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")?

  • 1
    $\begingroup$ See Section 2 (and other parts) here: projecteuclid.org/euclid.em/1227031895 $\endgroup$ Oct 4 '16 at 4:16
  • 1
    $\begingroup$ 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
    Oct 4 '16 at 7:07

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.