Let $N$ be a prime and let $\mathbb{T} \subset \mathrm{End}(J_0(N))$ be the Hecke algebra generated over $\mathbb{Z}$ by $U_N$ and the operators $T_p$ for primes $p \nmid N$. Fix a maximal ideal $\mathfrak{m} \subset \mathbb{T}$ and let $\mathbb{T}_{\mathfrak{m}}$ be the completion of $\mathbb{T}$ at $\mathfrak{m}$.

In the paper "Modular curves and the Eisenstein ideal" Mazur proves (Lemma II.15.1) that "multiplicity one" at $\mathfrak{m}$, i.e., $\mathrm{dim}_{\mathbb{T}/\mathfrak{m}}(J_0(N)[\mathfrak{m}](\overline{\mathbb{Q}})) = 2$, implies that the ring $\mathbb{T}_{\mathfrak{m}}$ is Gorenstein. Are there known examples where the converse implication fails, i.e., where $\mathbb{T}_{\mathfrak{m}}$ is Gorenstein but $\mathrm{dim}_{\mathbb{T}/\mathfrak{m}}(J_0(N)[\mathfrak{m}](\overline{\mathbb{Q}})) \neq 2$? (Or does the converse perhaps hold?) What about the analogous question in the case when $N$ is not necessarily prime (but, say, $\mathfrak{m}$ is "new")?

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    $\begingroup$ See Section 2 (and other parts) here: projecteuclid.org/euclid.em/1227031895 $\endgroup$ Commented Oct 4, 2016 at 4:16
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    $\begingroup$ My memory is that if the associated Galois rep is abs irred, and, crucially, if $T_p$ is not in $m$ (ordinarity) then Gross in his companion forms paper shows that the $m$-torsion in $J$ is an extension of $T/m$ (one-dimensional) by $T^\vee/m$, and hence by Nakayama $T$ is Gorenstein iff $J[m]$ is 2-dimensional. I don't know about other cases though. Note that Emerton and Calegari wrote a "Mazur revisited" paper once, this might be worth a look. $\endgroup$
    – znt
    Commented Oct 4, 2016 at 7:07

1 Answer 1


In the ordinary case, the argument is simple so let me recall it here.

The $p$-divisible group $J$ is an extension of an étale $p$-divisible group $J^{et}$ by a multiplicative $p$-divisible group $J^{m}$ and the Pontryagin dual $J^{et*}$ of $J^{et}$ is a free $\mathbb T_\mathfrak{m}$-module of rank 1 by the ordinarity assumption. Assume in addition that $\mathbb T_\mathfrak{m}$ is a Gorenstein ring. Then $\operatorname{Hom}(\mathbb T_\mathfrak{m},\mathbb Z_{p})/\mathfrak{m}\operatorname{Hom}(\mathbb T_\mathfrak{m},\mathbb Z_{p})$ is a $\mathbb T_\mathfrak{m}/\mathfrak{m}\mathbb T_\mathfrak{m}$-vector space of dimension 1 (here we use the fact that a ring $R$ is a Gorenstein ring if and only if its dualizing complex is concentrated in degree 0 and isomorphic to $R$).Then $J^{et}[\mathfrak m]$ is free of rank 1, by duality so is $J^{m*}[\mathfrak m]$ and finally $J[\mathfrak m]$ is a free $\mathbb T_\mathfrak{m}$-module of rank 2.

Note that this does not require the fact that $\bar{\rho}_{\mathfrak m}$ is absolutely irreducible.

If $T_p$ belongs to $\mathfrak m$, the argument is more involved but Proposition (14.2) of Modular curves and the Eisenstein Ideal asserts that $J[\mathfrak m]$ is always of dimension 2 in that case (even without assuming $\mathbb T_\mathfrak{m}$ to be Gorenstein).

So the question of the title definitely admits the answer yes and conversely for the question in the body of the text.


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