Pontrjagin duality for a topological ring

Question

Let $R$ be a locally compact topological ring, and let $S$ be its Pontrjagin dual under addition. For instance,

1. ℤ/nℤ is Pontrjagin dual to the ring ℤ/nℤ
2. $\mathbb{Q}$/ℤ is Pontrjagin dual to the ring $\hat{ℤ}$
3. ℤ($p^{∞}$) (the Prüfer group) is Pontrjagin dual to the ring ℤₚ
4. ℝ/ℤ is Pontrjagin dual to the ring ℤ
5. ℝ is Pontrjagin dual to the local field ℝ
6. ℚₚ is Pontrjagin dual to the local field ℚₚ
7. (the adeles) is Pontrjagin dual to itself.

Of these examples, the local fields and adeles are Pontrjagin dual to themselves.

I am trying to find a proof that $R$ and $S$ have a Haar measure which is tailored to the present case, where R is a ring (and its underlying additive group is commutative). Is it potentially easier to show that $R≅[[R,S¹],S¹]$ as a topological Abelian group in this situation, or when both $R$ and its Pontrjagin dual are rings?

Answer 1

Suppose $G$ is any locally compact abelian group. Then $\mathbb{Z}\times G$ is a locally compact ring with multiplication $(a,g)\cdot(b,h)=(ab,ah+bg)$ (in other words, take $1\in\mathbb{Z}$ as the unit and let $gh=0$ for all $g,h\in G$)). So, any proof of the existence of Haar measure for locally compact rings immediately implies its existence for locally compact abelian groups. Similarly, any proof of Pontrjagin duality for the additive group of an arbitrary locally compact ring will immediately imply Pontrjagin duality for an arbitrary locally compact abelian group. Thus you cannot expect there to be any easier proofs of these facts when restricted to rings.