Let $h^{ord}(N,\mathcal{O})$ be the $p$-ordinary Hecke algebra, and $\mathfrak{m}$ be a maximal ideal of the semi local ring $h^{ord}(N,\mathcal{O})$ corresponding to a residual representation $\bar{\rho}$ which is $p$-distinguish and minimal and such that the restriction of $\bar{\rho}$ to $G_{\mathbb{Q}(\sqrt{(-1)^{(p-1)/2}p)}}$ is absolutely irreducible.

let $T$ be the localisation of $h^{ord}(N,\mathcal{O})$ by $\mathfrak{m}$.

In the literature I see that we assume often that the tame level is a square free (since any semi-stable elliptic curves has a square free conductor)

There is some modularity lifting theorems $R^{ord}=T$ without the assumption that the tame level of $T$ is a square free?

I don't see immediately when we use this condition in the Horizontal control (in the Hecke side)?


1 Answer 1


To give a reference: in the article by Böckle referenced below, he proves an $R^{\mathrm{ord}}=T$ type theorem (Thm. 3.9) without assuming that the tame level is square free. In fact, he starts with the residual representation $\bar\rho$, then associates some conductor $N$ to $\bar\rho$ (which may not be square free) and then proves that there is an isomorphism between the corresponding local ring of the Hecke algebra of tame level $N$ and a deformation ring attached to $\bar\rho$.

In Böckle's article, the restricted Hecke algebra is used instead of the full one. However, for the $R^{\mathrm{ord}}=T$ statement, this does not make a difference. For this, see Theorems 10 and 12 in the article by Gouvêa referenced below.

Gebhard Böckle. "On the Density of Modular Points in Universal Deformation Spaces". American Journal of Mathematics 123.5 (2001), pp. 985-1007.

Fernando Q. Gouvêa. "Deforming Galois Representations: Controlling the Conductor". Journal of Number Theory 34 (1990), pp. 95-113.

  • $\begingroup$ There is a proof using Taylor-Wiles method? $\endgroup$ May 6, 2016 at 16:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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