The *strong multiplicity one theorem* for newforms says the following. Suppose that $f\_1 \in S\_k(\Gamma\_0(N\_1))$ and $f\_2 \in S\_k(\Gamma\_0(N\_2))$ are newforms, where $N\_1, N\_2 \geq 1$ are arbitrary integers. If $a\_{\ell}(f\_1) = a\_{\ell}(f\_2)$ for all primes $\ell$ not dividing a fixed integer $L$, then $f\_1 = f\_2$.


I'm looking for a "mod $p$ version" of this theorem:



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> Let $p$ be a prime. Suppose that $f\_1 \in S\_k(\Gamma\_0(N\_1))$ and $f\_2 \in S\_k(\Gamma\_0(N\_2))$ are newforms with Fourier coefficients in $\mathbf{Z}$, where $N\_1, N\_2 \geq 1$ are arbitrary integers. If $a\_{\ell}(f\_1) \equiv a\_{\ell}(f\_2)$ modulo $p$ for all primes $\ell$ not dividing a fixed integer $L$, then $f\_1 \equiv f\_2$ modulo $p$.
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Is this theorem true, and if so, does anyone know of a reference?



Multiplicity one for newforms modulo $p$
If by $f\_1\equiv f\_2$ modulo $p$, you mean that $a\_n(f\_1)\equiv a\_{n}(f\_2)$ modulo $p$ for all $n\in\mathbb N$ or maybe for all except finitely many, then this theorem cannot be true.


Let's start with a counterexample. Consider the two newforms
\begin{equation}
f(z)=\underset{n=1}{\overset{\infty}{\Sigma}}a\_{n}q^{n}\in S\_{2}(\Gamma\_0(11)),\ g(z)=\underset{n=1}{\overset{\infty}{\Sigma}}b\_{n}q^{n}\in S\_{2}(\Gamma\_0(77))
\end{equation}
attached respectively to the elliptic curves $X\_0(11)$ and $E:y^2+xy=x^3+x^2-51x+110$. Then $a\_{\ell}(f)\equiv a\_{\ell}(g)$ modulo 3 for all primes $\ell\neq 7$ yet $a\_7(f)=-2$ while $a\_7(g)=-1$ so $a\_7(f)\not\equiv a\_7(g)$ modulo 3, and in fact typically $a\_n(f)\not\equiv a\_n(g)$ modulo 3 if $n$ is a strictly positive integer which is divisible by 7.


What's going on? The problem is that $a\_\ell$ is the trace of the image of the Frobenius morphism at $\ell$ through the Galois representation attached to the newform provided $\ell$ does not divide the conductor of the newform. The fact that $a\_{\ell}(f)\equiv a\_\ell(g)$ for almost all primes implies that the residual representations $\bar{\rho}\_f$ and $\bar{\rho}\_g$ are isomorphic (by Cebotarev's density theorem). If $\ell$ divides neither the conductor of $f$ nor the conductor of $g$, then $a\_\ell(f)$ modulo $p$ and $a\_\ell(g)$ modulo $p$ are both the trace of the image of the Frobenius morphism, so they are equal. However, 7 divides the conductor of $g$ but not the Artin conductor of the residual representation $\bar{\rho}\_g$ nor the conductor of $f$ so the relation between $a\_7(g)$ and $a\_7(f)\equiv\mathrm{tr}\bar{\rho}\_g(\mathrm{Fr}(7))$ breaks down.


This will happen generally speaking whenever you have a level-raising (or equivalently level-lowering) phenomenon, that is to say congruences between forms which are new of different levels.


