[proofplan] We use the valence formula to control how many initial Fourier coefficients a nonzero modular form can have equal to zero. This gives the standard dimension formula for $M_k(SL_2(\mathbb Z))$. We then compare this dimension with the number of monomials $E_4^aE_6^b$ of weight $k$, prove those monomials are linearly independent by their vanishing orders at the elliptic point $\rho = e^{2\pi i/3}$, and conclude that they form a basis in every weight. Finally, the cusp-form statement follows from the fact that $\Delta$ has a simple zero at the cusp and no zeros on $\mathbb H$. [/proofplan]

[step:Record the valence formula and the forced elliptic vanishing orders]Let $\mathbb H := \{z \in \mathbb C : \operatorname{Im}(z)>0\}$ denote the complex upper half-plane. Let \begin{align*} q: \mathbb H &\to \mathbb C\\ z &\mapsto e^{2\pi i z} \end{align*} be the standard cusp parameter at infinity. For a nonzero modular form $f \in M_k(SL_2(\mathbb Z))$, let $\operatorname{ord}_\infty(f)$ denote the order of vanishing of the Fourier expansion of $f$ in the variable $q$ at $q=0$. For $z_0 \in \mathbb H$, let $\operatorname{ord}_{z_0}(f)$ denote the usual holomorphic vanishing order of $f$ at $z_0$. We use the following standard level one valence formula, not yet resolved to an Androma theorem ID: (citing a result not yet in the wiki: [Valence Formula for Level One Modular Forms](/theorems/4263)). If $f \in M_k(SL_2(\mathbb Z))$ is nonzero, then \begin{align*} \operatorname{ord}_\infty(f) + \frac{1}{2}\operatorname{ord}_i(f) + \frac{1}{3}\operatorname{ord}_\rho(f) + \sum_{z \in SL_2(\mathbb Z)\backslash \mathbb H,\ z\notin\{i,\rho\}} \operatorname{ord}_z(f) = \frac{k}{12}, \end{align*} where $\rho := e^{2\pi i/3}$, and the sum is over representatives for the non-elliptic orbits in the quotient. The elliptic fixed points impose congruence restrictions on possible vanishing orders. The matrix \begin{align*} S := \begin{pmatrix}0&-1\\1&0\end{pmatrix} \end{align*} fixes $i$, and the modular transformation law gives \begin{align*} f(i)=i^k f(i). \end{align*} More generally, the local parameter at $i$ changes sign under $S$, so $\operatorname{ord}_i(f)$ is congruent to $k/2$ modulo $2$ when $k$ is even. Similarly, the matrix \begin{align*} ST := \begin{pmatrix}0&-1\\1&1\end{pmatrix} \end{align*} fixes $\rho$, and the local parameter at $\rho$ is multiplied by a primitive third root of unity, so $\operatorname{ord}_\rho(f)$ is congruent to $-k/2$ modulo $3$ when $k$ is even. In particular, since the normalized Eisenstein series satisfy $E_4,E_6 \not\equiv 0$, the transformation law forces \begin{align*} E_4(\rho)=0,\qquad E_6(i)=0. \end{align*} The valence formula then gives \begin{align*} \operatorname{ord}_\rho(E_4)=1,\qquad \operatorname{ord}_i(E_6)=1, \end{align*} and shows that $E_4$ has no other zeros in $SL_2(\mathbb Z)\backslash(\mathbb H\cup\{\infty\})$ and $E_6$ has no other zeros there.[/step]

[guided]The main external input is the valence formula. It is the level one analogue of counting zeros of a [meromorphic function](/page/Meromorphic%20Function) on a compact Riemann surface. For a nonzero form $f \in M_k(SL_2(\mathbb Z))$, it says that the total weighted number of zeros in the modular curve equals $k/12$: \begin{align*} \operatorname{ord}_\infty(f) + \frac{1}{2}\operatorname{ord}_i(f) + \frac{1}{3}\operatorname{ord}_\rho(f) + \sum_{z \in SL_2(\mathbb Z)\backslash \mathbb H,\ z\notin\{i,\rho\}} \operatorname{ord}_z(f) = \frac{k}{12}. \end{align*} The factors $1/2$ and $1/3$ occur because $i$ and $\rho=e^{2\pi i/3}$ have nontrivial stabilizers in $SL_2(\mathbb Z)$. We also need the elliptic fixed points to identify the zeros of $E_4$ and $E_6$. The matrix \begin{align*} S := \begin{pmatrix}0&-1\\1&0\end{pmatrix} \end{align*} acts by $S z=-1/z$ and fixes $i$. If $f$ has weight $k$, then the modular transformation law gives \begin{align*} f(Sz)=(cz+d)^k f(z). \end{align*} At $z=i$, this becomes \begin{align*} f(i)=i^k f(i). \end{align*} For $k=6$, $i^6=-1$, so $E_6(i)=-E_6(i)$ and hence $E_6(i)=0$. Likewise, the matrix \begin{align*} ST := \begin{pmatrix}0&-1\\1&1\end{pmatrix} \end{align*} acts by $z\mapsto -1/(z+1)$ and fixes $\rho=e^{2\pi i/3}$. At $z=\rho$, the factor $cz+d$ is $\rho+1=e^{\pi i/3}$. For $E_4$ we obtain \begin{align*} E_4(\rho)=(\rho+1)^4 E_4(\rho)=e^{4\pi i/3}E_4(\rho). \end{align*} Since $e^{4\pi i/3}\neq 1$, this forces $E_4(\rho)=0$. Now apply the valence formula. The form $E_4$ has weight $4$, so the right-hand side is $4/12=1/3$. Since $\rho$ contributes with weight $1/3$ and $E_4(\rho)\neq 0$ is false, the zero at $\rho$ must have order exactly $1$, and there is no room for any other zero. Similarly, $E_6$ has weight $6$, so the right-hand side is $6/12=1/2$. The point $i$ contributes with weight $1/2$, so $\operatorname{ord}_i(E_6)=1$ and $E_6$ has no other zeros.[/guided]

[step:Compute the dimension of each space of level one modular forms]If $k$ is odd, then $M_k(SL_2(\mathbb Z))=0$, because $-I \in SL_2(\mathbb Z)$ gives \begin{align*} f(z)=(-1)^k f(z)=-f(z) \end{align*} for every $z \in \mathbb H$. Assume now that $k\geq 0$ is even. Write \begin{align*} k=12m+r,\qquad r\in\{0,2,4,6,8,10\}, \end{align*} with $m\in\mathbb Z_{\geq 0}$. The valence formula implies the dimension formula \begin{align*} \dim_{\mathbb C} M_k(SL_2(\mathbb Z))= \begin{cases} m, & r=2,\\ m+1, & r\in\{0,4,6,8,10\}. \end{cases} \end{align*} Indeed, if a nonzero form $f\in M_k(SL_2(\mathbb Z))$ has $\operatorname{ord}_\infty(f)\geq m+1$, then the valence formula gives \begin{align*} m+1\leq \operatorname{ord}_\infty(f)\leq \frac{k}{12}=m+\frac{r}{12}, \end{align*} which is impossible for every $r\in\{0,2,4,6,8,10\}$. Thus the first $m+1$ Fourier coefficients determine $f$, and \begin{align*} \dim_{\mathbb C} M_k(SL_2(\mathbb Z))\leq m+1. \end{align*} When $r=2$, a sharper bound holds. If $\operatorname{ord}_\infty(f)\geq m$, then the valence formula leaves only \begin{align*} \frac{k}{12}-m=\frac{1}{6} \end{align*} for all elliptic and interior contributions. But the elliptic congruence condition at $i$ gives $\operatorname{ord}_i(f)\equiv k/2\equiv 1 \pmod 2$, so $\operatorname{ord}_i(f)\geq 1$, and hence the contribution from $i$ is at least $1/2$, contradicting $1/6$. Therefore the first $m$ Fourier coefficients determine $f$ when $r=2$, and \begin{align*} \dim_{\mathbb C} M_k(SL_2(\mathbb Z))\leq m. \end{align*} The opposite inequalities will follow in the next step from the explicit independent monomials in $E_4$ and $E_6$.[/step]

[guided]The valence formula controls the order of vanishing at the cusp. If a nonzero modular form vanishes too many times at $q=0$, then the cusp contribution alone exceeds $k/12$, which is impossible because all other terms in the valence formula are nonnegative. Let $k=12m+r$ with $r\in\{0,2,4,6,8,10\}$. Suppose first that $f\in M_k(SL_2(\mathbb Z))$ is nonzero and $\operatorname{ord}_\infty(f)\geq m+1$. Then \begin{align*} m+1\leq \operatorname{ord}_\infty(f)\leq \frac{k}{12}=m+\frac{r}{12}. \end{align*} Since $r\leq 10$, we have $m+r/12<m+1$, a contradiction. Hence no nonzero form can have its first $m+1$ Fourier coefficients all equal to zero. Equivalently, the map sending $f$ to its first $m+1$ Fourier coefficients is injective, so \begin{align*} \dim_{\mathbb C} M_k(SL_2(\mathbb Z))\leq m+1. \end{align*} There is one exceptional residue class. When $k=12m+2$, we can improve the bound by one. Suppose $f\in M_k(SL_2(\mathbb Z))$ is nonzero and $\operatorname{ord}_\infty(f)\geq m$. The cusp contribution uses up $m$ of the total $m+1/6$, leaving only $1/6$ for all zeros in $\mathbb H$. But the elliptic fixed point $i$ forces extra vanishing. Since $k/2=6m+1$ is odd, the local transformation law at $i$ forces $\operatorname{ord}_i(f)$ to be odd, hence at least $1$. Therefore the $i$-term in the valence formula contributes at least \begin{align*} \frac{1}{2}\operatorname{ord}_i(f)\geq \frac{1}{2}, \end{align*} which is larger than the available $1/6$. This contradiction proves that no nonzero form in weight $12m+2$ can vanish to order at least $m$ at the cusp. Thus the first $m$ Fourier coefficients determine the form, and \begin{align*} \dim_{\mathbb C} M_{12m+2}(SL_2(\mathbb Z))\leq m. \end{align*}[/guided]

[step:Count and separate the monomials of each weight]For a fixed even integer $k\geq 0$, define \begin{align*} \mathcal B_k:=\{E_4^aE_6^b : a,b\in\mathbb Z_{\geq 0},\ 4a+6b=k\}. \end{align*} Each element of $\mathcal B_k$ lies in $M_k(SL_2(\mathbb Z))$ because products of modular forms are modular forms and weights add. The number of solutions to $4a+6b=k$ is \begin{align*} |\mathcal B_k|= \begin{cases} m, & k=12m+2,\\ m+1, & k=12m+r,\ r\in\{0,4,6,8,10\}. \end{cases} \end{align*} Indeed, dividing by $2$ gives $2a+3b=k/2$. The parity of $b$ is determined by $k/2$, and the possible nonnegative values of $b$ in that parity class give exactly the displayed count. The set $\mathcal B_k$ is linearly independent. If \begin{align*} \sum_{4a+6b=k} c_{a,b}E_4^aE_6^b=0 \end{align*} is a nontrivial linear relation, choose an index pair $(a_0,b_0)$ with $c_{a_0,b_0}\neq 0$ and $a_0$ minimal. Since $\operatorname{ord}_\rho(E_4)=1$ and $E_6(\rho)\neq 0$, the term $E_4^{a_0}E_6^{b_0}$ has vanishing order exactly $a_0$ at $\rho$, while every term with larger $a$ has vanishing order larger than $a_0$. Therefore the coefficient of the lowest nonzero power in the local expansion at $\rho$ is nonzero, contradicting the assumed relation. Hence all $c_{a,b}=0$.[/step]

[guided]We now build explicit modular forms. Since $E_4$ has weight $4$ and $E_6$ has weight $6$, the product $E_4^aE_6^b$ has weight $4a+6b$. Thus, for fixed $k$, the relevant monomials are \begin{align*} \mathcal B_k:=\{E_4^aE_6^b : a,b\in\mathbb Z_{\geq 0},\ 4a+6b=k\}. \end{align*} The count matches the dimension bound. Write $k=12m+r$ with $r\in\{0,2,4,6,8,10\}$. The equation $4a+6b=k$ is equivalent to \begin{align*} 2a+3b=\frac{k}{2}. \end{align*} Once $b$ is chosen, $a$ is forced by \begin{align*} a=\frac{k/2-3b}{2}. \end{align*} Thus $b$ must have the same parity as $k/2$ and must satisfy $0\leq 3b\leq k/2$. Counting those allowed $b$ gives $m+1$ possibilities except in the residue class $k\equiv 2\pmod{12}$, where it gives $m$ possibilities. It remains to prove that these monomials do not satisfy a linear relation. The point $\rho$ separates them. We already proved that $E_4$ has a simple zero at $\rho$ and $E_6$ does not vanish at $\rho$. Therefore \begin{align*} \operatorname{ord}_\rho(E_4^aE_6^b)=a. \end{align*} If a nonzero linear relation existed, choose among its nonzero terms one with minimal exponent $a_0$ of $E_4$. All other terms have vanishing order strictly larger than $a_0$ at $\rho$, so they cannot cancel the leading term of order $a_0$. This contradiction proves [linear independence](/page/Linear%20Independence).[/guided]

[step:Conclude that $E_4$ and $E_6$ generate the graded algebra] For each even $k\geq 0$, the linearly independent set $\mathcal B_k$ has cardinality equal to the upper bound for $\dim_{\mathbb C}M_k(SL_2(\mathbb Z))$ obtained from the valence formula. Hence $\mathcal B_k$ is a basis of $M_k(SL_2(\mathbb Z))$. For odd $k$, both $M_k(SL_2(\mathbb Z))$ and the weight-$k$ homogeneous part of $\mathbb C[E_4,E_6]$ are zero. Therefore every modular form of nonnegative weight is a unique homogeneous polynomial in $E_4$ and $E_6$, and multiplication corresponds to multiplication of polynomials. Thus \begin{align*} \bigoplus_{k\geq 0}M_k(SL_2(\mathbb Z))=\mathbb C[E_4,E_6] \end{align*} as graded $\mathbb C$-algebras. [/step]

[step:Show that every cusp form is divisible by $\Delta$]The Fourier expansions of the normalized Eisenstein series satisfy \begin{align*} E_4(q)&=1+240q+O(q^2),\\ E_6(q)&=1-504q+O(q^2), \end{align*} so \begin{align*} \Delta(q)=\frac{E_4(q)^3-E_6(q)^2}{1728}=q+O(q^2). \end{align*} Thus $\operatorname{ord}_\infty(\Delta)=1$. Since $\Delta\in M_{12}(SL_2(\mathbb Z))$, the valence formula gives \begin{align*} 1\leq \operatorname{ord}_\infty(\Delta)\leq \frac{12}{12}=1. \end{align*} All remaining terms in the valence formula must therefore be zero, so $\Delta$ has no zeros on $\mathbb H$. Let $k\geq 12$ and let $f\in S_k(SL_2(\mathbb Z))$. Define \begin{align*} g:\mathbb H &\to \mathbb C\\ z &\mapsto \frac{f(z)}{\Delta(z)}. \end{align*} Because $\Delta$ has no zeros on $\mathbb H$, the function $g$ is holomorphic on $\mathbb H$. Since $f$ has weight $k$ and $\Delta$ has weight $12$, the quotient $g$ has weight $k-12$. At the cusp, $f$ has $\operatorname{ord}_\infty(f)\geq 1$ and $\Delta$ has $\operatorname{ord}_\infty(\Delta)=1$, so \begin{align*} \operatorname{ord}_\infty(g)=\operatorname{ord}_\infty(f)-1\geq 0. \end{align*} Hence $g\in M_{k-12}(SL_2(\mathbb Z))$, and $f=\Delta g$. This proves \begin{align*} S_k(SL_2(\mathbb Z))\subset \Delta M_{k-12}(SL_2(\mathbb Z)). \end{align*} The reverse inclusion holds because if $g\in M_{k-12}(SL_2(\mathbb Z))$, then $\Delta g$ has weight $k$ and vanishes at the cusp since $\Delta$ vanishes there. Therefore \begin{align*} S_k(SL_2(\mathbb Z))=\Delta M_{k-12}(SL_2(\mathbb Z)) \end{align*} for $k\geq 12$.[/step]

[guided]The cusp-form statement is a divisibility statement. The role of $\Delta$ is that it is the unique basic modular form with a simple zero at the cusp and no zeros in the upper half-plane. First compute its first Fourier term. The normalized Eisenstein series have expansions \begin{align*} E_4(q)&=1+240q+O(q^2),\\ E_6(q)&=1-504q+O(q^2). \end{align*} Therefore \begin{align*} E_4(q)^3&=1+720q+O(q^2),\\ E_6(q)^2&=1-1008q+O(q^2), \end{align*} and hence \begin{align*} \Delta(q)=\frac{E_4(q)^3-E_6(q)^2}{1728}=q+O(q^2). \end{align*} So $\Delta$ has a simple zero at the cusp. Now apply the valence formula to $\Delta$, which has weight $12$. The right-hand side is $1$. Since the cusp already contributes $\operatorname{ord}_\infty(\Delta)=1$, every other nonnegative contribution must be zero. Hence $\Delta$ has no zeros on $\mathbb H$. Let $f\in S_k(SL_2(\mathbb Z))$ with $k\geq 12$. Because $f$ is a cusp form, its Fourier expansion has no constant term, so $\operatorname{ord}_\infty(f)\geq 1$. Define \begin{align*} g:\mathbb H &\to \mathbb C\\ z &\mapsto \frac{f(z)}{\Delta(z)}. \end{align*} This quotient is holomorphic on $\mathbb H$ because $\Delta$ has no zeros there. It is modular of weight $k-12$ because the transformation factors for $f$ and $\Delta$ divide. At the cusp, division by $\Delta$ subtracts exactly one order of vanishing: \begin{align*} \operatorname{ord}_\infty(g)=\operatorname{ord}_\infty(f)-1\geq 0. \end{align*} Thus $g$ is holomorphic at the cusp, so $g\in M_{k-12}(SL_2(\mathbb Z))$ and $f=\Delta g$. Conversely, if $g\in M_{k-12}(SL_2(\mathbb Z))$, then $\Delta g$ has weight $k$ and vanishes at the cusp because $\Delta$ has a zero there. Therefore $\Delta g\in S_k(SL_2(\mathbb Z))$.[/guided]

[step:Exclude cusp forms below weight twelve] Let $k<12$ and suppose $f\in S_k(SL_2(\mathbb Z))$ is nonzero. If $k$ is odd, then $M_k(SL_2(\mathbb Z))=0$, so no such $f$ exists. If $k$ is even, then $\operatorname{ord}_\infty(f)\geq 1$ because $f$ is cuspidal. The valence formula gives \begin{align*} 1\leq \operatorname{ord}_\infty(f)\leq \frac{k}{12}<1, \end{align*} a contradiction. Hence \begin{align*} S_k(SL_2(\mathbb Z))=0 \end{align*} for every $k<12$. This completes the proof. [/step]