$T$ is a set of some prime numbers. $S$ is the multiplictive set generated by $T$.
How to compute the Pontryagin dual of $S^{-1}Z$ ($Z$ is integal ring and $S^{-1}Z$ is localization of $Z$ at $S$).
If $T$ is an empty set, then the Pontryagin dual of $S^{-1}Z=Z$ is $R/Z$.
If $T$ is the set of all prime numbers, the Pontryagin dual of $S^{-1}Z=Q$ is naturally isomorphic to $A\_Q/Q$, where
$A\_Q$ is the adele ring of rational number $Q$.


Question: I guess the Pontryagin dual of $S^{-1}Z$ is isomorphic to $(R \times\prod\_{p\in T}Q\_{p})/(S^{-1}Z)$.
Is it right?



pontryagin dual of the group S^{-1}Z
Let $\mathbf A\_T$ denote the restricted direct product $\mathbf R\times \prod'\_{p\in T}\mathbf Q\_p$ (relative to the subgroups $\mathbf Z\_p$, $p\in T$).


The OP asked whether the Pontryagin dual of $S^{-1}\mathbf Z$ is isomorphic to $\mathbf A\_T/(S^{-1}\mathbf Z)$. Let's take it up in two steps:


Step 1: $\mathbf Q\_p/\mathbf Z\_p$ is canonically isomorphic (as an abstract group) to the subgroup of $\mathbf R/\mathbf Z$ consisting of elements whose order is a power of $p$ (mapping $p^{-n}\in \mathbf Q\_p$ to $p^{-n}\in \mathbf R/\mathbf Z$ determines this embedding). This gives rise to an element $\psi\_p\in \hat{\mathbf Q}\_p$ which vanishes on $\mathbf Z\_p$ but not on $p^{-1}\mathbf Z\_p$. Also define $\psi\_\infty$ to be the quotient map $\mathbf R\to \mathbf R/\mathbf Q$ composed with the inversion map $x\mapsto -x$.


Step 2: The formula $(a\_p)\mapsto ((x\_p)\mapsto \exp(2\pi i \sum\_{p\in T\cup \{\infty\}} \psi\_p(a\_p x\_p)))$ defines an isomorphism $\mathbf A\_T\to \widehat{\mathbf A\_T}$.
Under this isomorphism $S^{-1}\mathbf Z$ maps to precisely those characters which vanish on $S^{-1}\mathbf Z$, establishing the isomorphism $S^{-1}\mathbf Z=\widehat{\mathbf A\_T/S^{-1}\mathbf Z}$ (this can be seen with the help of partial fraction expansions of rational numbers).


