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 $M\_k(\mathrm{SL}\_2\mathbb{Z})$? 



Algebra of Hecke operators on $M_k(\mathrm{SL}_2\mathbb{Z})$ is an integral domain?
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 $\dim(S\_k(SL\_2(Z)))=1$.


