Is $\hom_{R}(M,T)$ locally compact whenever $M$ is, $R$ being a compact DVR? An important result in Harmonic analysis states that the Pontryagin dual $\widehat{A}$ of an locally compact abelian group $A$, defined as $\hom_{\mathbb{Z}}(A,T)$ where the hom is taken over continuous group homomorphisms and $T=\mathbb{R}/\mathbb{Z}$, is again locally compact.
Here and always I will always say "(locally) compact" when I mean "(locally) compact Hausdoff".
Recently I have been thinking a lot about how this sort of sitatuion generalizes when we examine morphisms of modules over rings other than $\mathbb{Z}$. In particular, I have been curious about the following:

If $R$ is a compact Discrete Valuation Ring does that imply that the dual $\widehat{M}$ of a locally compact module $M$ will be locally compact, where I define $\widehat{M}=\hom_R(M,T)$ with hom taken over continuous module homomorphisms and $T=\mathrm{Frac}(R)/R=K/R$?

The standard proofs for abelian groups use the theory of integration heavily and I'm not figuring out how to generalize that theory to the case of modules over other rings.
 A: Here is a partial answer.  Suppose that $M$ is not just locally compact, but has a neighborhood base of $0$ consisting of compact submodules.  Then I claim $\widehat{M}$ is locally compact.  To prove this, let $f\in\widehat{M}$.  Since $T$ is discrete, $\ker(f)$ is open, so it contains some compact open submodule $K\subseteq M$.  The set $U$ of continuous homomorphisms $M\to T$ that vanish on $K$ is then open in the compact-open topology.  I claim $U$ is compact, so it is a compact neighborhood of $f$ in $\widehat{M}$.
To prove this, note first that $U$ can naturally be identified with $\widehat{M/K}$.  Since $K$ is open, $M/K$ is discrete, so the compact-open topology on $\widehat{M/K}$ is just the product topology.  Also, $M/K$ must be a torsion $R$-module (if $x\in M/K$ were a non-torsion element then continuity of scalar multiplication would imply $Rx$ is not discrete).  So $\widehat{M/K}$ is actually a subspace of the product $\prod_{x\in M/K}\pi^{-n_x}R/R$, where $\pi$ is a uniformizer and $n_x$ is such that $\pi^{n_x}x=0$.  Since $R$ is compact, so is $\pi^{-n_x}R/R\cong R/(\pi^{n_x})$, and thus so is the product $\prod_{x\in M/K}\pi^{-n_x}R/R$.  The subset of this product consisting of the homomorphisms $M/K\to T$ is closed, and thus $\widehat{M/K}$ is compact.
A: Here is a solution to the problem, based of Eric Wofsey's partial response. Given any $f\in \widehat{M}$, $\ker(f)$ is clopen and so by local compactness there is an inclusion of subsets $U\leq K \leq \ker(f)$ with $U$ open and $K$ compact. Letting $V(A,B)$ denote the set consisting of maps in $\widehat{M}$ mapping $A$ into $B$, we obtain an inclusion of subsets
$$V(\ker(f),0)\leq V(K,0)\leq V(U,0).$$
Since $f\in V(\ker(f),0)$, we have that $V(K,0)$ is an open neighborhood of $f$. It remains to show that $V(U,0)$ is compact. To do this, we note that a morphism acts by $0$ on $U$ if and only if it acts by $0$ on the (open) submodule $\tilde{U}$ generated by $U$. Hence, $V(U,0)=V(\tilde{U},0)=\widehat{M/\tilde{U}}$ so by the second paragraph of Eric Wofsey's answer we are done.
