pontryagin dual of the group S^{-1}Z $T$ 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?
 A: 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).
