One dimensional $\ell$-adic representations of $\mathbb{Z}_p$

Question

I am reading the paper "Lectures on the Langlands program and conformal field theory" by E.Frenkel. In page 28, he computes the one-dimensional continuous representations of $\mathbb{Z}_p$ over $\mathbb{Q}_\ell$, here $\ell \neq p$.
He chooses any $\ell$-adic number $\mu$ such that $\mu-1 \in \ell \mathbb{Z}_\ell$. He claims that for any natural number $n$, $\mu^{p^n}-1 \in \ell^{p^n}\mathbb{Z}_\ell$ so it gives a continuous group homomorphism $\mathbb{Z}_p \to \mathbb{Q}_\ell^\times$. I do not know why it is true. I think that we only have $\mu^{p^n}-1 \in \ell\mathbb{Z}_\ell$.
So I want to know how to compute these one-dimensional representations actually?

Answer 1

Yes this is false as the example $\ell = 3$, $\mu = 4$, $p = 2$, $n = 1$ shows. Firstly, any continuous representation has to actually map into $\mathbb{Z}_\ell^\times$, because $\mathbb{Z}_p$ is compact. Now a one-dimensional representation is a continuous map of profinite groups $f: \mathbb{Z}_p \to \mathbb{Z}_\ell^\times$. The preimage of the open subgroup $V^{(n)} = 1 + \ell^n \mathbb{Z}_\ell$ is an open subgroup $U^{(n)}$ of $\mathbb{Z}_p$. Let $m = m(n)$ be minimal such that $p^m \mathbb{Z}_p \subseteq U^{(n)}$. Then $f$ is the same as a system of maps $f_n: \mathbb{Z} / p^m \mathbb{Z} \to \mathbb{Z}_\ell^\times / V^{(n)} = (\mathbb{Z} / \ell^n \mathbb{Z})^\times$ which is compatible with the inverse limits on both sides. You can look up the structure of the group on the right-hand side, and note that the image of $f_n$ has to be contained in its $p^m$-torsion. One sees that if $\ell = 2$, the only possibility for $f$ is the zero map. If $\ell \neq 2$ $f$ maps into the direct summand $\mathbb{Z} / (\ell -1) \mathbb{Z}$ of the right hand side, so factors through the quotient $\mathbb{Z}_p \to \mathbb{Z} / p^{v_p(\ell - 1)} \mathbb{Z}$, where $v_p$ is the $p$-adic valuation. Thus representations are in bijection with maps $\mathbb{Z} / p^{v_p(\ell -1)} \mathbb{Z} \to \mathbb{Z} / (\ell -1) \mathbb{Z}$, i. e. they all map into the roots of unity of $\mathbb{Z}_\ell$.