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!



Why is the natural map $\hom(A,\mathbb{R}/\mathbb{Z})\to K/A$ an isomorphism, $K/\mathbb{Q}_p$ unramified, $A=\mathcal{O}_K$?
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.


