I'm trying to compute $\text{Hom}_\text{continuous}(\hat{\mathbb Z}, (\mathbb C_p)^\ast)$. By $\hat{\mathbb Z}$ I mean the profinite completion of $\mathbb Z$ and by $(\mathbb C_p)^\ast$ I mean the invertible elements in $\mathbb C_p$.
I tried using the decomposition $\mathbb C_p ^\ast \cong p^\mathbb Q \times U(1)$.
$U(1)$ are the elements in $\mathbb C_p $ with absolute value $1$.
I'm not sure if that helps and I'm not even sure how to compute $\text{Hom}_\text{continuous}(\hat{\mathbb Z}, p^\mathbb Q)$. Could anyone help?
Note that $\hat{\mathbb{Z}}$ is the direct product of $\mathbb{Z}_p$ and $\mathbb{Z}^p$, where $\mathbb{Z}^p$ is again the infinite direct product $\prod_{l \neq p}\mathbb{Z}_l$. Therefore, one has a natural isomorphism $\operatorname{Hom}_{cont}(\hat{\mathbb{Z}}, \mathbb{C}_p^\times) \simeq\operatorname{Hom}_{cont}(\mathbb{Z}_p, \mathbb{C}_p^\times) \times \operatorname{Hom}_{cont}(\mathbb{Z}^p, \mathbb{C}_p^\times)$.
After this step, the problem decomposes into two parts.
For any real number $r$, we denote by $B(r)$ the subgroup $\{z \in \mathbb{C}_p^\times: |z - 1|_p<r\}$ of $\mathbb{C}_p^\times$.
For $\operatorname{Hom}_{cont}(\mathbb{Z}^p, \mathbb{C}_p^\times)$:
Let $\chi:\mathbb{Z}^p \rightarrow \mathbb{C}_p^\times$ be a continuous character. We consider the image $H$ of $\chi$.
Since $\mathbb{Z}^p$ is compact, $H$ must be a compact subgroup of $\mathbb{C}_p^\times$.
Since $H\cap B(p^{-1})$ is an open subgroup of $H$, we see that the index $[H:H\cap B(p^{-1})]$ is finite. This means that there exists an integer $M$ such that $H^M \subseteq B(p^{-1})$.
However, since the multiplication by $p$ map is an automorphism on $\mathbb{Z}^p$, we also have $H \subseteq H^p$, and therefore $H^M \subseteq H^{Mp} \subseteq B(p^{-p})$ (note that we used the fact $B(p^{-1})^p\subseteq B(p^{-p})$, which can be proved using the $p$-adic logarithm and exponential maps).
This procedure can be iterated and will give: $H^M \subset H^{Mp^k} \subseteq B(p^{-p^k})$ for all $k$. Letting $k$ tend to infinity, we see that $H^M$ must be equal to $\{1\}$.
Therefore, $\chi$ is a finite character and hence factor through $\mathbb{Z}/M\mathbb{Z}$ for some $M$ prime to $p$.
For $\operatorname{Hom}_{cont}(\mathbb{Z}_p, \mathbb{C}_p^\times)$:
Let $\chi:\mathbb{Z}_p \rightarrow \mathbb{C}_p^\times$ be a continuous character. By continuity, it is totally determined by its value at $1$. We write $a$ for the value of $\chi(1)$.
Apparantly, $a$ cannot be an arbitrary element of $\mathbb{C}_p^\times$. There is a restriction: since $\chi$ is continuous, the limit $\lim_{k \rightarrow \infty}\chi(p^k)$ should be equal to $\chi(0)$, which is $1$.
Hence $a$ must be a unit in $\mathbb{C}_p^\times$. Moreover, its image in the residual field must be equal to $1$. So $a$ is in the ball $B(1)$.
Conversely, for any element $a$ of $B(1)$, it is easy to see that there is a continuous character $\chi_a:\mathbb{Z}_p\rightarrow\mathbb{C}_p^\times$ which maps $1$ to $a$.
The map $a \mapsto \chi_a$ is then an isomorphism from $B(1)$ to $\operatorname{Hom}_{cont}(\mathbb{Z}_p, \mathbb{C}_p^\times)$.