LEC ALG3 8

Ring of fractions

A Commutative ring $R$ is always a subring of a larger ring $Q$, called the ring of fractions, in which every nonzero element of $R$ that is not a zero divisor is a unit in $Q$. If $R$ is an integral domain, $Q$ will be a field, called the field of fractions of $R$.

Theorem 113.1(Theorem).

Let $R$ be a commutative ring. Let $D$ be any nonempty subset of $R$ that does not contain $0$, does not contain any zero divisors and is closed under multiplication. Then there is a commutative ring $Q$ such that $Q$ contains $R$ as a subring and every element of $D$ is a unit in $Q$. Also,

1. every element of $Q$ is of the form $r{d}^{-1}$, for some $r\in R$ and $d\in D$. If $D=R\setminus \{0\}$ then $Q$ is a field.
2. $Q$ is the “smallest” ring containing $R$ in which all elements of $D$ become units. Precisely, $Q$ satisfies this universal property: If $h:R\to S$ in an injective ring homomorphism such that $h(d)$ is a unit for every $d\in D$, there exists a unique injective homomorphism $\overline{h}:Q\to S$ such that $\overline{h}|{}_R=h$.

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Localization

We now generalize the notion of ring of fractions so that $D$ can have zero divisors. Note that when $D$ only contained non-zero divisors, the defining equivalence relation on $D^{-1}R$ was

$$\frac{r}{s}=\frac{r^{\prime }}{s^{\prime }}\text{iff }r{s}^{\prime }=r^{\prime }s.$$

We cannot use the same definition here: suppose $r{s}^{\prime }-r^{\prime }s\ne 0$ and $u\in D$ is such that $u(r{s}^{\prime }-s^{\prime }r)=0$. Since $u$ is a unit in $D^{-1}R$, we have

$$\frac{r}{s}=\frac{r^{\prime }}{s^{\prime }}⟺\frac{ur}{s}=\frac{u{r}^{\prime }}{s^{\prime }},$$

forcing $r/s=r^{\prime }/s^{\prime }$. The following definition provides the obvious remedy:

Definition 113.2(Definition).

Let $R$ be a commutative ring and $S\subseteq R$ be a multiplicative subset. The localization $S^{-1}R$ is the ring of fractions $r/s$ with $r\in R$, $s\in S$, modulo the relation

$$\frac{r}{s}=\frac{r^{\prime }}{s^{\prime }}\text{ iff }\exists u\in S\text{ with }u(s^{\prime }r-s{r}^{\prime })=0.$$

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Standard examples of multiplicative sets include:

1. $S=R\setminus \mathfrak{p}$, for a prime ideal $\mathfrak{p}$
2. $S=\{1,f,f^2,\dots \}$
3. $S=R\setminus \{0\}$, where $R$ is an integral domain (localizing at this $S$ just gives us the field of fractions of $R$).

If $0\in S$, $S^{-1}R$ is the zero ring, so we usually exclude $0$ form $S$. There is a canonical embedding

$$\iota :R\to S^{-1}R,r\mapsto r/1,$$

which is injective iff $S$ does not contain any zero divisors.

There is a bijection between the set of prime ideals of $S^{-1}R$ and the set of prime ideals of $R$ that are disjoint from $S$ (stated here for integral domains, this apparently holds in general. Check this).

Localization of a module

Definition 113.3(Definition).

Let $M$ be an $R$-module and $S\subseteq R$ be a multiplicative set. The localization of $M$ at $S$ is denoted $S^{-1}M$ or $M_S$ and constructed analogously to Definition 2.

$S^{-1}M=0$ iff for every $m\in M$ there exists $s\in S$ with $sm=0$. For $S=R\setminus \mathfrak{p}$, this gives: $M_{\mathfrak{p}}=0$ iff for every $m\in M$ there is $s\notin p$ with $sm=0$. Equivalently, $M_{\mathfrak{p}}\ne 0$ iff there exists $m\in M$ with ${\text{Ann}}_R(m)\subseteq \mathfrak{p}$.