Return to Answer

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 and a cusp eigenform $f$, take some $n$ such that $\sigma_{k-1}(p)\ne a_p(f)$ then $T_p E_k= \sigma_{k-1}(n)E_k, T_pf=a_p(f)f$ 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 $\dim(S_k(SL_2(Z)))=1$.

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 $\dim(S_k(SL_2(Z)))=1$.

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 and a cusp eigenform $f$, take some $n$ such that $\sigma_{k-1}(p)\ne a_p(f)$ then $T_p E_k= \sigma_{k-1}(n)E_k, T_pf=a_p(f)f$ 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 $\sum_k M_k(SL_2(Z)) = \Bbb{C}[E_4,E_6]$ we know that $ M_k(SL_2(Z))=\sum_{4a+6b=k} \Bbb{C}E_4^a E_6^b$ is a $\Bbb{C}$-basis of $M_k(SL_2(Z)$ of modular forms with rational coefficients, since $T_n E_4^a E_6^b$ has rational coefficients too this implies the minimal polynomial of $T_p$ is in $\Bbb{Q}[X]$ thus $\Bbb{T(Q)},\Bbb{T(Z)}$ 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 cusp form $f$, 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 $\dim(S_k(SL_2(Z)))=1$.

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 and a cusp eigenform $f$, take some $n$ such that $\sigma_{k-1}(p)\ne a_p(f)$ then $T_p E_k= \sigma_{k-1}(n)E_k, T_pf=a_p(f)f$ 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 $\dim(S_k(SL_2(Z)))=1$.

You mean the 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 and a cusp eigenform $f$, take some $n$ such that $\sigma_{k-1}(p)\ne a_p(f)$ then $T_p E_k= \sigma_{k-1}(n)E_k, T_pf=a_p(f)f$ 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 $\sum_k M_k(SL_2(Z)) = \Bbb{C}[E_4,E_6]$ we know that $ M_k(SL_2(Z))=\sum_{4a+6b=k} \Bbb{C}E_4^a E_6^b$ is a $\Bbb{C}$-basis of $M_k(SL_2(Z)$ of modular forms with rational coefficients, since $T_n E_4^a E_6^b$ has rational coefficients too this implies the minimal polynomial of $T_p$ is in $\Bbb{Q}[X]$ thus $\Bbb{T(Q)},\Bbb{T(Z)}$ 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 cusp form $f$, in which case $\Bbb{T(Q)}|_{S_k}$ is isomorphic to $\Bbb{Q}(\{a_n(f)\})$ which 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 and a cusp eigenform $f$, take some $n$ such that $\sigma_{k-1}(p)\ne a_p(f)$ then $T_p E_k= \sigma_{k-1}(n)E_k, T_pf=a_p(f)f$ 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 $\sum_k M_k(SL_2(Z)) = \Bbb{C}[E_4,E_6]$ we know that $ M_k(SL_2(Z))=\sum_{4a+6b=k} \Bbb{C}E_4^a E_6^b$ is a $\Bbb{C}$-basis of $M_k(SL_2(Z)$ of modular forms with rational coefficients, since $T_n E_4^a E_6^b$ has rational coefficients too this implies the minimal polynomial of $T_p$ is in $\Bbb{Q}[X]$ thus $\Bbb{T(Q)},\Bbb{T(Z)}$ 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 cusp form $f$, 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 $\dim(S_k(SL_2(Z)))=1$.