Is $\hom_{R}(M,T)$ locally compact whenever $M$ is, $R$ being a compact DVR?

Question

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.

Answer 1

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.

Answer 2

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.