I'm quite familiar with the $p$-adic numbers and $p$-adic analysis, so I already know that any continuous group homomorphism $\varpi_{p}:\left(\mathbb{Q}_{p},+\right)\rightarrow\left(\mathbb{C}\backslash\left\{ 0\right\} ,\times\right)$ is of the form: $$\varpi_{p}\left(\mathfrak{y}\right)=e^{2\pi i\left\{ \mathfrak{s}\mathfrak{y}\right\} _{p}},\textrm{ }\forall\mathfrak{y}\in\mathbb{Q}_{p}$$ for some $\mathfrak{s}\in\mathbb{Q}_{p}$, where $\left\{ \cdot\right\} _{p}$ is the $p$-adic fractional part.
Now, let $\Bbbk$ be a number field, and let $\mathfrak{p}\in\overline{\mathbb{Q}}\backslash\mathbb{Q}$ be an irrational uniformizer of a prime ideal of $\Bbbk$. Let: $$d_{\mathfrak{p}}\overset{\textrm{def}}{=}\left|\prod_{\sigma\in\textrm{Gal}\left(\mathbb{Q}\left(\mathfrak{p}\right)/\mathbb{Q}\right)}\sigma\left(\mathfrak{p}\right)\right|$$ be the absolute value of the norm of $\mathfrak{p}$ over $\mathbb{Q}$— the smallest positive $\mathbb{Z}$-integer which is divisible by $\mathfrak{p}$. Then, letting $\Bbbk_{\mathfrak{p}}$ denote the $\mathfrak{p}$-adic completion of $\Bbbk$ (a local field) my guess is that every group homomorphism $\varpi_{\mathfrak{p}}:\left(\Bbbk_{\mathfrak{p}},+\right)\rightarrow\left(\mathbb{C}\backslash\left\{ 0\right\} ,\times\right)$ is of the form: $$\varpi_{\mathfrak{p}}\left(\mathfrak{y}\right)=e^{2\pi i\left\{ \mathfrak{s}\mathfrak{y}\right\} _{d_{\mathfrak{p}}}},\textrm{ }\forall\mathfrak{y}\in\Bbbk_{\mathfrak{p}}$$ for some $d_{\mathfrak{p}}$-adic rational number $\mathfrak{s}$, where $\left\{ \cdot\right\} _{d_{\mathfrak{p}}}$ is the $d_{\mathfrak{p}}$-adic fractional part. However, letting $s\in\mathbb{C}$ be arbitrary, isn't $\mathfrak{y}\mapsto e^{2\pi is\left\{ \mathfrak{s}\mathfrak{y}\right\} _{d_{\mathfrak{p}}}}$ then another such homomorphism? Or does there exist an $\mathfrak{s}^{\prime}\in\mathbb{Q}_{d_{\mathfrak{p}}}$ so that $\mathfrak{y}\mapsto e^{2\pi is\left\{ \mathfrak{s}\mathfrak{y}\right\} _{d_{\mathfrak{p}}}}$ is the same as $\mathfrak{y}\mapsto e^{2\pi i\left\{ \mathfrak{s}^{\prime}\mathfrak{y}\right\} _{d_{\mathfrak{p}}}}?$ Or, am I on the wrong track, and I should actually be writing: $$\varpi_{\mathfrak{p}}\left(\mathfrak{y}\right)=e^{2\pi i\left\{ \mathfrak{s}\mathfrak{y}\right\} _{\mathfrak{p}}}$$ for some $\mathfrak{s}\in\Bbbk_{\mathfrak{p}}$?
Basically: given a number field $\Bbbk$, and a completion $\Bbbk_{\mathfrak{p}}$ at a prime $\mathfrak{p}$ in the ring of $\Bbbk$-integers, let $\tilde{\Bbbk}_{\mathfrak{p}}$ denote the group of all continuous group homomorphisms $\varpi_{\mathfrak{p}}:\left(\Bbbk_{\mathfrak{p}},+\right)\rightarrow\left(\mathbb{C}\backslash\left\{ 0\right\} ,\times\right)$ (a sort of generalized Pontryagin dual of $\Bbbk_{\mathfrak{p}}$). What is $\tilde{\Bbbk}_{\mathfrak{p}}$ topological-group-isomorphic to (ex: does $\tilde{\Bbbk}_{\mathfrak{p}}\cong\Bbbk_{\mathfrak{p}}$)? And, what is the general explicit formula $f\left(x,y\right)\overset{\textrm{def}}{=}\left\langle x,y\right\rangle$ for the duality bracket, so that every element of $\tilde{\Bbbk}_{\mathfrak{p}}$ is of the form: $$\mathfrak{y}\mapsto\left\langle \mathfrak{s},\mathfrak{y}\right\rangle ,\textrm{ }\forall\mathfrak{y}\in\Bbbk_{\mathfrak{p}}$$ for some $\mathfrak{s}\in\tilde{\Bbbk}_{\mathfrak{p}}$?