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 $$PR_P=\left\{\frac{r}{s}\vert r\in P, s\notin P\right\}=\left\{\frac{0}{1},\frac{2}{1},\frac{4}{1},\frac{0}{3},\frac{2}{3},\frac{4}{3},\frac{0}{5},\frac{2}{5},\frac{4}{5}\right\}.$$ 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 $(0)$.

Answer 5

Consider $\newcommand\quotient[2]{{^{\Large #1}}/{_{ \Large #2}}} R = \quotient{\mathbb{Z}}{\langle 6 \rangle}$, $\mathfrak{p} = \quotient{\langle 2 \rangle}{\langle 6 \rangle} = \{0, 2, 4\}$ and $U = R \setminus \mathfrak{p} = \{1, 3, 5\}$. We want to compute the localization $U^{-1} R$ of $R$ at $U$.

According to the following questions, we can also construct $U^{-1} R$ as $\quotient{R[x_s \ : \ s \ \in \ U]}{\langle s x_s - 1 \ : \ s \ \in \ U\rangle}$.

• Can the localization $U^{-1} R$ also be constructed as $R[x_s : s \in U]/\langle s x_s - 1 : s \in U\rangle$?
• Localization as a quotient ring

This means that we have the following isomorphisms: $$ \begin{array}{rcl} \{1, 3, 5\}^{-1} \left(\quotient{\mathbb{Z}}{\langle 6 \rangle}\right) & \cong & \quotient{\left(\quotient{\mathbb{Z}}{\langle 6 \rangle}\right)[x_1, x_3, x_5]}{\langle x_1 - 1, 3x_3 - 1, 5x_5 - 1\rangle} \\ & \cong & \quotient{\left(\quotient{\mathbb{Z}}{\langle 6 \rangle}\right)[x, y, z]}{\langle x - 1, 3y - 1, 5z - 1\rangle} \\ & \cong & \quotient{\mathbb{Z}[x, y, z]}{\langle 6, x - 1, 3y - 1, 5z - 1\rangle} \\ & \cong & \quotient{\mathbb{Z}[y, z]}{\langle 6, 3y - 1, 5z - 1\rangle} \\ \end{array} $$ But now, consider the ideals $I = \langle 6, 3y - 1, 5z - 1\rangle$ and $J = \langle 2, y - 1, z - 1 \rangle$ of $\mathbb{Z}[y, z]$. We claim that $I = J$.

Indeed, we have:

• $2 = 2 - 6y + 6y = 2(1 - 3y) + 6y \in I$
• $y - 1 = 3y - 1 - 2y \in I$
• $z - 1 = 5z - 1 - 4z = 5z - 1 - 2(2z) \in I$

This shows that $J \subseteq I$.

On the other hand, we also have:

• $6 = 2 \cdot 3 \in J$
• $3y - 1 = y - 1 + 2y \in J$
• $5z - 1 = z - 1 + 4z = z - 1 + 2(2z) \in J$

This shows that $I \subseteq J$. Hence $I = J$ and we may continue our calculation: $$ \begin{array}{rcl} \quotient{\mathbb{Z}[y, z]}{\langle 6, 3y - 1, 5z - 1\rangle} & \cong & \quotient{\mathbb{Z}[y, z]}{\langle 2, y - 1, z - 1 \rangle} \\ & \cong & \quotient{\mathbb{Z}}{\langle 2 \rangle} \\ \end{array} $$ Putting everything together, we obtain: $$ \{1, 3, 5\}^{-1} \left(\quotient{\mathbb{Z}}{\langle 6 \rangle}\right) \cong \quotient{\mathbb{Z}}{\langle 2 \rangle} $$ In particular, $\{1, 3, 5\}^{-1} \left(\quotient{\mathbb{Z}}{\langle 6 \rangle}\right)$ is an integral domain.

The following questions are similar:

• How do we find localization of a ring?
• How to compute localization of module $\mathbb Z/n\mathbb Z$ at the prime ideal $(p)$?