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)?

We first recall the following well known result. 

Let $R$ be a commutative ring with unity, $I$ an ideal in $R$ and $S$ a multiplicative closed set in $R$. Let $\bar{S}$ denote the image of $S$ in the quotient ring $\bar{R}=R/I$. Then $\bar{S}^{-1}\bar{R} = S^{-1}R/IS^{-1}R$. (Check the book here.)

Now take $R=\mathbb{Z}$, $I=6\mathbb{Z}$ and $S=\mathbb{Z}-2\mathbb{Z}$. Then $\bar{\mathbb{Z}}_{(\bar{2})}= \bar{S}^{-1}\bar{R} = S^{-1}R/IS^{-1}R = \mathbb{Z}_{(2)}/ 6\mathbb{Z}_{(2)} = \mathbb{Z}_{(2)}/ 2\mathbb{Z}_{(2)}$ {since 3 is a unit in $\mathbb{Z}_{(2)}$; $2 \mathbb{Z}_{(2)}$ = $6 \mathbb{Z}_{(2)}$}  = $\mathbb{Z}/2\mathbb{Z}$.
