I am reading Fontaine and Yi's note "Theory of $p$ adic Galois Representations,
http://www.math.u-psud.fr/~fontaine/galoisrep.pdf
In section 1.1.4, (1), page 4, let $K$ be a field and $K^s$ an algebraic closure, and $G=\text{Gal}(K^s/K)$. If $l$ is a prime, define
\begin{equation} \mu_{l^n}(F)=\{a \in F|a^{l^n}=1\} \end{equation}
The Tate module of the multiplicative group $\mathbb{G}_m$ is defined as \begin{equation} T_l(\mathbb{G}_m)=\text{lim}\, \mu_{l^n}(K^s) \end{equation}
Where the limit is induced by the map
\begin{equation} \mu_{l^{n+1}}(K^s) \rightarrow \mu_{l^n}(K^s), a \rightarrow a^l \end{equation}
It is claimed that $T_l(\mathbb{G}_m)$ is a free $\mathbb{Z}_l$ module of rank 1, but I cannot see why, could anyone prove this?
I am really a beginner, and know very little about $p$ adic stuff.
$\Bbb Z_\ell\cong\varprojlim \Bbb Z/\ell^n\Bbb Z$, where the map $\Bbb Z/\ell^{n+1}\Bbb Z\to\Bbb Z/\ell^n\Bbb Z$ is the natural map taking $n\pmod{\ell^{n+1}}$ to $n\pmod{\ell^n}$ (maybe by definition, depending on how you define $\Bbb Z_\ell$). $\mu_{\ell^n}(K^s)\cong\Bbb Z/\ell^n\Bbb Z$ (but not canonically!) because $K^s$ is algebraically closed and hence contains all $\ell^n$-th roots of unity. You just need to check that the map $a\mapsto a^\ell$ from $\mu_{\ell^{n+1}}(K^s)\to\mu_{\ell^n}(K^s)$ induces the map $\Bbb Z/\ell^{n+1}\Bbb Z\to\Bbb Z/\ell^n\Bbb Z$ described above under an isomorphism $\mu_{\ell^n}(K^s)\cong\Bbb Z/\ell^n\Bbb Z$.