# Commutative Algebra, Notes 2: Localization

We will mostly adopt the Bourbaki definitions for the preliminary material.  An ideal $\mathfrak{p}$ in a ring $R$ is prime if $A/\mathfrak{p}$ is an integral domain.  Two ideals $\mathfrak{a},\mathfrak{b}$ are relatively prime if $\mathfrak{a}+\mathfrak{b}=R.$  A multiplicative subset of a set $X$ is a submagma $S$ of $X.$  In a ring, we assume it is a submagma with respect to multiplication of course.  We then have the canonical extension of $R$ by formal inverses of $S$ written $S^{-1}R:=R[S^{-1}]$ (although I will use the notation $R[S^{-1}]$ to reinforce the conceptualization of a ring extension).

Now suppose $M$ is an $R$-module.  A natural question to ask is “can we induce some $R[S^{-1}]$-module”?  The natural choice will be the $R[S^{-1}]$-module $M\otimes_R R[S^{-1}]$ where we define

$r(m\otimes r')=m\otimes rr'$

for $r,r'\in R[S^{-1}].$

Proposition 1.  Let $R$ be a ring and $S$ be a multiplicative subset.  Then there exists a ring $\hat{R}$ and homomorphism $f:R\to\hat{R}$ such that

1. Elements of $f(S)$ are invertible in $\hat{R}.$
2. For every homomorphism $g:R\to R'$ such that elements in $g(S)$ are invertible in $R',$ there exists a unique homomorphism $h:\hat{R}\to R'$ such that $g=h\circ f.$

It naturally turns out that $\hat{R}=R[S^{-1}]$ and $f(r)=rs/s$ for some $s\in S.$  Similarly we have:

Proposition 2.  Let $R$ be a ring, $S$ be a multiplicative subset, and $M$ be an $R$-module.  Then there exists an $R[S^{-1}]$-module $\hat{M}$  and homomorphism $f:M\to\hat{M}$ such that

1. For every $s\in S,$ the map $s:\hat{M}\to\hat{M}$ defined by $\hat{m}\mapsto s\hat{m}$ is bijective.
2. For every $R$-module $N$ such that the map $n\mapsto sn$ is bijective for every $s\in S$ and if we have a homomorphism $g:M\to N,$ there exists a unique homomorphism $h:\hat{M}\to M$ such that $g=h\circ f.$

Here we can show that $\hat{M}=M\otimes_R R[S^{-1}]$ and $f(m)=m\otimes 1.$  $R[S^{-1}]$ is called the ring of fractions of $R$ by $S,$ while $M\otimes_R R[S^{-1}],$ often denoted $S^{-1}M$ for short, is called the module of fractions of $M$ by $S.$

Definition 3.  Let $\mathfrak{a}$ be an ideal in $R.$  We define the radical of $\mathfrak{a}$ in $R$ as

$\sqrt[R]{\mathfrak{a}}=\sqrt{\mathfrak{a}}=\{r\in R:r^n\in\mathfrak{a}\mbox{~for some~}n\in\mathbb{N}\}.$

We call $\sqrt[R]{0}=\sqrt{0}$ the nilradical of $R.$  $R$ is a reduced ring if $\sqrt{0}=0.$

Let $A$ be an $R$-algebra and $S$ be a multiplicative subset of $R.$  Then since $A$ is an $R$-module, we can make $A$ into an $R[S^{-1}]$-module

$S^{-1}A=A\otimes_R R[S^{-1}].$

But we can also turn it into an $R[S^{-1}]$-algebra.  We need a homomorphism

$f:\left(A\otimes_R R[S^{-1}]\right)\otimes_{R[S^{-1}]}\left(A\otimes_R R[S^{-1}]\right)\to A\otimes_R R[S^{-1}].$

But the domain collapses to $(A\otimes_RA)\otimes_RR[S^{-1}]$ and then to $A\otimes_RR[S^{-1}]$ under the product of $A.$  So we have the canonical product we need.

[1]  Bourbaki, N.  Commutative Algebra, Ch 1-7.  Springer-Verlag.  1989.