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In certain cases a reason that causes the subring generated by a single Hecke operator $T_\ell$ in $\mathbb{T}_\mathbb{Z}$ to have index greater than one is the existence of exceptional primes for the normalised eigenforms $f\in S_k(\Gamma,\mathbb{C})$ (we shall say that $p$ is exceptional for $f$ if (one of) the $p$-adic Galois representation $\rho_{f,\lambda}$ associated to $f$ has small residual image, i.e., the image of $\bar\rho_{f,\lambda}$ does not contain $SL_2(F)$, for any finite extension $F/F_p$).

The special cases that I have in mind are instances of the inner twist phenomena that Kevin Buzzard has already mentioned in one of his comments, and you can see this also in level one, if I am right. Let me elaborate on that.

Let $p$ be a prime $\equiv 3$ mod $4$, and set $k=(p+1)/2$. Let $h$ be the class number of $Q(\sqrt{-p})$. The space $S_k(\Gamma(1))$ gives rise to $n=(h-1)/2$ distinct systems of mod $p$ Hecke eigenvalues so that the associated mod $p$ Galois representations $\rho_1,\ldots,\rho_n$ (which are unramified outside $p$) have dihedral image. This implies that if $\ell$ is a prime that is not a quadratic residue mod $p$, then the trace of $\rho_i$ at a Frobenius element at $\ell$ is zero.

In particular, if $n>1$ (i.e. $h>3$) then there are at least two DISTINCT systems of mod $p$ Hecke eigenvalues $(a_q)$ and $(b_q)$ such that their $\ell$-th members are equal. This implies that the integral Hecke ring $\mathbb{T}$ has more ring homomorphisms valued in $\bar{F_p}$ than the subring generated by $T_\ell$ alone does. In this case one can show that $p$ divides the index of the latter in the former.

With the previous argument you see that for almost all primes $p\equiv 3$ mod $4$ in weight $(p+1)/2$ half of the primes $\ell$ are such that $T_\ell$ does not generate $\mathbb{T}$. But what about the other half of the primes $\ell$?

By relating the traces of $\rho_i$ at Frobenius elements over primes $\ell$ that are split in $Q(\sqrt{-p})$ to the characters of the class group of $Q(\sqrt{-p})$ valued in $\bar{F_p}$ we get the following:

PROPOSITION: Let $A$ be the class group of $Q(\sqrt{-p})$. Assume that for every $a\in A$ there exists a pair of non trivial characters $\chi_i:A\rightarrow\bar{F_p}^*$, with $\chi_1\neq\chi_2$ and $\chi_1\neq \chi_2^{-1}$, such that $\chi_1(a)+\chi_1(a^{-1}) =\chi_2(a)+\chi_2(a^{-1})$. Then the integral Hecke ring $\mathbb{T}$ in weight $(p+1)/2$ and level one cannot be generated by a single $T_\ell$, for $\ell$ prime.

Few remarks: 1) The above does not say anything about the possibility of having a $T_n$ (with $n$ not prime) generating $\mathbb{T}$. In fact I think we cannot rule this out a priori.

2) Probably a little more group theoretic work can be done to reformulate the condition on the mod $p$ characters of $A$, and turn it into something nicer. I think if $A$ is not cyclic and $3$ does not divide its order then the assumption is satisfied.

3) Notice that we do not need to worry about $T_p$ generating $\mathbb{T}$ when $n>1$. In fact, since $P$ splits in $H/Q({\sqrt{-p}})$, where $H$ is the Hilbert Class Field of $Q(\sqrt{-p})$, all the $\rho_i$'s are the same locally at $p$ and one can show that the eigenvalue $a_p$ is $1$ for all of them.

An example. The cuspidal, integral Hecke ring of level one and weight $k=246=(491+1)/2$ is so that $p=491$ does not divide its discriminant. However every Hecke operators $T_n$, for $2\leq n\leq 153$ generates a subring of $\mathbb{T}$ of index divisible by $p$.

May be it could be nice to see whether the class group of $Q(\sqrt{491})$ satisfies the assumption of the proposition. I suspect it does!

[EDIT: The assumption of the proposition is NOT satisfied for $p=491$, unless I am wrong to an infinite amount. Therefore my suspicion is not confirmed. The proposition above, as far as I can tell, remains valid, although I have no example of a prime $p\equiv 3$ mod $4$ such that the class group $A$ of $Q(\sqrt{-p})$ satisfies the assumption of the proposition.

There still remains to give an explanation of the fact that $p=491$ seems to divide the index of the subring generated by $T_\ell$ in the integral, cuspidal Hecke ring of level $1$ and weight $246$. The arguments above explain such divisibility for primes $\ell$ that are non-quadratic residue mod $p$. For what concerns the other primes $\ell$, I am tempted to say that there should be a mod $491$ Galois representation arising from weight $246$ and level $1$ that is tamely ramified, reducible at $p$, and non-dihedral (may be with small image?): this in fact would cause the existence of a system of mod $p$ eigenvalues $(a_\ell)$, with $\ell\neq p$, arising from $S_{246} (\Gamma(1))$ such that its quadratic twist $(a_\ell \ell^{(p-1)/2})$ also arises from the same space.

Summarising

PROPOSITION: Let $p\equiv 3$ mod $4$. Assume that

1) the class number of $Q(\sqrt{-p})$ is > 3;

2) there exists a mod $p$ representation $\rho$ of $G_Q$ arising from $S_{(p+1)/2}(\Gamma(1))$ that is tamely ramified and reducible at $p$ and it is non-dihedral;

then for every prime $\ell$, $p$ divides the index of the subring generated by $T_\ell$ in the integral, cuspidal Hecke ring of weight $(p+1)/2$ and level $1$.

When $p=2083$ I learnt from some tables that there exists an odd, $A_5$ extension of $Q$ unramified outside $p$ that gives rise to a representation of the type we want in 2). Since condition 1) is also satisfied we see that the Hecke ring in weight $1044$ and level $1$ cannot be generated by a single $T_\ell$, for $\ell$ prime.]

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[EDIT: The assumption of the proposition is NOT satisfied for $p=491$, unless I am wrong to an infinite amount. Therefore my suspicion is not confirmed. The proposition above, as far as I can tell, remains valid, although I have no example of a prime $p\equiv 3$ mod $4$ such that the class group $A$ of $Q(\sqrt{-p})$ satisfies the assumption of the proposition.

There still remains to give an explanation of the fact that $p=491$ seems to divide the index of the subring generated by $T_\ell$ in the integral, cuspidal Hecke ring of level $1$ and weight $246$. The arguments above explain such divisibility for primes $\ell$ that are non-quadratic residue mod $p$. For what concerns the other primes $\ell$, I am tempted to say that there should be a mod $491$ Galois representation arising from weight $246$ and level $1$ that is tamely ramified, reducible at $p$, and non-dihedral (may be with small image?): this in fact would cause the existence of a system of mod $p$ eigenvalues $(a_\ell)$, with $\ell\neq p$, arising from $S_{246} (\Gamma(1))$ such that its quadratic twist $(a_\ell \ell^{(p-1)/2})$ also arises from the same space.

Summarising

PROPOSITION: Let $p\equiv 3$ mod $4$. Assume that

1) the class number of $Q(\sqrt{-p})$ is > 3;

2) there exists a mod $p$ representation $\rho$ of $G_Q$ arising $S_{(p+1)/2}(\Gamma(1))$ that is tamely ramified at $p$ and it is non-dihedral;

then for every prime $\ell$, $p$ divides the index of the subring generated by $T_\ell$ in the integral, cuspidal Hecke ring of weight $(p+1)/2$ and level $1$.

When $p=2083$ I learnt from some tables that there exists an odd, $A_5$ extension of $Q$ unramified outside $p$ that gives rise to a representation of the type we want in 2). Since condition 1) is also satisfied we see that the Hecke ring in weight $1044$ and level $1$ cannot be generated by a single $T_\ell$, for $\ell$ prime.]

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In certain cases a reason that causes the subring generated by a single Hecke operator $T_\ell$ in $\mathbb{T}_\mathbb{Z}$ to have index greater than one is the existence of exceptional primes for the normalised eigenforms $f\in S_k(\Gamma,\mathbb{C})$ (we shall say that $p$ is exceptional for $f$ if (one of) the $p$-adic Galois representation $\rho_{f,\lambda}$ associated to $f$ has small residual image, i.e., the image of $\bar\rho_{f,\lambda}$ does not contain $SL_2(F)$, for any finite extension $F/F_p$).

The special cases that I have in mind are instances of the inner twist phenomena that Kevin Buzzard has already mentioned in one of his comments, and you can see this also in level one, if I am right. Let me elaborate on that.

Let $p$ be a prime $\equiv 3$ mod $4$, and set $k=(p+1)/2$. Let $h$ be the class number of $Q(\sqrt{-p})$. The space $S_k(\Gamma(1))$ gives rise to $n=(h-1)/2$ distinct systems of mod $p$ Hecke eigenvalues so that the associated mod $p$ Galois representations $\rho_1,\ldots,\rho_n$ (which are unramified outside $p$) have dihedral image. This implies that if $\ell$ is a prime that is not a quadratic residue mod $p$, then the trace of $\rho_i$ at a Frobenius element at $\ell$ is zero.

In particular, if $n>1$ (i.e. $h>1$) h>3$) then there are at least two DISTINCT systems of mod$p$Hecke eigenvalues$(a_q)$and$(b_q)$such that their$\ell$-th members are equal. This implies that the integral Hecke ring$\mathbb{T}$has more ring homomorphisms valued in$\bar{F_p}$than the subring generated by$T_\ell$alone does. In this case one can show that$p$divides the index of the latter in the former. With the previous argument you see that for almost all primes$p\equiv 3$mod$4$in weight$(p+1)/2$half of the primes$\ell$are such that$T_\ell$does not generate$\mathbb{T}$. But what about the other half of the primes$\ell$? By relating the traces of$\rho_i$at Frobenius elements over primes$\ell$that are split in$Q(\sqrt{-p})$to the characters of the class group of$Q(\sqrt{-p})$valued in$\bar{F_p}$we get the following: PROPOSITION: Let$A$be the class group of$Q(\sqrt{-p})$. Assume that for every$a\in A$there exists a pair of non trivial characters$\chi_i:A\rightarrow\bar{F_p}^*$, with$\chi_1\neq\chi_2$and$\chi_1\neq \chi_2^{-1}$, such that$\chi_1(a)+\chi_1(a^{-1}) =\chi_2(a)+\chi_2(a^{-1})$. Then , if$A$is non trivial, the integral Hecke ring$\mathbb{T}$in weight$(p+1)/2$and level one cannot be generated by a single$T_\ell$, for$\ell$prime. Few remarks: 1) The above does not say anything about the possibility of having a$T_n$(with$n$not prime) generating$\mathbb{T}$. In fact I think we cannot rule this out a priori. 2) Probably a little more group theoretic work can be done to reformulate the condition on the mod$p$characters of$A$, and turn it into something nicer. I think if$A$is not cyclic and$3$does not divide its order then the assumption is satisfied. 3) Notice that we do not need to worry about$T_p$generating$\mathbb{T}$when$n>1$. In fact, since$P$splits in$H/Q({\sqrt{-p}})$, where$H$is the Hilbert Class Field of$Q(\sqrt{-p})$, all the$\rho_i$'s are the same locally at$p$and one can show that the eigenvalue$a_p$is$1$for all of them. An example. The cuspidal, integral Hecke ring of level one and weight$k=246=(491+1)/2$is so that$p=491$does not divide its discriminant. However every Hecke operators$T_n$, for$2\leq n\leq 153$generates a subring of$\mathbb{T}$of index divisible by$p$. May be it could be nice to see whether the class group of$Q(\sqrt{491})\$ satisfies the assumption of the proposition. I suspect it does!

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