Let $N \ge 1, k \ge 2$ be integers, and $M_k(\Gamma_1(N))$ the space of weight k modular forms of level $\Gamma_1(N)$. Let $\mathbb{T}$ be the $\mathbb{Z}$-subalgebra of $\operatorname{End} M_k(\Gamma_1(N))$ generated by the diamond operators $\langle d \rangle$ and the Hecke operators $T_\ell$ for $\ell \nmid N$. (This is sometimes called the "anemic Hecke algebra" because we've omitted the Hecke operators at primes dividing $N$).

If we make $\mathbb{T}$ still more anemic, by omitting the $T_\ell$'s for some finite subset $\Sigma$ of the primes $\ell \nmid N$, is the resulting subalgebra $\mathbb{T}^{\Sigma}$ the whole of $\mathbb{T}$?

(This is certainly true after tensoring with $\mathbb{Q}$ by the strong multiplicity one theorem, and I managed to convince myself it should be true integrally by a complicated argument using Galois representations and Chebotarev density; but is there a slick proof that doesn't use such heavy machinery?)

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    $\begingroup$ My first idea is that your proof works and cannot be substantially simplified. $\endgroup$
    – Joël
    Jul 1, 2013 at 19:22

1 Answer 1


It seems to me that the proof you describe is fine except maybe at the prime 2, the problem being that the argument you are sketching (presumably) relies on expressing the missing operators via traces of Frobenius morphisms modulo arbitrary powers of maximal ideals of the Hecke algebra, but this works only for Hecke operators $T_{\ell}$ with $\ell$ prime to the residual characteristic of the maximal ideal you are considering, so that one needs a supplementary argument at the residual characteristic of the maximal ideal. This is readily provided by $q$-expansion and the duality between Hecke operators and cusp forms except if the residual characteristic is 2.

Consequently, I'm quite sure that your $\mathbb T^\Sigma$ is equal to $\mathbb T$ if $2|N$ (in which case there are no operator at 2 to begin with) or if $2\notin \Sigma$ but otherwise the index of $\mathbb T^{\Sigma}$ in $\mathbb T$ might be a power of 2 (and actually is sometimes, see here for an example).

The argument with $q$-expansion can be found in

Multiplicities of $p$-finite mod $p$ Galois representations in $J_{0}(Np)$ (Ken Ribet) Bol. Soc. Brasil. Mat. (N.S.) 21 (1991), no. 2, 177–188.

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    $\begingroup$ So knowing a Galois representation mod $2$ does not tell you what the crystalline Frobenius eigenvalues are mod $2$? $\endgroup$ Jul 2, 2013 at 18:57
  • $\begingroup$ For the application I had in mind I'm going to be localizing at a prime p anyway, and I'm quite happy to assume p is not 2! But I'm also interested in the answer to @KevinVentullo 's question. $\endgroup$ Jul 3, 2013 at 13:10
  • $\begingroup$ @Kevin: "So knowing a Galois representation mod 2 does not tell you what the crystalline Frobenius eigenvalues are mod 2". I'm not sure exactly what you mean, but if ell=p, then the mod p Galois repn carries very little information about the Galois repn and the crystalline Frobenius. $\endgroup$ Jul 4, 2013 at 6:28

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