Localization at a prime ideal in $\mathbb{Z}/6\mathbb{Z}$ How can we compute the localization of the ring $\mathbb{Z}/6\mathbb{Z}$ at the prime ideal $2\mathbb{Z}/\mathbb{6Z}$? (or how do we see that this localization is an integral domain)?

Here's a much less computational approach to show that the localisation is an integral domain (even a field):
For any (commutative, unital) ring $R$ and prime ideal $P$, recall that $R_P$ is a local ring with unique maximal ideal $P_P$. 
In the given case, since there is an element $x \in R\setminus P$ (i.e. $3$) such that $x \in \operatorname{Ann}(P)$ it follows that $P_P$ is $(0)$. Thus $R_P$ is a local ring with unique maximal ideal $(0)$ and so is a field. 
Note that this actually gives us the stronger statement that $R_P$ is a field iff $P_P$ is $(0)$.
