Localization

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 29 March 2014

Let $R$ be a commutative ring.

A prime ideal is an ideal $𝔭$ of $R$ such that $R / 𝔭$ is an integral domain.
A maximal ideal is an ideal $𝔪$ of $R$ such that $R / 𝔪$ is a field.
A local ring is a ring with a unique maximal ideal.
A multiplicative subset is a subset $S \subseteq R$ which is closed under multiplication.

Exercise: An ideal $𝔭$ of $R$ is prime if and only if $R ∖ 𝔭$ is a multiplicative subset of $R$ .

Let $R$ be a commutative ring, let $S$ be a subset of $R$ and let $𝔭$ be a prime ideal of $R$ .

The multiplicative closure of $S$ is the smallest multiplicative subset $\overline{S}$ such that $\overline{S} \supseteq S$ .
The ring of fractions defined by $S$ is the ring $R [S^{- 1}] = {\frac{a}{s} | a \in R, s \in \overline{S}}$ $with \frac{a_{1}}{s_{1}} = \frac{a_{2}}{s_{2}} if there exists z \in \overline{S} such that z (a_{1} s_{2} - s_{1} a_{2}) = 0,$ with operations given by $\frac{a_{1}}{s_{1}} + \frac{a_{2}}{s_{2}} = \frac{a_{1} s_{2} + s_{1} a_{2}}{s_{1} s_{2}} and \frac{a_{1}}{s_{1}} \cdot \frac{a_{2}}{s_{2}} = \frac{a_{1} a_{2}}{s_{1} s_{2}},$ and with the map $R ↪ R [S^{- 1}] given by r \mapsto \frac{r}{1} .$
The localization of $R$ at $𝔭$ is $R_{𝔭} = R [S^{- 1}]$ where $S = R ∖ 𝔭 .$

Let $R$ be a ring, let $𝔭$ be a prime ideal of $R$ and let $R_{𝔭} = {\frac{a}{d} | a, d \in R, d \notin 𝔭}$ be the localization of $R$ at $𝔭$ .

The functor $\begin{array}{rccc} R_{𝔭} \otimes - : & {R -modules} & \to & {R_{𝔭} -modules}, \\ M & \mapsto & M_{𝔭} \end{array} where M_{𝔭} = R_{𝔭} \otimes_{R} M,$ is exact.
The map $\begin{array}{ccc} {proper ideals of R_{𝔭}} & \leftrightarrow & {proper ideals of R contained in 𝔭} \\ I & \mapsto & I \cap R \\ J_{𝔭} & \leftrightarrow & J \end{array}$ is a bijection which takes prime ideals to prime ideals and $𝔭_{𝔭}$ is the unique maximal ideal of $R_{𝔭} .$

		Proof.
	Let $𝔮$ be a prime ideal of $R$ contained in $𝔭$ . Then $R / 𝔮$ is an integral domain and $R_{𝔭} / 𝔮_{𝔭}$ is an integral domain??? a. ?? b. ?? This proof needs to be filled in. $□$

Notes and References

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