While looking at an analogue of Pontryagin duality for compact Discrete Valuation Rings (DVRs), I came about the observation that generally one should have an isomorphism of $A$-modules
$$\hom_{\mathbb{Z}}\left(A,\mathbb{R}/\mathbb{Z}\right)\xrightarrow{\sim} K/A,$$
where $A$ is a DVR that is compact with respect to its canonical topology, and $K$ is its field of fractions. The left hand side consists of continuous $\mathbb{Z}$-module homomorphisms from $A$ with its canonical metric topology to $\mathbb{R}/\mathbb{Z}$ with its usual topology. The $A$-module structure on the left is given by $(a\cdot \varphi)(x)=\varphi(a\cdot x)$.
For DVRs, being compact is equivalent to being complete and having finite residue field. This means that if the characteristic of $A$ is equal to the characteristic of its residue field then $A\cong \mathbb{F}_q[[t]]$ for some finite field $\mathbb{F}_q$. In this case, we can choose an $\mathbb{F}_q$-module isomorphism $i:\hom_{\mathbb{Z}}(\mathbb{F}_q,\mathbb{R}/\mathbb{Z})\xrightarrow{\sim}\mathbb{F}_q$ and define
\begin{align*}
\hom_{\mathbb{Z}}(\mathbb{F}_q[[t]],\mathbb{R}/\mathbb{Z})&\xrightarrow{} \mathbb{F}_q((t))/\mathbb{F}_q[[t]].\\
\varphi&\mapsto \sum_{n=0}^{\infty}i\left(\left.\varphi\right|_{\mathbb{F}_q \cdot x^n}\right) x^{-(n+1)}
\end{align*}
A tedious check of all the desired properties shows that this is an isomorphism of $A$-modules. One key step is the fact that $\varphi$ being continuous is equivalent to $\varphi$ acting by $0$ on $\mathbb{F}_q\cdot x^{n}$ for all large enough $n$, and hence giving a well defined element of $\mathbb{F}_q((t))/\mathbb{F}_q[[t]]$.
I now want to extend this result to the mixed characteristic case. I'm guessing that the correct first step is to do this when $A\cong W(\mathbb{F}_q)$ is the space of Witt vectors over its (finite) residue field. In particular, we can write $K/\mathbb{Q}_p$ to be the unique absolutely unramified extension of degree $\log_p(q)$ of $\mathbb{Q}_p$. My guess for defining a good map would be to use the fact that there is a multiplicative section $\omega:\mathbb{F}_q^{\times} \xrightarrow{} A^{\times}$ to the reduction map $A^{\times}\xrightarrow{} \mathbb{F}^{\times}_q$ given by sending any element $g\in \mathbb{F}_q^{\times}$ to $\lim_{n\to\infty}\left(\tilde{g}\right)^{q^n}$, where $\tilde{g}$ is any lift of $g$ to $A$. Namely, we construct
\begin{align*}
\hom_{\mathbb{Z}}\left(A,\mathbb{R}/\mathbb{Z}\right)&\xrightarrow{\ell} K/A.\\
\varphi &\mapsto \sum_{g\in \mathbb{F}_q^{\times}}\varphi(g)\cdot g^{-1}
\end{align*}
This map is well defined since $\varphi(g)$ must lie in the Prüfer $p$-group $\mathbb{Z}[1/p]/\mathbb{Z}$, since
$$\lim_{n\to\infty} p^{n}\varphi(g)=\varphi\left(\lim_{n\to\infty}p^n\cdot g\right)=0.$$
We can then naturally identify the groups $\mathbb{Z}[1/p]/\mathbb{Z}$ and $\mathbb{Q}_p/\mathbb{Z}_p$ and thus make sense of the expression $\sum_{g\in \mathbb{F}_q^{\times}}\varphi(g)\cdot g^{-1}$. Now, we check that $\ell$ is an $A$-module homomorphism. Since $p$ is a uniformizer, we have that $A$ is (topologically) generated by elements of the form $h\cdot p^n$ where $h\in \mathbb{F}_q^{\times}$. On elements of this form we see that
\begin{align*}
\ell\left((h\cdot p^n)\cdot \varphi\right)&=\sum_{g\in \mathbb{F}_q^{\times}}\varphi\left(\left(h\cdot p^n\right)\cdot g\right)\cdot g^{-1}\\
&=(h\cdot p^n)\cdot \sum_{g\in \mathbb{F}_q^{\times}}\varphi\left(hg\right)\cdot (hg)^{-1}\\
&=(h\cdot p^n)\cdot \ell(\varphi).
\end{align*}
Since $\ell$ clearly preserves addition, a short continuity argument shows we are done. Thus, what remains is this:
Is the above map $\ell$ bijective?
I have no idea why this map is injective. If $\ell(\varphi)\neq 0$ for some element $\varphi$ though, I can deduce that the map is surjective. This is done as follows. Given any $\mathbb{Z}$-module morphism $\varphi: A \xrightarrow{} \mathbb{R}/\mathbb{Z}$ I claim there is at least one $\mathbb{Z}$-module morphism $\varphi': A \xrightarrow{} \mathbb{R}/\mathbb{Z}$ such that $\varphi=p\cdot \varphi'$. From this subjectivity is obvious, since applying the claim to $\varphi$ repeatedly gives elements in the image of $\ell$ with arbitrarily low valuations and then multiplying by elements of $A$ and using the fact that $\ell$ is an $A$-module morphism gives the full codomain as the image.
To prove that such a map $\varphi'$ always exists, we do as follows. We define $\varphi'$ on $p\cdot A$ by $\varphi'(p\cdot x)=\varphi(x)$. By Baer's criterion, $\mathbb{R}/\mathbb{Z}$ is an injective object in the category of $\mathbb{Z}$-modules. This means we can lift the map $\varphi'$ to the full domain $A$. Carrying around slightly more data we can ensure that $\varphi'$ will be continuous, and hence we conclude the result.
Thank you!
Best Answer
This answer does not prove that the map described in the question is an isomorphism, but it does prove that an isomorphism between $\hom_{\mathbb{Z}}(A,\mathbb{R}/\mathbb{Z})$ and and $K/A$ exists.
To begin, we note that $M \mapsto \hom_{\mathbb{Z}}(A,M)$ is a right adjoint to the forgetful functor from the category of (locally compact) topological $A$-module to the category of (locally compact) topological groups. Hence, since $\mathbb{R}/\mathbb{Z}$ is a cogenerator in the target category we get that $\hom_{\mathbb{Z}}(A,\mathbb{R}/\mathbb{Z})$ is a cogenerator in the category of locally compact topological groups and thus there exists a nonzero morphism $f:K/A\xrightarrow{} \hom_{\mathbb{Z}}(A,\mathbb{R}/\mathbb{Z})$. Moreover, by factoring we get an injective map
$$f':(K/A)/\ker(f)\hookrightarrow{}\hom_{\mathbb{Z}}(A,\mathbb{R}/\mathbb{Z}).$$
Submodules of $K/A$ are all of the form $\ker_{n}=\left\{x\,\,\mathrm{s.t}\,\, \pi^n x=0\right\}$. multiplication by $\pi^n$ gives an isomorphism
$$K/A\xrightarrow{\sim} (K/A)/\ker_n,$$
and hence we can consider $f'$ as a morphism $K/A\hookrightarrow{} \hom_{\mathbb{Z}}(A,\mathbb{R}/\mathbb{Z})$. Both modules are clearly equal to their $p^{\infty}$-torsion. In particular, we can compare the size of their $p^n$-torsion for each $n$. Once we establish that they have the same number of elements for each $n$, then infectivity will imply surjectivity onto each $p^n$ and hence surjectivity onto the full module so we will be done.
Given any map $\varphi$ in $\hom_{\mathbb{Z}}(A,\mathbb{R}/\mathbb{Z})$, every element $x$ in the image of $\varphi$ must have the property that $\lim_{n\to\infty}p^n x$ goes to zero. The only such elements in $\mathbb{R}/\mathbb{Z}$ are those in the Prüfer $p$-group $\mathbb{Z}[1/p]/\mathbb{Z}$, and hence we get that
$$\hom_{\mathbb{Z}}(A,\mathbb{R}/\mathbb{Z})=\hom_{\mathbb{Z}}(A,\mathbb{Z}[1/p]/\mathbb{Z}).$$
Now, every module of a DVR is free and hence we have an (I believe topological) $\mathbb{Z}_p$-module isomorphism $A\xrightarrow{} \bigoplus \mathbb{Z}_p$ which commutes with the $\hom$. Pairing $\varphi: \mathbb{Z}_p\xrightarrow{} \mathbb{Z}[1/p]/\mathbb{Z}$ with $\varphi(1)$ and checking the module structure, we get an isomorphism between $\hom(\mathbb{Z}_p,\mathbb{Z}[1/p]/\mathbb{Z})$ and $\mathbb{Q}_p/\mathbb{Z}_p$. Counting $p$-torsion we get the correct number of elements and hence we conclude the result.
I am fairly sure the argument as stated here checks out, at least in the unramified case. I will be coming back in the next couple of days to flesh it out more.