[proofplan] We prove first that every modular form of level one is generated by $E_4$ and $E_6$. The main input is the level-one valence formula, together with the discriminant modular form $\Delta = (E_4^3 - E_6^2)/1728$, whose single zero at the cusp lets us divide cusp forms by $\Delta$ and lower the weight by $12$. An induction on the weight then proves surjectivity of the map $\mathbb{C}[X,Y] \to M_*(\Gamma)$. Finally, algebraic independence follows by reducing any homogeneous polynomial relation to a polynomial relation for the nonconstant modular function $E_6^2/E_4^3$. [/proofplan]

[step:Use the valence formula to record the basic low-weight consequences]We use the following standard level-one valence formula, which is not yet cited to an Androma theorem entry: if $f \in M_k(\Gamma)$ is nonzero, then \begin{align*} \operatorname{ord}_{\infty}(f) + \frac{1}{2}\operatorname{ord}_{i}(f) + \frac{1}{3}\operatorname{ord}_{\rho}(f) + \sum_{p \in \Gamma \backslash \mathbb{H},\ p \neq i,\rho} \operatorname{ord}_{p}(f) = \frac{k}{12}, \end{align*} where $\rho := e^{2\pi i/3}$, $\operatorname{ord}_{\infty}(f)$ denotes the order of vanishing of the $q$-expansion at the cusp, and $\operatorname{ord}_{p}(f)$ denotes the order of vanishing at $p \in \mathbb{H}$. For odd $k$, every $f \in M_k(\Gamma)$ is zero. Indeed, the matrix $-I \in \Gamma$ fixes every $\tau \in \mathbb{H}$ and acts on a weight-$k$ modular form by \begin{align*} f(\tau) = (-1)^k f(\tau). \end{align*} If $k$ is odd, this gives $f(\tau) = -f(\tau)$ for every $\tau \in \mathbb{H}$, hence $f = 0$. The same valence formula gives $M_0(\Gamma)=\mathbb{C}$ and $M_2(\Gamma)=0$. For $k=0$, the valence formula implies that a nonzero $f \in M_0(\Gamma)$ has no zeros on the compactified modular curve and no pole at the cusp; hence it is constant. For $k=2$, the right-hand side is $1/6$, while the left-hand side is a sum of nonnegative terms belonging to the additive semigroup generated by $1$, $1/2$, and $1/3$. The value $1/6$ cannot occur unless all terms vanish, which would give left-hand side $0$, so $M_2(\Gamma)=0$.[/step]

[guided]We first isolate the low-weight information that the induction will need. The standard level-one valence formula, not yet cited to an Androma theorem entry, says that every nonzero $f \in M_k(\Gamma)$ satisfies \begin{align*} \operatorname{ord}_{\infty}(f) + \frac{1}{2}\operatorname{ord}_{i}(f) + \frac{1}{3}\operatorname{ord}_{\rho}(f) + \sum_{p \in \Gamma \backslash \mathbb{H},\ p \neq i,\rho} \operatorname{ord}_{p}(f) = \frac{k}{12}, \end{align*} where $\rho := e^{2\pi i/3}$. Here $\operatorname{ord}_{\infty}(f)$ is the order of vanishing of the Fourier expansion at the cusp, and $\operatorname{ord}_{p}(f)$ is the ordinary holomorphic order of vanishing at the point $p \in \mathbb{H}$. The first consequence is that odd weights vanish. Let $k$ be odd and let $f \in M_k(\Gamma)$. Since $-I \in SL_2(\mathbb{Z})$ fixes every point of $\mathbb{H}$, the modular transformation law gives \begin{align*} f(\tau) = (-1)^k f(\tau) = -f(\tau) \end{align*} for every $\tau \in \mathbb{H}$. Thus $2f(\tau)=0$ for every $\tau$, so $f=0$. The second consequence is the base of the even-weight induction. For $k=0$, the valence formula says that a nonzero element of $M_0(\Gamma)$ has no zeros and no pole at the cusp. Equivalently, it is a [holomorphic function](/page/Holomorphic%20Function) on the compactified modular curve, so it is constant. Hence $M_0(\Gamma)=\mathbb{C}$. For $k=2$, the valence formula would force the nonnegative left-hand side to equal $1/6$. But the only possible positive contributions have sizes at least $1/3$, $1/2$, or $1$, and no combination of these nonnegative contributions equals $1/6$. Therefore no nonzero form of weight $2$ exists, and $M_2(\Gamma)=0$.[/guided]

[step:Define the discriminant form and the weight-lowering operation]Define the discriminant modular form \begin{align*} \Delta := \frac{E_4^3 - E_6^2}{1728} \in M_{12}(\Gamma). \end{align*} Its $q$-expansion begins \begin{align*} \Delta(\tau) = q + O(q^2), \qquad q := e^{2\pi i \tau}, \end{align*} so $\operatorname{ord}_{\infty}(\Delta)=1$. By the valence formula with $k=12$, the equality \begin{align*} \operatorname{ord}_{\infty}(\Delta) + \frac{1}{2}\operatorname{ord}_{i}(\Delta) + \frac{1}{3}\operatorname{ord}_{\rho}(\Delta) + \sum_{p \in \Gamma \backslash \mathbb{H},\ p \neq i,\rho} \operatorname{ord}_{p}(\Delta) = 1 \end{align*} forces every interior order of vanishing to be $0$. Hence $\Delta$ has no zeros on $\mathbb{H}$ and has a simple zero at the cusp. For each integer $m \geq 0$, define the cusp subspace \begin{align*} S_m(\Gamma) := \{u \in M_m(\Gamma) : \operatorname{ord}_{\infty}(u) \geq 1\}. \end{align*} Equivalently, $u \in S_m(\Gamma)$ exactly when the constant term of the $q$-expansion of $u$ at the cusp is $0$. Let $k \geq 12$ and let $g \in S_k(\Gamma)$. Define \begin{align*} h: \mathbb{H} &\to \mathbb{C}, \\ \tau &\mapsto \frac{g(\tau)}{\Delta(\tau)}. \end{align*} Since $\Delta$ has no zeros on $\mathbb{H}$, the function $h$ is holomorphic on $\mathbb{H}$. Since $g$ vanishes at the cusp and $\Delta$ has a simple zero at the cusp, the quotient $h$ is holomorphic at the cusp. The transformation laws of $g$ and $\Delta$ give \begin{align*} h(\gamma \tau) = (c\tau+d)^{k-12} h(\tau) \end{align*} for every $\gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \Gamma$. Thus $h \in M_{k-12}(\Gamma)$, and multiplication by $\Delta$ identifies $M_{k-12}(\Gamma)$ with the cusp subspace $S_k(\Gamma)$.[/step]

[guided]The induction will lower the weight by dividing by a fixed cusp form of weight $12$. The correct form is the discriminant \begin{align*} \Delta := \frac{E_4^3 - E_6^2}{1728} \in M_{12}(\Gamma). \end{align*} The normalization is chosen so that its Fourier expansion begins \begin{align*} \Delta(\tau) = q + O(q^2), \qquad q := e^{2\pi i \tau}. \end{align*} Thus $\Delta$ has a zero of order exactly $1$ at the cusp. We must also know that division by $\Delta$ introduces no poles inside $\mathbb{H}$. Apply the level-one valence formula to $\Delta$, whose weight is $12$. It gives \begin{align*} \operatorname{ord}_{\infty}(\Delta) + \frac{1}{2}\operatorname{ord}_{i}(\Delta) + \frac{1}{3}\operatorname{ord}_{\rho}(\Delta) + \sum_{p \in \Gamma \backslash \mathbb{H},\ p \neq i,\rho} \operatorname{ord}_{p}(\Delta) = 1. \end{align*} The first term is already $1$, because $\Delta(\tau)=q+O(q^2)$. Every other term is nonnegative, so all interior orders must be $0$. Therefore $\Delta$ has no zero in $\mathbb{H}$. We also name the cusp subspace before using it. For each integer $m \geq 0$, define \begin{align*} S_m(\Gamma) := \{u \in M_m(\Gamma) : \operatorname{ord}_{\infty}(u) \geq 1\}. \end{align*} Equivalently, $u \in S_m(\Gamma)$ exactly when the constant term of the $q$-expansion of $u$ at the cusp is $0$. This is the subspace of modular forms that vanish at the cusp. Now let $k \geq 12$ and let $g \in S_k(\Gamma)$. Define \begin{align*} h: \mathbb{H} &\to \mathbb{C}, \\ \tau &\mapsto \frac{g(\tau)}{\Delta(\tau)}. \end{align*} The absence of zeros of $\Delta$ on $\mathbb{H}$ makes $h$ holomorphic on $\mathbb{H}$. At the cusp, $g$ has vanishing order at least $1$ because it is a cusp form, and $\Delta$ has vanishing order exactly $1$, so the quotient has no pole at the cusp. Finally, the transformation law is computed directly: if $\gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \Gamma$, then \begin{align*} h(\gamma \tau) = \frac{g(\gamma \tau)}{\Delta(\gamma \tau)} = \frac{(c\tau+d)^k g(\tau)}{(c\tau+d)^{12}\Delta(\tau)} = (c\tau+d)^{k-12} h(\tau). \end{align*} Thus $h \in M_{k-12}(\Gamma)$. Conversely, if $h \in M_{k-12}(\Gamma)$, then $\Delta h \in S_k(\Gamma)$ because $\Delta$ is a cusp form. Hence multiplication by $\Delta$ identifies $M_{k-12}(\Gamma)$ with $S_k(\Gamma)$.[/guided]

[step:Express every modular form as a polynomial in $E_4$ and $E_6$]We prove by induction on $k \geq 0$ that every element of $M_k(\Gamma)$ lies in $\mathbb{C}[E_4,E_6]$. The cases of odd $k$, $k=0$, and $k=2$ were handled above. Let $k \geq 4$ be even. Choose nonnegative integers $a,b \in \mathbb{Z}_{\geq 0}$ such that \begin{align*} 4a + 6b = k. \end{align*} Such $a,b$ exist because $k/2 \geq 2$ and every integer at least $2$ is a nonnegative integer combination of $2$ and $3$. Let $f \in M_k(\Gamma)$. Write its Fourier expansion at the cusp as \begin{align*} f(\tau) = c_0 + \sum_{n=1}^{\infty} c_n q^n, \qquad q := e^{2\pi i\tau}, \end{align*} where $c_n \in \mathbb{C}$. Since both $E_4$ and $E_6$ have constant term $1$, the modular form $E_4^aE_6^b$ has constant term $1$. Therefore \begin{align*} g := f - c_0 E_4^aE_6^b \end{align*} belongs to $S_k(\Gamma)$. If $k<12$, the valence formula gives $g=0$, because a nonzero cusp form would have $\operatorname{ord}_{\infty}(g) \geq 1$ and hence \begin{align*} 1 \leq \frac{k}{12} < 1, \end{align*} a contradiction. Thus $f=c_0E_4^aE_6^b$. If $k \geq 12$, the previous step gives a modular form $h \in M_{k-12}(\Gamma)$ such that \begin{align*} g = \Delta h. \end{align*} By the induction hypothesis, $h \in \mathbb{C}[E_4,E_6]$. Since $\Delta=(E_4^3-E_6^2)/1728$ also belongs to $\mathbb{C}[E_4,E_6]$, we get $g \in \mathbb{C}[E_4,E_6]$, and hence \begin{align*} f = c_0E_4^aE_6^b + g \in \mathbb{C}[E_4,E_6]. \end{align*} This proves that $\Phi$ is surjective.[/step]

[guided]We now prove generation by induction on the weight. The assertion is that every $f \in M_k(\Gamma)$ is a polynomial in $E_4$ and $E_6$. The base cases have already been established. If $k$ is odd, then $M_k(\Gamma)=0$. If $k=0$, then $M_0(\Gamma)=\mathbb{C}$, which is generated by the constant polynomial. If $k=2$, then $M_2(\Gamma)=0$. Now assume $k \geq 4$ is even and that all modular forms of weights strictly smaller than $k$ have already been shown to lie in $\mathbb{C}[E_4,E_6]$. We first choose one monomial of the correct weight. We need nonnegative integers $a,b$ satisfying \begin{align*} 4a+6b=k. \end{align*} Dividing by $2$, this is equivalent to writing $k/2$ as $2a+3b$. Since $k/2 \geq 2$, and every integer at least $2$ is a nonnegative integer combination of $2$ and $3$, such $a,b$ exist. Let $f \in M_k(\Gamma)$. Its Fourier expansion at the cusp has the form \begin{align*} f(\tau) = c_0 + \sum_{n=1}^{\infty} c_n q^n, \qquad q := e^{2\pi i\tau}, \end{align*} with coefficients $c_n \in \mathbb{C}$. The normalized Eisenstein series satisfy $E_4(\tau)=1+O(q)$ and $E_6(\tau)=1+O(q)$, so the product $E_4^aE_6^b$ has constant term $1$. Therefore the modular form \begin{align*} g := f - c_0E_4^aE_6^b \end{align*} has constant term $0$. This means exactly that $g$ is a cusp form of weight $k$. If $k<12$, then $g$ must vanish. Indeed, if $g$ were nonzero, then $\operatorname{ord}_{\infty}(g)\geq 1$ because $g$ is cuspidal. The valence formula would then imply \begin{align*} 1 \leq \operatorname{ord}_{\infty}(g) \leq \frac{k}{12} < 1, \end{align*} which is impossible. Hence $g=0$, and $f=c_0E_4^aE_6^b$ is already a polynomial in $E_4$ and $E_6$. If $k \geq 12$, we use the weight-lowering step. Since $g \in S_k(\Gamma)$, there exists $h \in M_{k-12}(\Gamma)$ such that \begin{align*} g = \Delta h. \end{align*} By the induction hypothesis, $h$ is a polynomial in $E_4$ and $E_6$. Also \begin{align*} \Delta = \frac{E_4^3-E_6^2}{1728} \end{align*} is itself a polynomial in $E_4$ and $E_6$. Therefore $g=\Delta h$ is a polynomial in $E_4$ and $E_6$, and so \begin{align*} f = c_0E_4^aE_6^b + g \end{align*} is also a polynomial in $E_4$ and $E_6$. This completes the induction and proves that $\Phi$ is surjective.[/guided]

[step:Show that no nonzero polynomial relation exists between $E_4$ and $E_6$]It remains to prove that $\Phi$ is injective. Since $\Phi$ is a graded homomorphism into the direct sum $\bigoplus_{k \geq 0} M_k(\Gamma)$, it suffices to rule out nonzero weighted-homogeneous relations. Let $P(X,Y) \in \mathbb{C}[X,Y]$ be weighted homogeneous of weight $K$ with $\deg X=4$ and $\deg Y=6$, and suppose \begin{align*} P(E_4,E_6)=0. \end{align*} All monomials $X^aY^b$ appearing in $P$ satisfy $4a+6b=K$, so the parity of $b$ is fixed. Let $\varepsilon \in \{0,1\}$ be this common parity. Then there exist an integer $A \geq 0$ and a polynomial $R(T)\in\mathbb{C}[T]$ such that \begin{align*} P(X,Y)=X^A Y^\varepsilon R\left(\frac{Y^2}{X^3}\right) \end{align*} after multiplying through by the evident common power of $X$; equivalently, on the [open set](/page/Open%20Set) where $E_4 \neq 0$ and $E_6 \neq 0$, the relation implies \begin{align*} R\left(\frac{E_6^2}{E_4^3}\right)=0. \end{align*} Define the meromorphic modular function \begin{align*} t: \mathbb{H} \setminus \{E_4=0\} &\to \mathbb{C}, \\ \tau &\mapsto \frac{E_6(\tau)^2}{E_4(\tau)^3}. \end{align*} Using \begin{align*} \Delta = \frac{E_4^3-E_6^2}{1728}, \end{align*} we have \begin{align*} t = 1 - 1728\,\frac{\Delta}{E_4^3}. \end{align*} The Fourier expansions $E_4(\tau)=1+240q+O(q^2)$, $E_6(\tau)=1-504q+O(q^2)$, and $\Delta(\tau)=q+O(q^2)$ give \begin{align*} t(\tau)=1-1728q+O(q^2), \qquad q=e^{2\pi i\tau}. \end{align*} Thus $t$ is not constant. If $R$ were nonzero, the image of the connected open set on which the above relation holds would be contained in the finite zero set of $R$. Since $t$ is holomorphic there, this would force $t$ to be locally constant and hence constant by [analytic continuation](/page/Analytic%20Continuation), contradicting its $q$-expansion. Therefore $R=0$, so $P=0$. Hence no nonzero weighted-homogeneous relation exists, and $\Phi$ is injective.[/step]

[guided]We now prove algebraic independence. Because the target is the graded direct sum \begin{align*} M_*(\Gamma)=\bigoplus_{k\geq 0}M_k(\Gamma), \end{align*} a polynomial relation decomposes into weighted-homogeneous parts. Thus it is enough to show that a weighted-homogeneous polynomial relation cannot occur. Let $P(X,Y)\in\mathbb{C}[X,Y]$ be weighted homogeneous for the grading $\deg X=4$ and $\deg Y=6$, and suppose \begin{align*} P(E_4,E_6)=0. \end{align*} If a monomial $X^aY^b$ occurs in $P$, then $4a+6b=K$ for one fixed weight $K$. Reducing this congruence modulo $4$ shows that the parity of $b$ is fixed among all monomials. Let $\varepsilon\in\{0,1\}$ denote this common parity. After factoring out the common parity contribution $Y^\varepsilon$ and the smallest occurring power of $X$, the remaining monomials are powers of the ratio $Y^2/X^3$. Therefore there is a polynomial $R(T)\in\mathbb{C}[T]$ such that, on the open set where the denominators do not vanish, the relation $P(E_4,E_6)=0$ implies \begin{align*} R\left(\frac{E_6^2}{E_4^3}\right)=0. \end{align*} Define \begin{align*} t: \mathbb{H}\setminus\{E_4=0\} &\to \mathbb{C}, \\ \tau &\mapsto \frac{E_6(\tau)^2}{E_4(\tau)^3}. \end{align*} This is holomorphic on its domain. The identity \begin{align*} \Delta=\frac{E_4^3-E_6^2}{1728} \end{align*} rewrites it as \begin{align*} t=1-1728\,\frac{\Delta}{E_4^3}. \end{align*} Near the cusp, the Fourier expansions \begin{align*} E_4(\tau)&=1+240q+O(q^2),\\ E_6(\tau)&=1-504q+O(q^2),\\ \Delta(\tau)&=q+O(q^2), \qquad q=e^{2\pi i\tau}, \end{align*} give \begin{align*} t(\tau)=1-1728q+O(q^2). \end{align*} Hence $t$ is nonconstant. Now suppose $R$ were nonzero. Its zero set in $\mathbb{C}$ is finite. The relation \begin{align*} R(t(\tau))=0 \end{align*} would force the image of $t$ on a nonempty connected open subset of $\mathbb{H}\setminus\{E_4=0\}$ to lie in this finite set. A holomorphic function from a connected domain to a finite set is constant, so $t$ would be constant on that open subset. By analytic continuation, $t$ would be constant wherever it is holomorphic, contradicting the nonconstant Fourier expansion just computed. Therefore $R=0$, and hence $P=0$. Thus there is no nonzero weighted-homogeneous polynomial relation between $E_4$ and $E_6$. Since every relation decomposes into weighted-homogeneous relations, the homomorphism $\Phi$ is injective.[/guided]

[step:Conclude the graded ring identification] The homomorphism \begin{align*} \Phi:\mathbb{C}[X,Y]\to M_*(\Gamma), \qquad X\mapsto E_4,\quad Y\mapsto E_6, \end{align*} is surjective by the induction argument and injective by algebraic independence. Therefore $\Phi$ is an isomorphism of graded $\mathbb{C}$-algebras, with $\deg E_4=4$ and $\deg E_6=6$. Hence \begin{align*} M_*(SL_2(\mathbb{Z}))=\mathbb{C}[E_4,E_6]. \end{align*} [/step]