## More Localisation

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 $A[S-1] = {as | a∈A, s∈S_}$ with and with $as + bt = at+sb st and as ⋅ bt = ab st .$

Let $M$ be an $A-$module. The module of fractions with denominators in $S$ is the $A\left[{S}^{-1}\right]-$module $S-1M = {ms | m∈M, s∈S_}$ with and $ms + nt = tm+sn st and as ⋅ nt = an st .$

Let $f:M\to N$ be an $A-$module homomorphism. Define $S-1f: S-1M → S-1N by (S-1f) (ms) = f(m) s .$

HW: Let $A$ be a ring and let $S\subseteq A.$ Show that $S-1: { A-modules } → { S-1A-modules }$ 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 $A=ℤ$ and $S=\left\{s\in ℤ\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}s\ne 0\right\}$ then $S=\stackrel{_}{S}$ and $A\left[{S}^{-1}\right]=ℚ.$
2. Show that if and $S=\left\{s\in ℤ\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}s\notin pℤ\right\}$ then $S=\stackrel{_}{S}$ and $A\left[{S}^{-1}\right]={ℤ}_{\left(p\right)}.$
3. Show that if and $S=\left\{{p}^{k}\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}k\in {ℤ}_{\ge 0}\right\}$ then $S=\stackrel{_}{S}$ and $A\left[{S}^{-1}\right]=ℤ\left[\frac{1}{p}\right].$

## 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?