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Let $\mathbb{T}_{\mathbb{Z}}$ be a $\mathbb{Z}$-module generated by Hecke operators $T_n$ acting on the space of cups forms $S_{k}(\Gamma,\mathbb{C})$ for the congruent subgroup satisfying $\Gamma_1(N)\subset\Gamma$ for some positive $N$. We know that the $\mathbb{Z}$-module $\mathbb{T}_{\mathbb{Z}}$ is finitely generated and the upper bound for the number of generators is given by the Sturm bound. When we compute some examples like for $\Gamma=\Gamma_0(33)$ and $k=2$ we get $\mathbb{T}_{\mathbb{Z}} = \mathbb{Z}[T_3]/(T_3+1)(T_3^2+T_3+3)$. In all other cases I have computed it always happens that the algebra looks similarly,i.e. $\mathbb{T}_{\mathbb{Z}} = \mathbb{Z}[T_k]/f(T_k)$ for some polynomial with integer coefficients. Is it always the case or are there any know examples that $\mathbb{T}_{\mathbb{Z}}$ as a ring is generated minimally by more than one Hecke operator ?

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This will not always be the case, but instead of just looking round at low level, if you want to find an example you perhaps need a more targetted approach, going for specific phenomena that will catch this. I believe there are examples of modular forms with "inner twists", possibly around level 512 or so, which have the property that the coefficient field is not of the form $\mathbf{Q}(a_n)$ for any Hecke eigenvalue $a_n$ -- I remember talking to William Stein about this years ago when he was writing his programs and thinking about ways to implement coefficient fields.... – Kevin Buzzard Jun 11 '11 at 21:50
...and the question arose as to whether one could always use one eigenvalue and the answer was "no -- here is an explicit example where one can prove that one eigenvalue isn't enough". If I'm remembering correctly then even $\mathbb{T}_{\mathbf{Q}}$ will not be generated by one Hecke operator in this situation. – Kevin Buzzard Jun 11 '11 at 21:52
I wonder if there's an example where no element of $\mathbb{T}_{\mathbb{Z}}$ generates it as a $\mathbb{Z}$-algebra -- not just no $\mathbb{T}_\ell$, but no $\mathbb{Z}$-linear combination of them? – David Loeffler Jun 12 '11 at 8:57
@David: I think I know how to come up with such an example. I think $(\Z/2Z)[x,y]/(x^2,y^2)$ cannot be generated by a single element, and using oldforms I don't think it's hard to come up with a Hecke algebra that surjects onto this -- take a newform not congruent to any other newforms of the same weight/level mod 2, and then bump up the level by $p^3q^3$ where $p$ and $q$ are distinct primes for which the Frobenii are non-scalar and where you can't raise the level of the associated mod 2 representation. The cubes push the nilpotents in. – Kevin Buzzard Jun 12 '11 at 10:13
Consider $S_2(\Gamma_0(40))$. The Hecke algebra $\mathbb{T}_\mathbb{Z}$ is generated by two operators $T_2$ and $T_3$ as a $\mathbb{Z}$-module. They satisfy relations $T_2^2=0$, $T_3(T_3+2)=0$ and $T_2 T_3 = -2 T_2$. With respect to the basis $f_1=q+q^5+O(q^6)$, $f_2=q^2+O(q^6)$, $f_3=q^3+q^5+O(q^6)$ the operators can be represented by suitable matrices. If you consider an element $T(a,b,c)=a+b T_2+c T_3$ and assume it generates the whole algebra over $\mathbb{Z}$ its minimal polynomial must be of degree 3 which is only possible when $a\neq 0$ and $b\neq 0$. – B. Naskrecki Jun 12 '11 at 11:18

[I took the time to chase this up so may as well post it as an answer.]

There is a (cuspidal) modular (eigen)form of level $\Gamma_0(512)$ and weight 2, which if I remember correctly was shown to me by Luis Dieulefait, and which has the following property: its coefficient field is the biquadratic extension $\mathbf{Q}(\sqrt{2},\sqrt{3})$ of $\mathbf{Q}$, but writing $f=\sum_n a_n q^n=q+\ldots$, each $a_n$ has the property that $a_n^2\in\mathbf{Z}$!

Hence the Hecke algebra at this level, even with $\mathbf{Q}$ coefficients, is not generated by one Hecke operator.

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Thanks! Inspired by this answer I did an extensive search and found also that the smallest case (at least in weight 2 and $\Gamma=\Gamma_0 (N)$) when the algebra is not generated by a single operator is for $N=40$ - we need $T_2$ and $T_3$. This complicates very much the way I can draw a "picture" of $Spec\mathbb{T}_{\mathbb{Z}}$ ! – B. Naskrecki Jun 12 '11 at 8:50

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.


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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Do you think there can be any generalization of the proposition to the case when $N>1$ ? It might be interesting to see class field theory interrelating this! – B. Naskrecki Jun 12 '11 at 8:53
@Naskrecki: I think there are indeed generalisations of the above to higher levels. However the situation is more complicated (the integral Hecke ring need not be reduced for example). Perhaps you want to consider the new part of the integral Hecke ring, then you can try and investigate what are the mod p dihedral representations arising from that level, and in a given weight. – Tommaso Centeleghe Jun 12 '11 at 9:54
For $p=3299$, $Cl(\mathbb{Q}(\sqrt{-p}))\cong C_3\times C_9$, which satisfies the hypothesis. – Dror Speiser Jun 13 '11 at 4:19
@Dror: thanks for checking it, great. I suspect there is no such example when the class group is cyclic. – Tommaso Centeleghe Jun 13 '11 at 7:48

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