[proofplan] We prove the prime-to-level statement, since the full Hecke algebra including the bad-prime operators $U_p:=T_p$ for primes $p$ with $p\mid N$ need not act semisimply on oldforms. First we realize the prime-to-level Hecke algebra on the finite-dimensional rational modular-form space coming from $q$-expansions, which gives finite-dimensionality and a faithful scalar extension to $\mathbb{C}$. The main input is the simultaneous spectral theorem for the commuting normal prime-to-level Hecke operators for the Petersson inner product. This identifies the complexified prime-to-level Hecke algebra with a subalgebra of a finite product of copies of $\mathbb{C}$, hence shows it is reduced; reducedness then descends from $\mathbb{C}$ to $\mathbb{Q}$, and the Artinian radical criterion gives semisimplicity. [/proofplan]

[step:Realize the prime-to-level rational Hecke algebra on the rational modular-form space] Let \begin{align*} V := S_2(\Gamma_0(N)) \end{align*} be the finite-dimensional complex [vector space](/page/Vector%20Space) of weight-$2$ cusp forms of level $\Gamma_0(N)$. Let $V_{\mathbb{Q}}$ denote the finite-dimensional $\mathbb{Q}$-vector space of cusp forms in $V$ whose $q$-expansions have rational coefficients, so that the scalar-extension map gives \begin{align*} V_{\mathbb{Q}} \otimes_{\mathbb{Q}} \mathbb{C} &\cong V. \end{align*} For every integer $n\geq 1$ with $\gcd(n,N)=1$, the Hecke operator $T_n$ preserves rational $q$-expansions, hence defines a $\mathbb{Q}$-linear endomorphism of $V_{\mathbb{Q}}$. Let $\mathbb{T}^{(N)}$ be the $\mathbb{Z}$-subalgebra of $\operatorname{End}_{\mathbb{Q}}(V_{\mathbb{Q}})$ generated by these operators, and define \begin{align*} \mathbb{T}^{(N)}_{\mathbb{Q}} := \mathbb{T}^{(N)} \otimes_{\mathbb{Z}} \mathbb{Q}. \end{align*} Equivalently, $\mathbb{T}^{(N)}_{\mathbb{Q}}$ is the $\mathbb{Q}$-subalgebra of $\operatorname{End}_{\mathbb{Q}}(V_{\mathbb{Q}})$ generated by the operators $T_n$ with $\gcd(n,N)=1$. Since $\operatorname{End}_{\mathbb{Q}}(V_{\mathbb{Q}})$ has $\mathbb{Q}$-dimension $(\dim_{\mathbb{Q}}V_{\mathbb{Q}})^2$, the subspace $\mathbb{T}^{(N)}_{\mathbb{Q}}$ is finite-dimensional over $\mathbb{Q}$. The Hecke operators $T_n$ with $\gcd(n,N)=1$ commute with one another, so the algebra they generate is commutative. Thus $\mathbb{T}^{(N)}_{\mathbb{Q}}$ is a finite-dimensional commutative $\mathbb{Q}$-algebra. Moreover, scalar extension gives an injective $\mathbb{C}$-algebra homomorphism \begin{align*} \iota_{\mathbb{C}}: \mathbb{T}^{(N)}_{\mathbb{Q}}\otimes_{\mathbb{Q}}\mathbb{C} &\to \operatorname{End}_{\mathbb{C}}(V), \end{align*} because extension of scalars from $\mathbb{Q}$ to $\mathbb{C}$ identifies $\operatorname{End}_{\mathbb{Q}}(V_{\mathbb{Q}})\otimes_{\mathbb{Q}}\mathbb{C}$ with $\operatorname{End}_{\mathbb{C}}(V)$ and preserves injective maps. [/step]

[step:Diagonalize the complexified prime-to-level Hecke action]Define the complexified prime-to-level Hecke algebra \begin{align*} \mathbb{T}^{(N)}_{\mathbb{C}} := \mathbb{T}^{(N)}_{\mathbb{Q}} \otimes_{\mathbb{Q}} \mathbb{C}. \end{align*} The action of $\mathbb{T}^{(N)}_{\mathbb{Q}}$ on $V$ extends uniquely to a $\mathbb{C}$-algebra action of $\mathbb{T}^{(N)}_{\mathbb{C}}$ on $V$. We use the Petersson inner product on $V$. For every integer $n\geq 1$ with $\gcd(n,N)=1$, the standard prime-to-level Hecke adjointness relation gives $T_n^*=T_n$ for this inner product; hence each $T_n$ is normal. These operators commute with one another. Since $V$ is finite-dimensional over $\mathbb{C}$, the [simultaneous spectral theorem for commuting normal operators](/page/Spectral%20Theorem), applied to this possibly infinite commuting family, gives a finite set $\Lambda$ of $\mathbb{C}$-algebra characters \begin{align*} \lambda: \mathbb{T}^{(N)}_{\mathbb{C}} &\to \mathbb{C} \end{align*} and a direct sum decomposition \begin{align*} V = \bigoplus_{\lambda \in \Lambda} V_{\lambda}, \end{align*} where $V_{\lambda}$ is the nonzero simultaneous eigenspace \begin{align*} V_{\lambda}:=\{v \in V : A v = \lambda(A)v \text{ for every } A \in \mathbb{T}^{(N)}_{\mathbb{C}}\}. \end{align*} Thus, for every $\lambda \in \Lambda$, every $v \in V_{\lambda}$, and every $A \in \mathbb{T}^{(N)}_{\mathbb{C}}$, one has \begin{align*} A v = \lambda(A)v. \end{align*}[/step]

[guided]The point of restricting to the prime-to-level Hecke algebra is that the available diagonalisation theorem applies exactly to those operators. On the finite-dimensional complex vector space \begin{align*} V := S_2(\Gamma_0(N)), \end{align*} the Petersson inner product is a positive definite Hermitian inner product. For every integer $n \geq 1$ with $\gcd(n,N)=1$, the standard prime-to-level Hecke adjointness relation gives $T_n^*=T_n$, so $T_n$ is self-adjoint and therefore normal. The operators $T_n$ commute with one another. Thus the hypotheses of the [simultaneous spectral theorem for commuting normal operators](/page/Spectral%20Theorem) are satisfied for the possibly infinite family generating $\mathbb{T}^{(N)}_{\mathbb{C}}$: the space is finite-dimensional, each operator is normal, and all operators commute. The theorem gives a direct sum decomposition of $V$ into common eigenspaces for the prime-to-level Hecke action. Equivalently, there is a finite set $\Lambda$ of characters \begin{align*} \lambda: \mathbb{T}^{(N)}_{\mathbb{C}} &\to \mathbb{C} \end{align*} and nonzero subspaces \begin{align*} V_{\lambda}:=\{v \in V : A v = \lambda(A)v \text{ for every } A \in \mathbb{T}^{(N)}_{\mathbb{C}}\} \end{align*} such that \begin{align*} V = \bigoplus_{\lambda \in \Lambda} V_{\lambda}. \end{align*} Thus, for every prime-to-level Hecke operator $A \in \mathbb{T}^{(N)}_{\mathbb{C}}$ and every vector $v \in V_{\lambda}$, the action is scalar: \begin{align*} A v = \lambda(A)v. \end{align*} This avoids the bad-prime operators $U_p$ with $p \mid N$, whose action on oldforms is not supplied by this normal-operator argument.[/guided]

[step:Identify the complexified prime-to-level Hecke algebra with a reduced algebra]Define the evaluation homomorphism \begin{align*} \Phi: \mathbb{T}^{(N)}_{\mathbb{C}} &\to \prod_{\lambda \in \Lambda} \mathbb{C} \\ A &\mapsto \bigl(\lambda(A)\bigr)_{\lambda \in \Lambda}. \end{align*} This is a $\mathbb{C}$-algebra homomorphism because each $\lambda$ is a $\mathbb{C}$-algebra character. We show that $\Phi$ is injective. Let $A \in \ker \Phi$. Then $\lambda(A)=0$ for every $\lambda \in \Lambda$. Therefore $A$ acts as the zero operator on each summand $V_{\lambda}$. Since \begin{align*} V = \bigoplus_{\lambda \in \Lambda} V_{\lambda}, \end{align*} the operator $A$ acts as zero on all of $V$. The action of $\mathbb{T}^{(N)}_{\mathbb{C}}$ on $V$ is faithful by the injectivity of the scalar-extension map $\iota_{\mathbb{C}}$ constructed in the first step. Hence $A = 0$, so $\Phi$ is injective. Thus $\mathbb{T}^{(N)}_{\mathbb{C}}$ is isomorphic to a $\mathbb{C}$-subalgebra of the product algebra $\prod_{\lambda \in \Lambda}\mathbb{C}$. Since a finite product of fields has no nonzero nilpotent elements, every subalgebra of it has no nonzero nilpotent elements. Therefore $\mathbb{T}^{(N)}_{\mathbb{C}}$ is reduced.[/step]

[guided]The purpose of the map $\Phi$ is to record how a Hecke operator acts on each simultaneous eigenspace. If two operators have the same scalar on every eigenspace, then they act in exactly the same way on every summand of the direct sum decomposition, and hence they are the same operator. Define \begin{align*} \Phi: \mathbb{T}^{(N)}_{\mathbb{C}} &\to \prod_{\lambda \in \Lambda} \mathbb{C} \\ A &\mapsto \bigl(\lambda(A)\bigr)_{\lambda \in \Lambda}. \end{align*} Because $\lambda(AB)=\lambda(A)\lambda(B)$ and $\lambda(A+B)=\lambda(A)+\lambda(B)$ for each character $\lambda$, this is a $\mathbb{C}$-algebra homomorphism. Now suppose $A \in \ker \Phi$. Then every coordinate of $\Phi(A)$ is zero, so $\lambda(A)=0$ for each $\lambda \in \Lambda$. If $v \in V_{\lambda}$, the simultaneous eigenspace property gives \begin{align*} A v = \lambda(A)v = 0. \end{align*} Since every vector $v \in V$ decomposes uniquely as a finite sum \begin{align*} v = \sum_{\lambda \in \Lambda} v_{\lambda}, \qquad v_{\lambda} \in V_{\lambda}, \end{align*} we get \begin{align*} A v = \sum_{\lambda \in \Lambda} A v_{\lambda} = 0. \end{align*} So $A$ is the zero endomorphism of $V$. This implies $A=0$ in $\mathbb{T}^{(N)}_{\mathbb{C}}$ because the first step constructed an injective scalar-extension action \begin{align*} \iota_{\mathbb{C}}: \mathbb{T}^{(N)}_{\mathbb{Q}}\otimes_{\mathbb{Q}}\mathbb{C} &\to \operatorname{End}_{\mathbb{C}}(V). \end{align*} The injectivity ultimately comes from the rational $q$-expansion model $V_{\mathbb{Q}}$ and scalar extension from $\mathbb{Q}$ to $\mathbb{C}$. Hence $\Phi$ is injective. The product algebra $\prod_{\lambda \in \Lambda}\mathbb{C}$ is reduced: if $(z_{\lambda})_{\lambda \in \Lambda}^m=0$, then $z_{\lambda}^m=0$ in $\mathbb{C}$ for every $\lambda$, and hence $z_{\lambda}=0$ for every $\lambda$. Since $\mathbb{T}^{(N)}_{\mathbb{C}}$ embeds into this reduced algebra, it also has no nonzero nilpotent elements.[/guided]

[step:Descend reducedness from $\mathbb{C}$ to $\mathbb{Q}$]We show that $\mathbb{T}^{(N)}_{\mathbb{Q}}$ is reduced. Let $a \in \mathbb{T}^{(N)}_{\mathbb{Q}}$ satisfy $a^m=0$ for some integer $m \geq 1$. Then \begin{align*} (a \otimes 1)^m = a^m \otimes 1 = 0 \end{align*} in $\mathbb{T}^{(N)}_{\mathbb{C}}$. Since $\mathbb{T}^{(N)}_{\mathbb{C}}$ is reduced, this implies \begin{align*} a \otimes 1 = 0. \end{align*} The [field extension](/page/Field%20Extension) $\mathbb{Q} \subset \mathbb{C}$ is [faithfully flat](/page/Flat%20Module), so the canonical map \begin{align*} \mathbb{T}^{(N)}_{\mathbb{Q}} &\to \mathbb{T}^{(N)}_{\mathbb{Q}} \otimes_{\mathbb{Q}} \mathbb{C} \\ a &\mapsto a \otimes 1 \end{align*} is injective. Therefore $a=0$. Hence $\mathbb{T}^{(N)}_{\mathbb{Q}}$ is reduced.[/step]

[guided]We must show that nilpotents cannot appear over $\mathbb{Q}$ if they disappear after extending scalars to $\mathbb{C}$. Let $a \in \mathbb{T}^{(N)}_{\mathbb{Q}}$ and suppose that $a^m=0$ for some integer $m\geq 1$. In the scalar extension $\mathbb{T}^{(N)}_{\mathbb{C}}=\mathbb{T}^{(N)}_{\mathbb{Q}}\otimes_{\mathbb{Q}}\mathbb{C}$, multiplication is compatible with tensor products, so \begin{align*} (a\otimes 1)^m = a^m\otimes 1 = 0. \end{align*} The previous step proved that $\mathbb{T}^{(N)}_{\mathbb{C}}$ is reduced, hence $a\otimes 1=0$. It remains to return from $a\otimes 1=0$ to $a=0$. The field extension $\mathbb{Q}\subset\mathbb{C}$ is [faithfully flat](/page/Flat%20Module), and faithful flatness implies that the scalar-extension map \begin{align*} \mathbb{T}^{(N)}_{\mathbb{Q}} &\to \mathbb{T}^{(N)}_{\mathbb{Q}}\otimes_{\mathbb{Q}}\mathbb{C} \\ a &\mapsto a\otimes 1 \end{align*} is injective. Therefore $a=0$. Since every nilpotent element of $\mathbb{T}^{(N)}_{\mathbb{Q}}$ is zero, the algebra $\mathbb{T}^{(N)}_{\mathbb{Q}}$ is reduced.[/guided]

[step:Conclude that the reduced finite-dimensional algebra is semisimple]It remains to prove semisimplicity. Let $A := \mathbb{T}^{(N)}_{\mathbb{Q}}$. We have shown that $A$ is a finite-dimensional commutative reduced algebra over the field $\mathbb{Q}$. For every maximal ideal $\mathfrak{m} \subset A$, the quotient $A/\mathfrak{m}$ is a finite field extension of $\mathbb{Q}$. Since $A$ is finite-dimensional over $\mathbb{Q}$, it is an [Artinian algebra](/page/Artinian%20Ring), so its nilradical equals its [Jacobson radical](/page/Jacobson%20Radical). Because $A$ is reduced, its nilradical is zero. Hence the Jacobson radical of $A$ is zero. By the [Jacobson radical criterion for semisimplicity](/page/Semisimple%20Algebra), a finite-dimensional algebra over a field is semisimple if and only if its Jacobson radical is zero. Therefore $A=\mathbb{T}^{(N)}_{\mathbb{Q}}$ is semisimple. Combining this with the first step, $\mathbb{T}^{(N)}_{\mathbb{Q}}$ is a finite-dimensional commutative semisimple $\mathbb{Q}$-algebra.[/step]

[guided]Set \begin{align*} A := \mathbb{T}^{(N)}_{\mathbb{Q}}. \end{align*} The first step showed that $A$ is finite-dimensional and commutative over $\mathbb{Q}$, and the previous step showed that $A$ is reduced. We now convert reducedness into semisimplicity. Because $A$ is finite-dimensional over the field $\mathbb{Q}$, it is an [Artinian algebra](/page/Artinian%20Ring). In an Artinian commutative ring, the nilradical agrees with the [Jacobson radical](/page/Jacobson%20Radical). The nilradical of $A$ is zero because $A$ is reduced, so the Jacobson radical of $A$ is zero: \begin{align*} \operatorname{Jac}(A)=0. \end{align*} The [Jacobson radical criterion for semisimplicity](/page/Semisimple%20Algebra) states that a finite-dimensional algebra over a field is semisimple exactly when its Jacobson radical is zero. Applying this criterion to the finite-dimensional $\mathbb{Q}$-algebra $A$ gives that $A$ is semisimple. Substituting back $A=\mathbb{T}^{(N)}_{\mathbb{Q}}$, we conclude that $\mathbb{T}^{(N)}_{\mathbb{Q}}$ is a finite-dimensional commutative semisimple $\mathbb{Q}$-algebra.[/guided]