I am trying to find a way to easily compute localisations of the ring $\mathbb{Z}/n\mathbb{Z}$ ($n>1$). Is there any general result for this? I found here that when the multiplicative subset is $S=\{1,b,b^2,\dots\}$ and the prime factorization of $n$ is $n = \prod p_i^{l_i}$ then $$S^{−1}(\mathbb{Z}/n\mathbb{Z}) \cong \prod_{p_i \nmid b} \mathbb{Z}/p_i^{l_i}\mathbb{Z}$$
Does this formula somehow generalize to an arbitrary multiplicative subset $S \subset \mathbb{Z}/n\mathbb{Z}$?
By Chinese remainder theorem, we have a ring homomorphism $$\Bbb Z/n\Bbb Z\xrightarrow\sim\prod_i\Bbb Z/p_i^{l_i}\Bbb Z$$ By exactness of localization, we get an isomorphism: $$S^{-1}(\Bbb Z/n\Bbb Z)\xrightarrow\sim\prod_iS^{-1}(\Bbb Z/p_i^{l_i}\Bbb Z)$$ If $p_i$ divide some element in $S$, then $S^{-1}(\Bbb Z/p_i^{l_i}\Bbb Z)=\{0\}$, otherwise $S^{-1}(\Bbb Z/p_i^{l_i}\Bbb Z)\cong\Bbb Z/p_i^{l_i}\Bbb Z$.
If $T$ is the saturation of $S$, that's the set of all divisors of elements in $S$, then $$S^{-1}(\Bbb Z/n\Bbb Z)\cong\prod_{p_i\notin T}S^{-1}(\Bbb Z/p_i^{l_i}\Bbb Z)$$
