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.

Answer 2

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 $\mathbb{Z}_{(2)}$; $2 \mathbb{Z}_{(2)}$ = $6 \mathbb{Z}_{(2)}$} = $\mathbb{Z}/2\mathbb{Z}$.

Answer 3

One simple way to compute this is to exploit the universal property of localization. By definition $\rm\, L = \mathbb Z/6_{\,(2)} = S^{-1}\, \mathbb Z/6\ $ where $\rm\ S\ =\ \mathbb Z/6 \ \backslash\ 2\ \mathbb Z/6\ =\ \{\bar 1, \bar 3, \bar 5\}.\,$ Since the natural map $\rm\ \mathbb Z/6\ \to\ \mathbb Z/2\ $ maps $\rm\:S\:$ to units, by said universality property it must factor through $\rm\,L,\, $ i.e. $\rm\ \mathbb Z/6\ \to\ L\ \to\ \mathbb Z/2,\, $ so either $\rm\, L = \mathbb Z/6\, $ or $\rm\, L = \mathbb Z/2,\,$ but $\rm\ \bar 3\in S,\ \ {\bar3}^{-1}\!\!\not\in \mathbb Z/6,\, $ so $\rm\,L \ne \mathbb Z/6$.

Answer 4

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 $(0)$.