(This self-answer post was extracted from an answer I posted in this question; see this meta thread.)
Is there a general rule for computing the localization of $\mathbb{Z}/n\mathbb{Z}$ at a prime ideal?
This computation is useful, for example, in coming up with counterexamples that localization is a local property, providing examples of localizing at a submonoid containing zero divisors, and in general when seeing examples of localizations.
$\def\Z{\mathbb{Z}}$
The prime ideals of $\Z/n\Z$ are of the form $(p) = p\Z/n\Z$ where $p$ is a prime divisor of $n$. In general, we have: $$(\Z/n\Z)_{(p)} \cong \Z/p^{\mathrm{ord}_p(n)}\Z$$ where $\mathrm{ord}_p(n)$ denotes the multiplicity of $p$ in the prime factorization of $n$.
We can see this in two steps. Let $S$ be the multiplicatively closed set $\Z/n\Z \setminus (p)$. Elements of this localization are of the form $\frac{a}{b}$ where $a, b \in \Z$ and $p \nmid b$ (by abuse of notation we use integers to denote the corresponding image in $\Z/n\Z$). Let $m = \mathrm{ord}_p(n)$. Then:
We can rewrite any of these fractions as a fraction with denominator $1$, as $\frac{a}{b} = \frac{ab^*}{1}$ where $b^*$ denotes a multiplicative inverse of $b$ modulo $p^m$ (which exists because $b$ is coprime to $p^m$). Indeed, $a\cdot 1 - (ab^*)\cdot b \equiv 0 \mod p^m$ so we can multiply it by the integer $\frac{n}{p^m}$, which is in $S$ by definition of $m$, to get $0$ in $\Z/n\Z$.
Fractions with equal denominator (which we can now assume is 1) are equal iff their numerators are congruent modulo $p^m$. Indeed, $\frac{a}{1} = \frac{a'}{1}$ iff there exists an integer $k$ coprime to $p$ such that $k(a - a') \equiv 0 \mod n$. By unique factorization, this happens iff $a \equiv a' \mod p^m$.
Therefore the localization is a ring with the $p^m$ elements $\frac01,\frac{1}{1}$, $\frac{2}{1}, \ldots, \frac{p^m-1}{1}$, and its additive group is cyclic, so it must be isomorphic to $\Z/p^m\Z$.
So, for example:
