so $f=0$ when $k$ is odd. If $k=0$, then $M_0(SL_2(\mathbb{Z})) = \mathbb{C}$, so every weight-$0$ form is the image of a constant polynomial. If $k=2$, then $M_2(SL_2(\mathbb{Z})) = \{0\}$ by the standard dimension formula for modular forms on $SL_2(\mathbb{Z})$; this dimension formula is a consequence of the valence formula, cited here as a result not yet in the wiki. Now let $k \geq 4$ be even and assume the assertion has been proved for all smaller nonnegative weights. Since every even integer $k \geq 4$ can be written as \begin{align*} k = 4a + 6b \end{align*} for some $a,b \in \mathbb{Z}_{\geq 0}$, choose such a pair $(a,b)$ and define \begin{align*} G: \mathbb{H} &\to \mathbb{C} \\ \tau &\mapsto E_4(\tau)^a E_6(\tau)^b. \end{align*} Then $G \in M_k(SL_2(\mathbb{Z}))$, and its $q$-expansion has constant term $1$. Let $f \in M_k(SL_2(\mathbb{Z}))$, and let $c \in \mathbb{C}$ denote the constant term of the $q$-expansion of $f$ at the cusp. Define \begin{align*} h: \mathbb{H} &\to \mathbb{C} \\ \tau &\mapsto f(\tau) - cG(\tau). \end{align*} Then $h \in M_k(SL_2(\mathbb{Z}))$, and the constant term of $h$ is zero. Hence $h \in S_k(SL_2(\mathbb{Z}))$. By the discriminant divisor step, there exists $F \in M_{k-12}(SL_2(\mathbb{Z}))$ such that \begin{align*} h = \Delta F. \end{align*} If $k < 12$, then $M_{k-12}(SL_2(\mathbb{Z})) = \{0\}$, so $h=0$ and $f=cG$ lies in the image of $\Phi$. If $k \geq 12$, the induction hypothesis gives a polynomial $P \in \mathbb{C}[X,Y]$ of weighted degree $k-12$ such that \begin{align*} F = P(E_4,E_6). \end{align*} Since \begin{align*} \Delta = \frac{E_4^3 - E_6^2}{1728}, \end{align*} we obtain \begin{align*} f = cE_4^aE_6^b + \Delta F = cE_4^aE_6^b + \frac{E_4^3 - E_6^2}{1728}P(E_4,E_6). \end{align*} Thus $f$ lies in the image of $\Phi$. This proves surjectivity.[/step]

[guided]The induction is built around one simple idea: first remove the constant term, then divide by the discriminant. Let $k \geq 0$. If $k$ is odd, the modular transformation law for the element $-I \in SL_2(\mathbb{Z})$ gives \begin{align*} f(\tau) = (-1)^k f(\tau) \end{align*} for every $f \in M_k(SL_2(\mathbb{Z}))$. Since $(-1)^k=-1$ for odd $k$, this forces $f=0$. Thus odd weights contribute nothing. The weight $0$ space consists of constants, and those are already in the image of $\Phi$. The exceptional weight $2$ space is zero by the standard dimension formula for modular forms on $SL_2(\mathbb{Z})$, which is a consequence of the valence formula; this is cited here as a result not yet in the wiki. Now assume $k \geq 4$ is even and that every modular form of smaller nonnegative weight is already known to be a polynomial in $E_4$ and $E_6$. We need one polynomial modular form of weight $k$ whose constant term is nonzero. Since every even integer $k \geq 4$ has the form \begin{align*} k = 4a + 6b \end{align*} for some $a,b \in \mathbb{Z}_{\geq 0}$, define \begin{align*} G: \mathbb{H} &\to \mathbb{C} \\ \tau &\mapsto E_4(\tau)^aE_6(\tau)^b. \end{align*} Then $G$ has weight $k$. Because both $E_4$ and $E_6$ have constant term $1$, the constant term of $G$ is also $1$. Let $f \in M_k(SL_2(\mathbb{Z}))$, and let $c \in \mathbb{C}$ be the constant term of its $q$-expansion. The form \begin{align*} h: \mathbb{H} &\to \mathbb{C} \\ \tau &\mapsto f(\tau)-cG(\tau) \end{align*} still has weight $k$, but its constant term is now zero. Therefore $h$ is a cusp form. This is where $\Delta$ enters. The discriminant divisor statement says that every weight-$k$ cusp form is uniquely divisible by $\Delta$: \begin{align*} S_k(SL_2(\mathbb{Z})) = \Delta M_{k-12}(SL_2(\mathbb{Z})). \end{align*} Hence there exists $F \in M_{k-12}(SL_2(\mathbb{Z}))$ with \begin{align*} h = \Delta F. \end{align*} If $k<12$, then the weight $k-12$ is negative, so $F=0$ and hence $h=0$. Thus $f=cG$ is already a polynomial in $E_4$ and $E_6$. If $k \geq 12$, then $k-12 < k$, so the induction hypothesis applies to $F$. There is a weighted homogeneous polynomial $P \in \mathbb{C}[X,Y]$ of degree $k-12$ such that \begin{align*} F = P(E_4,E_6). \end{align*} Using \begin{align*} \Delta = \frac{E_4^3-E_6^2}{1728}, \end{align*} we get \begin{align*} f &= cG + \Delta F \\ &= cE_4^aE_6^b + \frac{E_4^3-E_6^2}{1728}P(E_4,E_6). \end{align*} The right-hand side is a polynomial expression in $E_4$ and $E_6$. This completes the induction and proves surjectivity of $\Phi$.[/guided]

[step:Count weighted monomials and compare with the modular form dimension formula] For each integer $k \geq 0$, let $V_k \subset \mathbb{C}[X,Y]$ denote the finite-dimensional $\mathbb{C}$-[vector space](/page/Vector%20Space) of weighted homogeneous polynomials of degree $k$, where $\deg X=4$ and $\deg Y=6$. A basis of $V_k$ is given by the monomials $X^aY^b$ satisfying \begin{align*} 4a + 6b = k. \end{align*} Let \begin{align*} N_k := \#\{(a,b) \in \mathbb{Z}_{\geq 0}^2 : 4a+6b=k\}. \end{align*} Then $\dim_{\mathbb{C}} V_k = N_k$. If $k$ is odd, then $N_k=0$. If $k=12m+r$ with $m \in \mathbb{Z}_{\geq 0}$ and $r \in \{0,2,4,6,8,10\}$, direct counting of the allowed values of $b$ gives \begin{align*} N_k = \begin{cases} m+1, & r \in \{0,4,6,8,10\}, \\ m, & r=2. \end{cases} \end{align*} For $k=2$, this gives $N_2=0$. By the standard dimension formula for modular forms on $SL_2(\mathbb{Z})$, \begin{align*} \dim_{\mathbb{C}} M_k(SL_2(\mathbb{Z})) = \begin{cases} 0, & k \text{ odd}, \\ 0, & k=2, \\ 1, & k=0, \\ \left\lfloor \frac{k}{12} \right\rfloor + 1, & k \geq 0 \text{ even and } k \not\equiv 2 \pmod{12}, \\ \left\lfloor \frac{k}{12} \right\rfloor, & k \geq 0 \text{ even and } k \equiv 2 \pmod{12}. \end{cases} \end{align*} This is precisely the same number $N_k$. Hence \begin{align*} \dim_{\mathbb{C}} V_k = \dim_{\mathbb{C}} M_k(SL_2(\mathbb{Z})) \end{align*} for every $k \geq 0$. Here we are citing a result not yet in the wiki: the dimension formula for modular forms on $SL_2(\mathbb{Z})$ derived from the valence formula. [/step]

[step:Deduce injectivity and conclude that the graded map is an isomorphism] For every $k \geq 0$, the restriction \begin{align*} \Phi_k: V_k &\to M_k(SL_2(\mathbb{Z})) \\ P(X,Y) &\mapsto P(E_4,E_6) \end{align*} is a [linear map](/page/Linear%20Map). The surjectivity proved above says that $\Phi_k$ is surjective for every $k$. The dimension comparison gives \begin{align*} \dim_{\mathbb{C}} V_k = \dim_{\mathbb{C}} M_k(SL_2(\mathbb{Z})). \end{align*} Therefore each $\Phi_k$ is an isomorphism of complex vector spaces. Since \begin{align*} \mathbb{C}[X,Y] = \bigoplus_{k \geq 0} V_k \end{align*} as a graded vector space and \begin{align*} R = \bigoplus_{k \geq 0} M_k(SL_2(\mathbb{Z})), \end{align*} the map $\Phi$ is bijective. It is already a graded $\mathbb{C}$-algebra homomorphism by construction, so $\Phi$ is an isomorphism of graded $\mathbb{C}$-algebras. Hence \begin{align*} \bigoplus_{k \geq 0}M_k(SL_2(\mathbb{Z})) = \mathbb{C}[E_4,E_6], \end{align*} with $E_4$ in degree $4$ and $E_6$ in degree $6$. [/step]