More Localisation

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

Last update: 15 June 2012

The functor $\left[S^{-1}\right]$

Let $A$ be a ring. A set $\stackrel{\_}{S}\subseteq A$ is multiplicative, if $\stackrel{\_}{S}$ satisfies

1. $1\in \stackrel{\_}{S}$
2. If $s_1,s_2\in \stackrel{\_}{S}$ then $s_1{s}_2\in \stackrel{\_}{S}.$

Let $A$ be a ring and $S\subseteq A.$ The multiplicative closure of $S$ is a subset $\stackrel{\_}{S}\subseteq A$ such that

1. $\stackrel{\_}{S}$ is multiplicative and $\stackrel{\_}{S}\supseteq S.$
2. If $T\subseteq A$ is multiplicative and $T\supseteq S$ then $T\supseteq \stackrel{\_}{S}.$

The ring of fractions with denominators in $S$ is $$A\left[S^{-1}\right]=\left\{\frac{a}{s}\right|a\in A,s\in \stackrel{\_}{S}\}$$ with $$\frac{a}{s}=\frac{a\prime }{s\prime }if\; there\; exists\; t\in \stackrel{\_}{S} \; with\; t(as\prime -sa\prime )=0$$ and with $$\frac{a}{s}+\frac{b}{t}=\frac{at+sb}{st}and\frac{a}{s}\cdot \frac{b}{t}=\frac{ab}{st}.$$

Let $M$ be an $A-$module. The module of fractions with denominators in $S$ is the $A\left[S^{-1}\right]-$module $$S^{-1}M=\left\{\frac{m}{s}\right|m\in M,s\in \stackrel{\_}{S}\}$$ with $$\frac{m}{s}=\frac{m\prime }{s\prime }if\; there\; exists\; t\in \stackrel{\_}{S} \; with\; t(sm\prime -s\prime m)=0$$ and $$\frac{m}{s}+\frac{n}{t}=\frac{tm+sn}{st}and\frac{a}{s}\cdot \frac{n}{t}=\frac{an}{st}.$$

Let $f:M\to N$ be an $A-$module homomorphism. Define $$S^{-1}f:S^{-1}M\to S^{-1}Nby\left(S^{-1}f\right)\left(\frac{m}{s}\right)=\frac{f\left(m\right)}{s}.$$

HW: Let $A$ be a ring and let $S\subseteq A.$ Show that $$S^{-1}:\left\{A-modules\right\}\to \left\{S^{-1}A-modules\right\}$$ is an exact functor.

HW: Formulate and prove an appropriate universal property for $A\left[S^{-1}\right].$

HW: Formulate and prove an appropriate universal property for $S^{-1}M.$

HW: Show that $S^{-1}M=M{\otimes }_{A}A\left[S^{-1}\right].$

HW:

1. Show that if $A=\mathbb{Z}$ and $S=\{s\in \mathbb{Z}|s\ne 0\}$ then $S=\stackrel{\_}{S}$ and $A\left[S^{-1}\right]=\mathbb{Q}.$
2. Show that if $A=\mathbb{Z}, p\in \mathbb{Z},$ and $S=\{s\in \mathbb{Z}|s\notin p\mathbb{Z}\}$ then $S=\stackrel{\_}{S}$ and $A\left[S^{-1}\right]={\mathbb{Z}}_{\left(p\right)}.$
3. Show that if $A=\mathbb{Z}, p\in \mathbb{Z}$ and $S=\left\{p^k\right|k\in {\mathbb{Z}}_{\ge 0}\}$ then $S=\stackrel{\_}{S}$ and $A\left[S^{-1}\right]=\mathbb{Z}\left[\frac{1}{p}\right].$

Notes and References

Localisations are covered in [AM, Ch.3] and [Bou, Comm. Alg., Ch.II§2]. In particular the solution to HW2 is found in [AM, Prop. 3.1 and Cor. 3.2] and [Bou, Comm. Alg., Ch.II Prop. 1 and Def. 2]. The solution to HW3 is discussed in [Bou, Comm. Alg., Prop 3 and following 3 paragraphs]. The solution to HW1 is found in [AM, Prop. 3.3].

References

References?

page history