[proofplan] The ring identity expresses every level $1$ modular form as a weighted polynomial in $E_4$ and $E_6$, and multiplication by $\Delta$ then identifies cusp forms with lower-weight modular forms. The Hecke coefficient formula turns each $T_n$ into an explicitly computable finite matrix after choosing enough initial Fourier coefficients to distinguish forms of a fixed weight. The simultaneous Hecke eigenbasis theorem diagonalizes these matrices on $S_k$. Finally, the Hecke relations for a normalized eigenform imply multiplicativity and the prime-power recurrence for the Fourier coefficients, and these two identities are exactly the Euler product expansion of $L(f,s)$. [/proofplan] [step:Use the graded ring identity to reduce modular forms to weighted polynomials] Let $R := \mathbb{C}[X,Y]$ be the [polynomial ring](/page/Polynomial%20Ring) graded by assigning $\deg X = 4$ and $\deg Y = 6$. Define the graded $\mathbb{C}$-algebra homomorphism \begin{align*} \Phi: R &\to M_* \\ P(X,Y) &\mapsto P(E_4,E_6). \end{align*} By the standard structure theorem for level $1$ modular forms, namely the graded ring theorem $M_*=\mathbb{C}[E_4,E_6]$ (citing a result not yet in the wiki: [Structure Theorem for Level One Modular Forms](/theorems/4264)), the map $\Phi$ is an isomorphism of graded $\mathbb{C}$-algebras. Therefore, for each integer $k$, the homogeneous component $M_k$ is spanned by the monomials \begin{align*} E_4^a E_6^b \end{align*} with $a,b \in \mathbb{Z}_{\geq 0}$ and $4a+6b=k$. Since there are only finitely many such pairs $(a,b)$, the construction of $M_k$ is reduced to finite-dimensional linear algebra in these monomials. [/step] [step:Use multiplication by $\Delta$ to construct the cusp-form spaces] For each integer $k$, define the multiplication map \begin{align*} \mu_{\Delta,k}: M_{k-12} &\to S_k \\ g &\mapsto \Delta g, \end{align*} with $M_{k-12}=\{0\}$ when $k-12<0$. By the standard cusp-form structure theorem for level $1$ modular forms (citing a result not yet in the wiki: Cusp Forms Are Multiples of the Discriminant), the map $\mu_{\Delta,k}$ is an isomorphism onto $S_k$. Hence the finite spanning set for $M_{k-12}$ obtained from the preceding step gives a finite spanning set for $S_k$, namely the forms \begin{align*} \Delta E_4^a E_6^b \end{align*} with $a,b \in \mathbb{Z}_{\geq 0}$ and $4a+6b=k-12$. [/step] [step:Compute Hecke matrices from finitely many Fourier coefficients] Fix $k \in \mathbb{Z}$ and let $V$ denote either $M_k$ or $S_k$. Since $V$ is finite-dimensional, choose a basis $v_1,\dots,v_r$ of $V$. For each $i \in \{1,\dots,r\}$, write the Fourier expansion \begin{align*} v_i(z)=\sum_{m=0}^{\infty} b_{i,m}q^m. \end{align*} The $q$-expansion map \begin{align*} Q: V &\to \mathbb{C}[[q]] \\ h &\mapsto \sum_{m=0}^{\infty} c_m q^m \end{align*} is injective, because a modular form whose Fourier expansion at the cusp is identically zero vanishes on a punctured neighbourhood of the cusp and hence vanishes identically by the identity theorem for holomorphic functions. Since $V$ is finite-dimensional and $Q(v_1),\dots,Q(v_r)$ are linearly independent in $\mathbb{C}[[q]]$, there exists $N \in \mathbb{Z}_{\geq 0}$ such that the truncated coefficient map \begin{align*} Q_N: V &\to \mathbb{C}^{N+1} \\ h=\sum_{m=0}^{\infty} c_m q^m &\mapsto (c_0,c_1,\dots,c_N) \end{align*} is injective. Now fix $n \in \mathbb{N}$. For a form \begin{align*} h(z)=\sum_{m=0}^{\infty} c_mq^m \end{align*} in $V$, the Hecke coefficient formula gives \begin{align*} T_n h = \sum_{m=0}^{\infty} \left( \sum_{d\mid \gcd(m,n)} d^{k-1}c_{mn/d^2} \right)q^m. \end{align*} Thus the first $N+1$ coefficients of $T_n h$ are determined by finitely many coefficients of $h$. Since each basis vector $v_i$ is explicitly computable from the polynomial expressions in $E_4$, $E_6$, and possibly $\Delta$, the vectors $Q_N(T_n v_i)$ can be computed finitely. Injectivity of $Q_N$ then determines the unique coordinates of $T_n v_i$ in the basis $v_1,\dots,v_r$, and hence determines the matrix of $T_n$ on $V$. [/step] [step:Diagonalize the Hecke action on cusp forms] By the simultaneous Hecke eigenbasis theorem for level $1$ cusp forms (citing a result not yet in the wiki: Simultaneous Diagonalization of Hecke Operators on Cusp Forms), for every integer $k$ the finite-dimensional complex [vector space](/page/Vector%20Space) $S_k$ has a basis consisting of common eigenvectors for all operators $T_n$. Equivalently, there are forms $f_1,\dots,f_d \in S_k$, where $d=\dim_{\mathbb{C}}S_k$, such that each $f_j$ is nonzero and for every $n \in \mathbb{N}$ there exists $\lambda_{j,n}\in\mathbb{C}$ satisfying \begin{align*} T_n f_j = \lambda_{j,n} f_j. \end{align*} After scaling any eigenvector with nonzero first Fourier coefficient, one obtains a normalized eigenform. The preceding step gives the matrices of the $T_n$, so this eigenbasis is obtained by finite-dimensional diagonalization. [/step] [step:Identify the Hecke eigenvalues with Fourier coefficients] Let \begin{align*} f: \mathbb{H} &\to \mathbb{C} \\ z &\mapsto \sum_{m=1}^{\infty} a_mq^m \end{align*} be a normalized simultaneous Hecke eigenform in $S_k$, so $a_1=1$. For each $n\in\mathbb{N}$, let $\lambda_n\in\mathbb{C}$ be the scalar defined by \begin{align*} T_n f=\lambda_n f. \end{align*} Taking the coefficient of $q^1$ in the Hecke formula gives \begin{align*} [T_n f]_{q^1} = \sum_{d\mid \gcd(1,n)} d^{k-1}a_{n/d^2} = a_n. \end{align*} On the other hand, the coefficient of $q^1$ in $\lambda_n f$ is $\lambda_n a_1=\lambda_n$. Since $a_1=1$, we obtain \begin{align*} \lambda_n=a_n \end{align*} for every $n\in\mathbb{N}$. [/step] [step:Derive multiplicativity and the prime-power recurrence] The normalized Hecke operators satisfy the standard Hecke relation \begin{align*} T_mT_n=\sum_{d\mid \gcd(m,n)} d^{k-1}T_{mn/d^2} \end{align*} on $S_k$ (citing a result not yet in the wiki: Hecke Operator Multiplication Formula). Applying this identity to the common eigenform $f$ and using $\lambda_j=a_j$ gives \begin{align*} a_ma_n=\sum_{d\mid \gcd(m,n)} d^{k-1}a_{mn/d^2}. \end{align*} If $\gcd(m,n)=1$, the only divisor is $d=1$, and hence \begin{align*} a_{mn}=a_ma_n. \end{align*} Thus the coefficient sequence $(a_n)_{n\geq 1}$ is multiplicative. Now fix a prime number $p$ and an integer $r\geq 1$. Taking $m=p$ and $n=p^r$ in the same Hecke relation gives \begin{align*} a_pa_{p^r}=a_{p^{r+1}}+p^{k-1}a_{p^{r-1}}. \end{align*} Equivalently, \begin{align*} a_{p^{r+1}}=a_pa_{p^r}-p^{k-1}a_{p^{r-1}}. \end{align*} This is the prime-power recurrence. [guided] We first translate the operator identities into coefficient identities. The Hecke multiplication formula says that, on weight $k$ cusp forms, \begin{align*} T_mT_n=\sum_{d\mid \gcd(m,n)} d^{k-1}T_{mn/d^2}. \end{align*} This formula applies to $f$ because $f\in S_k$ and $f$ is a simultaneous eigenvector for every $T_n$. Applying both sides to $f$ gives \begin{align*} T_mT_n f = \sum_{d\mid \gcd(m,n)} d^{k-1}T_{mn/d^2}f. \end{align*} Since $T_jf=a_jf$ for every $j\in\mathbb{N}$, the left-hand side is \begin{align*} T_m(a_nf)=a_nT_mf=a_na_mf, \end{align*} and the right-hand side is \begin{align*} \sum_{d\mid \gcd(m,n)} d^{k-1}a_{mn/d^2}f. \end{align*} Because $f\neq 0$, the scalar coefficients are equal: \begin{align*} a_ma_n=\sum_{d\mid \gcd(m,n)} d^{k-1}a_{mn/d^2}. \end{align*} When $\gcd(m,n)=1$, the divisor set contains only $d=1$. Therefore the preceding identity reduces to \begin{align*} a_ma_n=a_{mn}. \end{align*} This is the multiplicativity needed to factor the Dirichlet series into local prime factors. For a prime $p$ and an integer $r\geq 1$, the common divisors of $p$ and $p^r$ are $1$ and $p$. Substituting $m=p$ and $n=p^r$ gives \begin{align*} a_pa_{p^r} = a_{p^{r+1}}+p^{k-1}a_{p^{r-1}}. \end{align*} Solving for $a_{p^{r+1}}$ yields \begin{align*} a_{p^{r+1}}=a_pa_{p^r}-p^{k-1}a_{p^{r-1}}. \end{align*} This recurrence determines all prime-power coefficients from $a_p$ and $a_1=1$. [/guided] [/step] [step:Convert the coefficient identities into the Euler product] Let $s\in\mathbb{C}$ lie in the half-plane where \begin{align*} \sum_{n=1}^{\infty} |a_n n^{-s}| \end{align*} converges. Absolute convergence permits rearrangement of the Dirichlet series by unique factorization and multiplicativity: \begin{align*} L(f,s) = \sum_{n=1}^{\infty}a_nn^{-s} = \prod_p\left(\sum_{r=0}^{\infty}a_{p^r}p^{-rs}\right). \end{align*} For a fixed prime $p$, define \begin{align*} A_p: \{X\in\mathbb{C}: |X|\ \text{is sufficiently small}\} &\to \mathbb{C} \\ X &\mapsto \sum_{r=0}^{\infty}a_{p^r}X^r. \end{align*} Using $a_1=1$ and the recurrence $a_{p^{r+1}}=a_pa_{p^r}-p^{k-1}a_{p^{r-1}}$, we compute \begin{align*} (1-a_pX+p^{k-1}X^2)A_p(X)=1. \end{align*} Therefore \begin{align*} A_p(X)=\left(1-a_pX+p^{k-1}X^2\right)^{-1}. \end{align*} Substituting $X=p^{-s}$ gives \begin{align*} \sum_{r=0}^{\infty}a_{p^r}p^{-rs} = \left(1-a_pp^{-s}+p^{k-1-2s}\right)^{-1}. \end{align*} Hence, in the half-plane of absolute convergence, \begin{align*} L(f,s) = \prod_p\left(1-a_pp^{-s}+p^{k-1-2s}\right)^{-1}. \end{align*} This is the asserted Euler product, and the proof is complete. [guided] The goal is to pass from the coefficient relations to a product over primes. We are allowed to rearrange the series because $s$ is assumed to lie in the half-plane of absolute convergence: \begin{align*} \sum_{n=1}^{\infty}|a_nn^{-s}|<\infty. \end{align*} By unique factorization, every $n\in\mathbb{N}$ has a unique expression as a finite product of prime powers. Since the coefficients are multiplicative, the absolutely convergent Dirichlet series factors as \begin{align*} L(f,s) = \sum_{n=1}^{\infty}a_nn^{-s} = \prod_p\left(\sum_{r=0}^{\infty}a_{p^r}p^{-rs}\right). \end{align*} It remains to compute each local factor. Fix a prime $p$ and define the local generating function \begin{align*} A_p: \{X\in\mathbb{C}: |X|\ \text{is sufficiently small}\} &\to \mathbb{C} \\ X &\mapsto \sum_{r=0}^{\infty}a_{p^r}X^r. \end{align*} The recurrence \begin{align*} a_{p^{r+1}}=a_pa_{p^r}-p^{k-1}a_{p^{r-1}} \end{align*} is designed precisely to make this generating function rational. Multiplying by $1-a_pX+p^{k-1}X^2$, we get \begin{align*} (1-a_pX+p^{k-1}X^2)A_p(X) &= \sum_{r=0}^{\infty}a_{p^r}X^r - a_p\sum_{r=0}^{\infty}a_{p^r}X^{r+1} + p^{k-1}\sum_{r=0}^{\infty}a_{p^r}X^{r+2}. \end{align*} The constant term is $a_1=1$. The coefficient of $X$ is $a_p-a_pa_1=0$. For every power $X^{r+1}$ with $r\geq 1$, the coefficient is \begin{align*} a_{p^{r+1}}-a_pa_{p^r}+p^{k-1}a_{p^{r-1}}, \end{align*} which is $0$ by the prime-power recurrence. Thus \begin{align*} (1-a_pX+p^{k-1}X^2)A_p(X)=1, \end{align*} and therefore \begin{align*} A_p(X)=\left(1-a_pX+p^{k-1}X^2\right)^{-1}. \end{align*} Putting $X=p^{-s}$ gives the local factor \begin{align*} \sum_{r=0}^{\infty}a_{p^r}p^{-rs} = \left(1-a_pp^{-s}+p^{k-1-2s}\right)^{-1}. \end{align*} Combining the local factors over all primes yields \begin{align*} L(f,s) = \prod_p\left(1-a_pp^{-s}+p^{k-1-2s}\right)^{-1}. \end{align*} [/guided] [/step]