(Sorry for my poor english..)

Let $F(z)\in S_{2k}(SL_2(\mathbb{Z})$) be a newform and $\ell$ be a prime larger than $3$. Let $K$ be a some number field and $v$ be a prime of $K$ over $\ell$. Let $K_v$ be a $v$-adic completion of $K$ and $O_{K,v}$ be a ring of integers of $K_v$. Let $v=(\pi)$. I already know that Serre and Deligne proved that there exists a Galois representation $\rho_{F,v}$ such that \begin{equation} \rho_{F,v} : Gal(\overline{\mathbb{Q}}/\mathbb{Q})\to GL_2(\mathcal{O}_{K,v}) \end{equation} with for primes $p\neq \ell$, \begin{equation} \text{Tr}(\text{Frob}_p)=\lambda_p ,\quad \det(\text{Frob}_p)=p^{2k-1} \end{equation} where $\text{Frob}_p$ is a Frobenius element.

Let $\rho_{F,v}^{m}$ be a reduction of $\rho_{F,v}$ modulo $(\pi^{m})$. In other words, \begin{equation} \rho_{F,v}^{m} : Gal(\overline{\mathbb{Q}}/\mathbb{Q})\to GL_2(\mathcal{O}_{K,v}/(\pi^m)) \end{equation} with for primes $p\neq \ell$, \begin{equation} \text{Tr}(\text{Frob}_p)\equiv \lambda_p, \quad \det(\text{Frob}_p)\equiv p^{2k-1} \pmod{(\pi^m)}. \end{equation} Does there exist $N\in \mathbb{N}$ such that for $p_1\equiv p_2 \pmod{N}$ implies that $\rho_{F,v}^m(\text{Frob}_{p_1})\equiv \rho_{F,v}^m(\text{Frob}_{p_2})$?


1 Answer 1


Write $L$ for the finite Galois extension of $\mathbb Q$ with Galois group $G_{\mathbb Q}/\operatorname{Ker}\rho_{F,v}^m$. Then $\rho_{F,v}^m(\operatorname{Frob}_p)$ is the identity in $\operatorname{GL}_2(\mathcal O_{K,v}/\pi^m)$ if and only if $\operatorname{Frob}_p$ is the identity in $\operatorname{Gal}(L/\mathbb Q)$ if and only if $p$ splits completely in $L/\mathbb Q$.

If an $N$ as in your question exists, then choosing it large enough so as to eliminate at most finitely many exceptions, we can arrange that there exists $a$ such that $p$ splits completely in $L/\mathbb Q$ if $p\equiv a$ modulo $N$. This implies that the extension $L/\mathbb Q$ is abelian, or equivalently that $\rho_{F,v}^m$ has abelian image. For $m$ large enough, this is never true for $\rho_{F,v}^m$. So the $N$ in your question does not exist.

The statement that $L/\mathbb Q$ must be abelian if $p\equiv a$ modulo $N$ implies that $p$ splits completely in $L/\mathbb Q$ is closely related to classical questions in class field theory, but is not quite stated in the usual form (for instance, it is much easier to prove that $L/\mathbb Q$ must be abelian if $p\equiv a$ modulo $N$ is equivalent to the statement that $p$ splits completely in $L/\mathbb Q$). Nevertheless, a complete proof can be found for instance at the following MO answer.

Why do congruence conditions not suffice to determine which primes split in non-abelian extensions?


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