Suppose \(R\) is a commutative ring and \(S \subseteq R\) is a subset such that
\(1 \in S\text{,}\)
\(S\) is closed under multiplication (i.e., if \(x,y \in S\text{,}\) then \(xy \in S\)), and
\(S\) does not contain \(0\) nor any zerodivisors.
Such a subset \(S\) is called a multiplicatively closed subset of non zerodivisors of \(R\text{.}\)
Example11.2.Examples of Multaplicatively Closed Sets.
Two types of multiplicatively closed sets are most commonly used in practice:
If \(R\) is a field then \(S=R\setminus \{0\}\) is a multiplicatively closed set of non zerodivisors.
If \(R\) is an arbitrary ring with \(1\neq 0\) and \(x\in R\) is a non zerodivisor then the set of non negative powers of \(x\text{,}\)\(S=\{x^n \mid n\in \mathbb{Z}, n\geq 0\}\text{,}\) is multiplicatively closed.
Definition11.3.Field of Fractions.
If \(R\) is an integral domain and \(S\) is a multiplicatively closed subset of nonzerodivisors, the field of fractions\(S^{-1}R\) is the set of equivalence classes
If \(R\) is an integral domain and \(S\) is a multiplicatively closed subset of nonzerodivisors, the rules given in the above definition for \(+\) and \(\cdot\) make \(S^{-1} R\) into a field. Moreover, the function \(R\to S^{-1}R\) sending \(r\) to \(\frac{r}{1}\) is an injective ring homomorphism.
Proof.
There is a lot of small things to check and we’ll just do a few. Right off the bat we need to be sure the given equivalence relation really is one. The reflexive and symmetric properties are clear. But the proof of transitivity illustrates a key point: Say \(\frac{r}{s} \sim \frac{r'}{s'} \sim \frac{r''}{s''}\text{.}\) Then \(rs'= r's\) and \(r's'' = r''s'\text{.}\) We need to deduce that \(rs'' = r''s\text{.}\) The given equations imply \(rs's'' = r'ss'' = r''ss'\) and since \(s'\) is a nonzerodivisor we conclude \(rs'' = r''s\text{.}\) This is in fact the only time that the fact that \(S\) consists of nonzerodivisors is used.
We also need to be sure our rules for \(+\) and \(\cdot\) make sense and are independent of representation. They “make sense” since we assume \(S\) is closed under \(\cdot\text{.}\) To show \(+\) is independent of representations, say \(\frac{r}{s} \sim \frac{r''}{s''}\text{,}\) so that \(rs'' = r''s\text{.}\) Then
and so we need to show \((rs' + r's)(s''s') \sim (r''s' + r's'')(ss')\text{.}\) This is clear upon expaning out both sides and using \(rs'' = r''s\text{.}\) In a similar way one shows \(\cdot\) is well-defined.
From now on we just write \(=\) instead of \(\sim\) when dealing with fractions.
The associative and distributive axioms involve a straightfoward but tedious check, and we skip them entirely. The fact that \(+\) and \(\cdot\) are commutative is clear from their definitions. The set \(S^{-1}R\) is a group under addition since it has a \(0\) element, namely \(\frac{0}{1}\text{,}\) and \(\frac{r}{s} +\frac{-r}{s} = \frac{rs - rs}{s^2} = \frac{0}{s^2} = \frac{0}{1}\text{,}\) with the last equality holding since \(0 \cdot 1 = s^2 \cdot 0\text{.}\) The \(1\) element is \(\frac{1}{1}\text{.}\) (Note that we have used that \(1 \in S\) a couple times here — indeed, without this assumption \(S\) could be empty and then \(S^{-1}R\) would be the empty set.)
\(R\) being a domain means \(xy=0\) implies (\(x=0\) or \(y=0\)). The contrapositive to this statement is: if \(x\neq 0\) and \(y\neq 0\) then \(xy\neq 0\text{,}\) which shows \(S\) is a multiplicatively closed set of nonzerodivisors.
It remains only to show every non-zero element of \(S^{-1}R\) is a unit. Given \(\frac{r}{s} \ne 0\text{,}\) note that \(r \ne 0\) and hence \(r \in S\text{.}\) So \(\frac{s}{r}\) is also an element of \(S^{-1}R\text{.}\) We have \(\frac{r}{s} \frac{s}{r} = \frac{sr}{sr} = \frac{1}{1}\text{,}\) where the last equation holds by the definition of \(\sim\text{.}\)
The fact that \(r \mapsto \frac{r}{1}\) is a ring homomorphism is straightforward to check. It’s injective since \(\frac{r}{1} = \frac{0}{1}\) implies \(r = 0\text{.}\)
Example11.5.Examples of Fields of Fractions.
For a specific example, the field of fractions of \(\mathbb{Z}\) is \(\mathbb{Q}\text{.}\)
or another, if \(d\) is a squarefree integer and \(R = \mathbb{Z}[\sqrt{d}]\) is an integral domain and we will show soon that its field of fractions is (isomorphic to) the field \(\mathbb{Q}(\sqrt{d})\text{.}\)
For yet another, \({\mathbb{R}}[x]\) is an integral domain. Its field of fractions, usually denoted \({\mathbb{R}}(x)\) constists of all rational functions. This last example could be generalized by replacing \({\mathbb{R}}\) with any field and also by using any number of variables.
