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## Serre’s conjecture for mod-p^n representations?

I think this may be a silly question, but here goes. Let $\rho:\mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})\to \mathrm{GL}_2(\overline{\mathbf{F}_p})$ be a representation; say $\rho$ is of S-type if it is continuous, unramified almost everywhere, and the determinant of complex conjugation is $-1$. Serre's conjecture, now a theorem of Khare-Wintenberger, states that every $\rho$ of S-type arises from some modular form $f=\sum a_n e(nz)$ in the sense that $\mathrm{tr}\rho(\mathrm{Frob}_l)=a_l\;( \mathrm{mod}\;p)$ for (almost all) primes $l$.

Question: Are S-type representations $\rho:\mathrm{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})\to \mathrm{GL}_2(\mathbf{Z}/p^n\mathbf{Z})$ for $n\geq 2$ also expected/known to be modular?

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In what sense? If you mean "come from the reduction of $\rho_f$ for some Hecke eigenform $f$'', no, they are not.

If you mean "come from the reduction of $\rho$ where $\rho:G_{\mathbb Q} \to GL_2(\mathbb T)$ is the Galois rep'n attached to the Hecke algebra $\mathbb T$ acting on modular forms of some sufficiently large level, then the answer is known to be yes in most cases (i.e. with comparitively minor technical restrictions on $\rho$). This is the content of so-called big $R =$ big $\mathbb T$ theorems, due to Gouvea--Mazur, Boeckle, and others (combined with Serre's conjecture to know that $\overline{\rho}$ is modular).

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By "modular" I just meant whether the traces on Frobenii agree mod $p^n$ with modular forms coefficients. – David Hansen Nov 30 2010 at 14:26
Then, as I wrote, the answer to your question is "no, this is not expected". – Emerton Nov 30 2010 at 16:44
@David: Emerton's answer is correct but your clarification, where you define "modular", is unclear. The traces of Frobenii will agree mod $p^n$ with coefficients of modular forms, but not with coefficients of eigenforms. The point is that the forms you have to use will not be eigenforms, but they will be eigenforms mod $p^n$ (i.e. $Tf=\lambda f+p^ng$ with $g$ having integral coefficients). – Kevin Buzzard Nov 30 2010 at 18:08
Thanks, Kevin! That clears things up for me perfectly. – David Hansen Nov 30 2010 at 19:26
Dear Kevin, Thanks for that. I had implicitly understood "modular forms" in David's initial comment to mean "eigenforms", hence the "no" in my reply. – Emerton Dec 1 2010 at 3:47