Does Ribet's level lowering theorem hold for prime powers? I often use the following theorem (that one can state more generally) in my research.
Let E/Q be an elliptic curve of conductor N corresponding to a modular form f(E), l a prime of good or multiplicative reduction for E,  and \rho(l) the 2 dimensional mod l Galois representation given by the action on the l-torsion points. Suppose that the torsion subscheme E[l] extends to a finite flat group scheme over Z_l, and let p be a prime of multiplicative reduction for E such that \rho(l) is unramified at p (e.g. the number field (Q(E[l]) generated by the coordinates of the l-torsion points is unramified at p). Then there exists a modular form f of conductor N/p  such that f is congruent to f(E) mod l (when f has Fourier coefficients over Z then this means that all but finitely many of the coefficents are congruent mod l); one can `lower the level' from N to N/p.
Does such a result hold for powers of primes? E.g. if this holds for the mod l^n representation (instead of the mod l) does one get a congruence mod l^n?
 A: There's some slides from a talk by Ian Kiming here which discuss this question. He states a theorem (on slide number 8) corresponding to the existence of the map from a Hecke algebra at level N/(p^u) (where p^u is the largest power of p dividing N) to Z/ell^n Z. As buzzard says, it's not clear that this map will lift, but Kiming speculates that if you allow the weight of your modular form to vary you can find a char 0 lift. 
A: If you put yourself in a position where an R=T theorem holds at level N/p (e.g.E[ell] irreducible, big image, ell>2), then you'll get a map from a Hecke algebra at level N/p to Z/ell^nZ. But in general I don't see why this ring homomorphism should lift to a homomorphism from T to a char 0 integral domain and would bet on counterexamples if I were a betting man.
A: One case that you can say a bit more is if the congruence number at level $N/p$ is coprime to $l$. Specifically, if $f(E)$ is congruent to a unique modular form $g$ at level $N/p$ modulo $l$, then you can say that $f(E)$ is congruent to $g$ modulo $l^n$.
