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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:

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 $p$.

Is this theorem true, and if so, does anyone know of a reference?

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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.

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  • $\begingroup$ Thanks for this example, it really clears things up! $\endgroup$ Commented Jun 13, 2023 at 12:46
  • $\begingroup$ I have a related but soft question. Are there results that quantify how many level raised $g$ exist for a given $f$ under Ribet's theorem? $\endgroup$
    – Hodge-Tate
    Commented Aug 6, 2023 at 23:35
  • $\begingroup$ @Hodge-Tate Check out the paper "Quantitative Level Lowering" by Lundell - some of the results could be helpful. $\endgroup$ Commented Aug 7, 2023 at 1:29

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