[Math] Localization at a prime ideal in $\mathbb{Z}/6\mathbb{Z}$

Question

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)?

Answer 1

The localization of a ring $R$ at a prime $P$ is the local ring $R_P$, having maximal ideal $PR_P$. In this case, $R=\mathbb{Z}/6\mathbb{Z}$ and $P=2\mathbb{Z}/6\mathbb{Z}$, so that the maximal ideal $$PR_P=\{\textstyle\frac{r}{s}\mid r\in P, s\notin P\}=\{\frac{0}{1},\frac{2}{1},\frac{4}{1},\frac{0}{3},\frac{2}{3},\frac{4}{3},\frac{0}{5},\frac{2}{5},\frac{4}{5}\}$$ However, note that $\frac{r_1}{s_1}=\frac{r_2}{s_2}$ iff there is a $u\notin P$ such that $u(r_1s_2-r_2s_1)=0$. Thus, for example $\frac{2}{1}=\frac{0}{1}$ because $3(2\cdot1-0\cdot 1)=0$. In fact every element of $PR_P$ is 0, by a similar computation. Thus $PR_P=0$, and a local ring whose maximal ideal is the 0 ideal is a field (and hence in particular an integral domain). Thus $R_P$ is an integral domain.

EDIT: We really also have to check that $R_P$ is not in fact the zero ring (which is not an integral domain or field). We can do this directly, by checking that $\frac{1}{1}\neq\frac{0}{1}$ (because there is no $s\notin P$ such that $s(1\cdot1-0\cdot 1)=s=0$), or we can do it as follows: The canonical map $f:R\rightarrow R_P$ defined by $f(r)=\frac{r}{1}$ can easily be seen to have kernel $\ker(f)=\{r\in R\mid \exists s\notin P: sr=0\}$. For $R=\mathbb{Z}/6\mathbb{Z}$ and $P=2\mathbb{Z}/6\mathbb{Z}$, note that $\ker(f)=P$, so that by the first isomorphism theorem $R/P$ injects into $R_P$, so that $R_P$ is not the zero ring.