Algebra of Hecke operators on $M_k(\mathrm{SL}_2\mathbb{Z})$ is an integral domain?

Question

Let $M_k(\mathrm{SL}_2\mathbb{Z})$ be the space of modular forms of (integer) weight for the full modular group. Let $\mathbf{H}$ denote the Algebra generated by the Hecke operators $T_n$. Is $\mathbf{H}$ an integral domain? Specifically, let $T_mT_n=0,$ can we conclude either $T_m=0$ or $T_n=0$ on $M_k(\mathrm{SL}_2\mathbb{Z})$?

Answer 1

You mean the (commutative, normal for the Petersson inner product thus diagonalizable) complex algebra $\Bbb{T(C)}$ of endomorphisms of the complex vector space $M_k(SL_2(Z))$ ($k$ even) generated by the identity and the $T_n$.

If $\dim(M_k(SL_2(Z))=1$ then it is $= \Bbb{C}$, otherwise it is not an integral domain, as it contains $E_k$ plus a cusp eigenform $f$, take some $p$ such that $\sigma_{k-1}(p)\ne a_p(f)$ then $T_p E_k= \sigma_{k-1}(p)E_k, T_pf=a_p(f)f$ ie. the minimal polynomial of $T_p $ is $(X-\sigma_{k-1}(p))(X-a_p(f))g(X)$ so that $(T_p-a_p(f)) (T_p-\sigma_{k-1}(p))g(T_p)=0$.

From that $E_4^3-E_6^2=1728\Delta$ has only one simple zero at $i\infty$ we get the $\Bbb{C}$-basis of modular forms with rational coefficients $$ M_k(SL_2(Z))=\sum_{4a+6b=k} \Bbb{C}E_4^a E_6^b$$ Since $T_n E_4^a E_6^b$ has rational coefficients too this implies the matrix of $T_n$ in this basis has rational entries so that the minimal polynomial of $T_n$ is in $\Bbb{Q}[X]$ and hence $\Bbb{T(Q)},\Bbb{T(Z)}$ are not integral domains neither.

On the other hand $T_nT_m\ne 0$ because $T_nT_m E_k = \sigma_{k-1}(n)\sigma_{k-1}(m)E_k$.

Maeda's conjecture is saying $S_k(SL_2(Z))$ is generated by the Galois orbit of a single eigenform $f$, in which case $\Bbb{T(Q)}|_{S_k(SL_2(Z))}$ is isomorphic to $\Bbb{Q}(\{a_n(f)\})$ which is an integral domain. $\Bbb{T(C)}|_{S_k(SL_2(Z))}$ is an integral domain iff $\dim(S_k(SL_2(Z)))=1$.