Let $S \subset \mathbb{Z}$ denote a multiplicative set, i.e. $1 \in S$ and if $a,b \in S$ then $ab \in S$.
Let $p_i \in \mathbb{Z}$ be a prime, and let $e_i \in \mathbb{Z}^+$. Furthermore, let $S$ be such that $\forall s \in S: \text{gcd}(s,p_i) = 1$.
I want to show that it follows that $S^{-1}(\mathbb{Z}/(p_i^{e_i})) \cong \mathbb{Z}/(p_i^{e_i})$, that is, the localization of $\mathbb{Z}/(p_i^{e_i})$ at the multiplicative set $S$ is isomorphic to $\mathbb{Z}/(p_i^{e_i})$ itself.
I am having a hard time showing that this is the case. I suppose I would have to show that every element on the form $\frac{x}{s}$ for $x \in \mathbb{Z}/(p_i^{e_i})$ is equivalent to some $x' \in \mathbb{Z}/(p_i^{e_i})$.
I´d be interested to se either an explicit isomorphism or an argument.