Consider the ring $A = \mathbb{Z}/60\mathbb{Z}$ and the prime ideal $\mathfrak{p} = (2)$. Describe the localization $A_{\mathfrak{p}}.$
Apparrently this should be $\mathbb{Z}/4\mathbb{Z}$, but I don't see how this is gotten. We have that $$(\mathbb{Z}/60\mathbb{Z})_{(2)} = (\mathbb{Z}/60\mathbb{Z} \setminus (2))^{-1}\mathbb{Z}/60\mathbb{Z} = \left\{ \frac{a}{b} \mid a \in \mathbb{Z}/60\mathbb{Z}, b \notin (2) \right\}$$ but I don't see how taking the denominators to not be multiples of $2$ reduces this set to only $4$ elements?
It helps to think of $\mathbb{Z}/60 \mathbb{Z}$ as the product $\mathbb{Z}/4 \mathbb{Z} \times \mathbb{Z}/3 \mathbb{Z} \times \mathbb{Z}/5 \mathbb{Z}$, as per the Chinese Remainder Theorem. If we localize at $(2)$, then the equivalence classes 3 and 5 are in the multiplicative system $\mathbb{Z}/60 \mathbb{Z} \setminus (2)$, and those annihilate the elements found in the latter factors. In general for a commutative ring $A$ and multiplicative system $S \subseteq A$, $ a/s \in S^{-1} A$ is 0 iff $ta = 0$ for some $t \in S$. This does not happen for $\mathbb{Z}/4 \mathbb{Z}$, since the only elements $x$ of $\mathbb{Z}/60 \mathbb{Z}$ such that there is a $0 \neq y \in \mathbb{Z}/4 \mathbb{Z}$ with $xy = 0$ are in $(2)$.