---- > [!definition] Definition. ([[field of fractions]]) > > The **field of fractions** $K(R)$, or $\text{Frac }R$, of an [[integral domain]] $R$ is the 'smallest [[field]] containing $R, in the sense that it satisfies the following [[universal property]]: there is an [[injection|injective]] [[ring homomorphism]] $i:R \to K(R)$ such that any [[injection|injective]] [[ring homomorphism]] $j:R \to L$ factors through $K(R)$ as $j=\alpha \circ i$ for unique $\alpha:K(R) \to L$. > > Note that by [[every nonzero ring homomorphism out of a field is injective]], $\alpha$ is an [[injection]] just like $i$ and $j$. > > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAGkQBfU9TXfIRQAmclVqMWbADLdeIDNjwEiARlKrx9Zq0QgASt3EwoAc3hFQAMwBOEALZJRIHBCRkXdLAzYALCBAA1iDU2lJ6WHLWdo6Izq5I6p7efgGBjDghEjpsAFZZDHQARjAMAAr8ykIgNlimvpk80Q7u1AmISWG6IAA6PTAAHlhwOHAAhAAEfYxovnRGXEA > \begin{tikzcd} > K(R) \arrow[rr, "\exists! \hat{j}"] & & L \\ > & R \arrow[lu, "i", hook] \arrow[ru, "j"', hook'] & > \end{tikzcd} > \end{document} > ``` > > This defines $K(R)$ [[terminal objects are unique up to a unique isomorphism|up to]] [[isomorphism]], should it exist. > > And exist it should! Consider the set $R \times (R^{*})$ of pairs $(a,r)$ of elements of $R$, where $r \neq 0$. The pair $(a,r)$ will be associated with a 'fraction' $\frac{a}{r}$, defined as the [[equivalence class]] of $(a,r) \in R \times (R^{*})$ under the [[equivalence relation]] $(a,r) \sim (b,s) \iff as-br=0.$ > Then $K(R)$ is defined to be the set $K(R):=\left\{ \frac{a}{r}: a \in R , r \in R , r \neq 0\right\}.$ > endowed with operations $\begin{align} > \frac{a}{r}+\frac{b}{s}=& \frac{as+br}{rs} \\ > \frac{a}{r} \cdot \frac{b}{s}= &\frac{ab}{rs} > \end{align}$ > which turn $K(R)$ into a [[field]]. > [!basicexample] > - $K(F)=F$ for any [[field]] $F$ ($F \xrightarrow{\id} F$ factors uniquely through $K(F)$, meaning $\hat{j}\left( \frac{a}{r} \right)=ar^{-1}=\hat{j} \circ i(a r ^{-1})$; since $\hat{j}$ [[every nonzero ring homomorphism out of a field is injective|is injective]] this means $\frac{a}{r}=i(ar ^{-1})$ so that $i$ is a [[surjection]].) > - $K(\mathbb{Z})=\mathbb{Q}$ > - ^basic-example > [!justification] > To begin, any time the term [[universal property]] is used there must be a [[category]] and [[terminal object]] therein lurking. Here, that [[category]] is the [[subcategory]] of the [[coslice category]] $\mathsf{Ring}^{R}$ obtained by keeping as objects just the [[ring homomorphism|ring]] [[injection|embeddings]] of $R$ into a [[field]]. The claim, then, is that this [[category]] has an [[terminal object|initial object]] that is to be called $K(R)$, should it exist. > > And exist it does. > > Consider the set $R \times (R^{*})$ of pairs $(a,r)$ of elements of $R$, where $r \neq 0$. The pair $(a,r)$ will be associated with a 'fraction' $\frac{a}{r}$, defined as the [[equivalence class]] of $(a,r) \in R \times (R^{*})$ under the [[equivalence relation]] $(a,r) \sim (b,s) \iff as-br=0.$ > As a set, $K(R)$ is defined $K(R):=\left\{ \frac{a}{r}: a \in R , r \in R , r \neq 0\right\}.$ > Note that its elements behave like ordinary fractions, for example, $\frac{as}{rs}=\frac{a}{r}$ > if $s \neq 0$: indeed, $(as)r=a(rs)$ by associativity and [[commutative ring|commutativity]] in $R$, hence $(as, rs) \sim (a, r)$. > > It is easy to check that the operations $\begin{align} > \frac{a}{r}+\frac{b}{s}=& \frac{as+br}{rs} \\ > \frac{a}{r} \cdot \frac{b}{s}= &\frac{ab}{rs} > \end{align}$ > are well-defined (here we need the [[integral domain]] assumption, to ensure $rs \neq 0$) and turn $K(R)$ into a [[field]]. > > We claim that the [[injection|injective]] [[ring homomorphism]] $\begin{align} > R \xrightarrow{i}& K(R) \\ > a \xmapsto{i} & \frac{a}{1} > \end{align}$ > is [[terminal object|initial]]. Let $j:R \to L$ be any [[injection|injective]] [[ring homomorphism]]; we need to define $\hat{j}:K(R) \to L$ making the diagram > > > ```tikz > \usepackage{tikz-cd} > \usepackage{amsmath} > \begin{document} > % https://tikzcd.yichuanshen.de/#N4Igdg9gJgpgziAXAbVABwnAlgFyxMJZABgBpiBdUkANwEMAbAVxiRAGkQBfU9TXfIRQAmclVqMWbADLdeIDNjwEiARlKrx9Zq0QgASt3EwoAc3hFQAMwBOEALZJRIHBCRkXdLAzYALCBAA1iDU2lJ6WHLWdo6Izq5I6p7efgGBjDghEjpsAFZZDHQARjAMAAr8ykIgNlimvpk80Q7u1AmISWG6IAA6PTAAHlhwOHAAhAAEfYxovnRGXEA > \begin{tikzcd} > K(R) \arrow[rr, "\hat{j}"] & & L \\ > & R \arrow[lu, "i", hook] \arrow[ru, "j"', hook'] & > \end{tikzcd} > \end{document} > ``` > commute, and we must show $\hat{j}$ is unique. > > Uniqueness follows because the definition of $\hat{j}$ is forced upon us: $\begin{align} > \hat{j}\left( \frac{a}{r} \right) = & \hat{j}\left( \frac{a}{1} \cdot \left( \frac{r}{1} \right) ^{-1} \right) \\ > = & \hat{j}\left( \frac{a}{1} \right) \cdot \hat{j}\left( \frac{r}{1} \right) ^{-1} \\ > = & ( \hat{j} \circ i(a) ) (\hat{j} \circ i(r)) ^{-1} \\ > = & j(a) j(r)^{-1}. > \end{align}$ > It just has to be shown that $\hat{j}$ exists — i.e., that it is well-defined and is a [[ring homomorphism]]. For well-definition, assume $(a,r) \sim (b,s)$, then $as=br$ in $R$, hence $j(a)j(s)=j(b)j(r)$ in $L$, and $j(a)j(r)^{-1}=j(b)j(s)^{-1}$ > which shows $\hat{j}$ is well-defined. That it is a [[ring homomorphism]] is straightforward to see. ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```