Let $f=\sum_{n=1}^{\infty} a_n(f)q^n$ be a normalized cuspidal newform of weight $k\geq 2$, level $N$, nebentypus character $\varepsilon$, and coefficient field $K_f$. Let $\mathcal{O}_f$ be the ring of integers of $K_f$. For every rational prime $\ell$ and every prime ideal $\lambda \subset \mathcal{O}_f$ lying above $\ell$, let $K_{f,\lambda}$ be the completion of $K_f$ at $\lambda$, let $\mathcal{O}_{f,\lambda}$ be its valuation ring, and let $k_\lambda=\mathcal{O}_{f,\lambda}/\lambda\mathcal{O}_{f,\lambda}$ be its residue field.

Then there exists a continuous semisimple representation

\begin{align*} \bar{\rho}_{f,\lambda}:G_{\mathbb{Q}}\longrightarrow GL_2(k_\lambda) \end{align*}

which is unramified outside the primes dividing $N\ell$ and such that, for every rational prime $p\nmid N\ell$,

\begin{align*} \operatorname{tr}\bar{\rho}_{f,\lambda}(\operatorname{Frob}_p)&\equiv a_p(f)\pmod{\lambda},\\ \det\bar{\rho}_{f,\lambda}(\operatorname{Frob}_p)&\equiv \varepsilon(p)p^{k-1}\pmod{\lambda}. \end{align*}

Here $\operatorname{Frob}_p$ denotes an arithmetic Frobenius element at $p$, and the congruences mean reduction in $k_\lambda$ through the embedding $K_f\hookrightarrow K_{f,\lambda}$.