§1P. Fields, Integral Domains, Fields of Fractions

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 24 February 2011

Fields, integral domains, fields of fractions

Let $F$ be a commutative ring. Then $F$ is a filed if and only if the only ideals of $F$ are $\left(0\right)$ and $F$.

		Proof.
	$\Rightarrow$ Assume $F$ is a field. To show: The only ideals of $F$ are $\left(0\right)$ and $F$. Let $I$ be an ideal of $F$. Suppose $I\ne \left(0\right)$. Then there is an element $x\in I$, $x\ne 0$. Since $F$ is a field, there is an element $x^{-1}\in F$ such that $x{x}^{-1}=1.$ So $1=x{x}^{-1}\in I.$ So, if $y\in F$, then $y=y\cdot 1\in \in I$. So $F\subseteq I\subseteq F.$ So $I=F$. So the only ideals of $F$ are $\left(0\right)$ and $F$. $\Leftarrow$ Assume that the only ideals of $F$ are $\left(0\right)$ and $F$. To show: $F$ is a field. Let $x\in F$, $x\ne 0$. Then, since $\left(x\right)\ne 0,$ we must have $\left(x\right)=F$ So there is is some $y\in F$ such that $xy=1.$ So $F$ is a field. $\square$

Let $R$ be a commutative ring and let $M$ be an ideal of $R$. Then $R/M$ is a field if and only if $M$ is a maximal ideal.

		Proof.
	$\Rightarrow$ Assume $R/M$ is a field To show: $M$ is a maximal ideal. By Lemma 1.1, the only ideals of $R/M$ are $\left(0\right)$ and $R/M$. The correspondence theorem, Ex 2.1.5c), says that there is a one-to-one correspondence between ideals of $R/M$ and ideals of $R$ containing $M$. Thus the only ideals of $R$ containing $M$ are $M$ and $R$. So $M$ is a maximal ideal. $\Leftarrow$ Assume $M$ is a maximal ideal. To show: $R/M$ is a field. The only ideals of $R$ containing $M$ are $M$ and $R$. The correspondence theorem, Ex 2.1.5c), says that there is a one-to-one correspondence between ideals of $R/M$ and ideals of $R$ containing $M$. Thus the only ideals of $R/M$ are $\left(0\right)$ and $R/M$. So, by Lemma 1.1, $R/M$ is a field. $\square$

(Cancellation Law) Let $R$ be an integral domain. If $a,b,c\in R$, $c\ne 0$, and $ac=bc$, then $a=b$.

		Proof.
	Assume $a,b,c\in R$, $c\ne 0$, and $ac=bc$. Then $0=ac-bc=\left(a-b\right)c.$ Since $R$ is an integral domain and $c\ne 0,$ $a-b=0.$ So $a=b$. $\square$

Let $R$ be a commutative ring and let $P$ be an ideal of $R$. Then $R/P$ is an integral domain if and only if $P$ is prime ideal.

		Proof.
	$\Rightarrow$ Assume $R/P$ is an integral domain. To show: $P$ is a prime ideal. Let $a,b\in R$, and suppose that $ab\in P$. To show: Either $a\in P$ or $b\in P$. Since $ab\in P$, we have $\left(a+P\right)\left(b+P\right)=ab+P=0+P,$ in $R/P$. Since $R/P$ is an integral domain, either $a+P=0+P$ or $b+P=0+P.$ Thus either $a\in P$ or $b\in P$. So $P$ is a prime ideal. $\Leftarrow$ Assume $P$ is a prime ideal. To show: $R/P$ is an integral domain. Let $a,b\in R$, such that $\left(a+P\right)\left(b+P\right)=0+P.$ To show: Either $a+P=0+P$ or $b+P=0+P.$ Then $ab+P=0+P$. So $ab\in P$. Since $P$ is prime, either $a\in P$ or $b\in P$. So either $a+P=0+P$ or $b+P=0+P.$ So $R/P$ is an integral domain. $\square$

Let $R$ be an integral domain. Let $F_R=\left\{\frac{a}{b}\mid a,b\in R,b\ne 0\right\}$ be its set of fractions. Then equality of fractions is an equivalence relation.

		Proof.
	To show: $a/b=a/b$. If $a/b=c/d$ then $c/d=a/b$. If $a/b=c/d$ and $c/d=e/f$ then $a/b=e/f$. Since $ab=ba,$ $a/b=a/b.$ Assume $a/b=c/d$. Then $ad=bc$. Since $R$ is commutative, $cb=da$. So $c/d=a/b$. Assume $a/b=c/d$ and $c/d=e/f$. Then $ad=bc$ and $cf=de$. To show: $af=be$. Since $ad=bc$ and $cf=de$, and $adcf=bcde$. Thus, by commutativity, $afcd=becd.$ Then, by the cancellation law for an integral domain, Proposition 1.3, $af=be$. So $a/b=e/f$. $\square$

Let $R$ be an integral domain. Let $F_R=\left\{\frac{a}{b}\mid a,b\in R,b\ne 0\right\}$ be its set of fractions. Let equality of fractions be as defined in (1.1). Then the operations $+:F_R\hspace{0.17em}\times \hspace{0.17em}{F}_R\to F_R$ and $\times :F_R\hspace{0.17em}\times \hspace{0.17em}{F}_R\to F_R$ given by $$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\text{and}\frac{a}{b}\cdot \frac{c}{d}=\frac{ac}{bd}$$ are well defined.

		Proof.
	Assume ${\displaystyle \frac{a}{b}}={\displaystyle \frac{a\text{'}}{a\text{'}}}$ and ${\displaystyle \frac{c}{d}}={\displaystyle \frac{c\text{'}}{d\text{'}}}$ To show: ${\displaystyle \frac{a}{b}}+{\displaystyle \frac{c}{d}}={\displaystyle \frac{a\text{'}}{b\text{'}}}+{\displaystyle \frac{c\text{'}}{d\text{'}}.}$ ${\displaystyle \frac{a}{b}}\cdot {\displaystyle \frac{c}{d}}={\displaystyle \frac{a\text{'}}{b\text{'}}}\cdot {\displaystyle \frac{c\text{'}}{d\text{'}}}$ To show: ${\displaystyle \frac{ad+bc}{bd}}={\displaystyle \frac{a\text{'}d\text{'}+b\text{'}c\text{'}}{b\text{'}d\text{'}}}$ To show: $\left(ad+bc\right)b\text{'}d\text{'}=\left(a\text{'}d\text{'}+b\text{'}c\text{'}\right)bd.$ We know that $ab\text{'}=ba\text{'}$ and $cd\text{'}=dc\text{'}.$ So $$\underbrace{adb\text{'}}d\text{'}+b\underbrace{cb\text{'}d\text{'}}=a\text{'}bdd\text{'}+bdb\text{'}c\text{'}=\left(a\text{'}d\text{'}+b\text{'}c\text{'}\right)bd.$$ So ${\displaystyle \frac{a}{b}}+{\displaystyle \frac{c}{d}}={\displaystyle \frac{a\text{'}}{b\text{'}}}+{\displaystyle \frac{c\text{'}}{d\text{'}}.}$ To show ${\displaystyle \frac{ac}{bd}}={\displaystyle \frac{a\text{'}c\text{'}}{b\text{'}d\text{'}}.}$ To show: $acb\text{'}d\text{'}=a\text{'}c\text{'}bd.$ We know that $ab\text{'}=ba\text{'}$ and $cd\text{'}=dc\text{'}.$ So $$acb\text{'}d\text{'}=ba\text{'}cd\text{'}=ba\text{'}dc\text{'}=a\text{'}c\text{'}bd.$$ So ${\displaystyle \frac{ac}{bd}}={\displaystyle \frac{a\text{'}c\text{'}}{b\text{'}d\text{'}}}.$ $\square$

Let $R$ be an integral domain and let Let $R$ be an integral domain. Let $F_R=\left\{\frac{a}{b}\mid a,b\in R,b\ne 0\right\}$ be its set of fractions. Let equality of fractions be as defined in (1.1) and let operations $+:F_R\hspace{0.17em}\times \hspace{0.17em}{F}_R\to F_R$ and $\times :F_R\hspace{0.17em}\times \hspace{0.17em}{F}_R\to F_R$ be as given in (1.2). Then $F_R$ is a field.

		Proof.
	To show: $F_R$ is a ring. $F_R$ is commutative. If $x\in F_R$ and $x\ne 0$ then there exists $x^{-1}\in F_R$ such that $x{x}^{-1}=1$. To show: $+:F_R\hspace{0.17em}\times \hspace{0.17em}{F}_R\to F_R$ is well defined. $\times :F_R\hspace{0.17em}\times \hspace{0.17em}{F}_R\to F_R$ is well defined. If $p/q,m/n,r/s\in F_R$ then $\left(p/q+m/n\right)+r/s=p/q+\left(m/n+r/s\right).$ If $p/q,m/n\in F_R$ then $p/q+m/n=m/n+p/q.$ There is an element $0\in F_R$ such that $0+m/n=m/n$ for all $m/n\in F_R$. If $x\in F_R$ there is an element $-x\in F_R$ such that $x+\left(-x\right)=0.$ If $p/q,m/n,r/s\in F_R$ then $p/q\cdot \left(m/n\cdot r/s\right)=\left(p/q\cdot m/n\right)\cdot r/s.$ There is an element $1\in F_R$ such that $1\cdot x=x$ for all $x\in F_R$. If $m/n,p/q,r/s\in F_R$ then $m/n\left(p/q+r/s\right)=m/n\cdot p/q+m/n\cdot r/s$ and $\left(p/q+r/s\right)m/n=p/q\cdot m/n+r/s\cdot m/n.$ and ab. have been proved in Proposition 1.6. Assume $p/q,m/n,r/s\in F_R.$ To show: $\left(p/q+m/n\right)+r/s=p/q+\left(m/n+r/s\right).$ By the definition of the operation $+:F_R\hspace{0.17em}\times \hspace{0.17em}{F}_R\to F_R,$ $$\begin{array}{rcl}\left(\frac{p}{q}+\frac{m}{n}\right)+\frac{r}{s}& =& \frac{pn+mq}{qn}+\frac{r}{s}\\ & =& \frac{\left(pn+mq\right)s+qnr}{qns}\\ & =& \frac{pns+mqs+qnr}{qns}.\end{array}$$ By the definition of the operation $+:F_R\hspace{0.17em}\times \hspace{0.17em}{F}_R\to F_R,$ $$\begin{array}{rcl}\frac{p}{q}+\left(\frac{m}{n}+\frac{r}{s}\right)& =& \frac{p}{q}+\left(\frac{ms+nr}{ns}\right)\\ & =& \frac{pns+q\left(ms+nr\right)}{qns}\\ & =& \frac{pns+qms+qnr}{qns}.\end{array}$$ Since $R$ is commutative ($R$ is an integral domain), $$\frac{pns+mqs+qnr}{qns}=\frac{pns+qms+qnr}{qns}.$$ So $$\left(\frac{p}{q}+\frac{m}{n}\right)+\frac{r}{s}=\frac{p}{q}+\left(\frac{m}{n}+\frac{r}{s}\right).$$ Assume $p/q,m/n\in F_R$. To show: $pq+m/n=m/n+p/q.$ By the definition of $+:F_R\hspace{0.17em}\times \hspace{0.17em}{F}_R\to F_R,$ $$\frac{p}{q}+\frac{m}{n}=\frac{pn+qm}{qn}.$$ By the definition of $+:F_R\hspace{0.17em}\times \hspace{0.17em}{F}_R\to F_R,$ $$\frac{m}{n}+\frac{p}{q}=\frac{mq+np}{nq}.$$ Since $R$ is commutative, $$\frac{pn+qm}{qn}=\frac{mq+np}{nq}.$$ So $$\frac{p}{q}+\frac{m}{n}=\frac{m}{n}+\frac{p}{q}.$$ To show: There is an element $0\in F_R$ such that $0+m/n=m/n$ for all $m/n\in F_R.$ Assume $m/n\in F_R.$ Then $$\frac{0}{1}+\frac{m}{n}=\frac{0\cdot n+m}{1\cdot n}=\frac{0+m}{n}=\frac{m}{n}.$$ So $0\in F_R$ such that $0+m/n=m/n$ for all $m/n\in F_R.$ So $0/1$ is an identity for $+:F_R\hspace{0.17em}\times \hspace{0.17em}{F}_R\to F_R.$ Assume $m/n\in F_R.$ Then $$\frac{m}{n}+\frac{\left(-m\right)}{n}=\frac{mn+\left(-nn\right)}{n^2}=\frac{0}{n^2.}$$ To show: $0/n^2=0/1.$ Since $0=0\cdot 1=0\cdot n^2=0,$ $0/n^2=0/1.$ So $$\frac{m}{n}+\frac{\left(-m\right)}{n}=\frac{0}{1}.$$ Assume $p/q,m/n,r/s\in F_R.$ To show: $p/q\cdot \left(m/n\cdot r/s\right)=\left(p/q\cdot m/n\right)\cdot r/s=pmr/qns.$ By the definition of the operation $\times :F_R\hspace{0.17em}\times \hspace{0.17em}{F}_R\to F_R,$ $$\frac{p}{q}\left(\frac{m}{n}\cdot \frac{r}{s}\right)=\frac{p}{q}\cdot \left(\frac{mr}{ns}\right)=\frac{pmr}{qns}.$$ By the definition of the operation $\times :F_R\hspace{0.17em}\times \hspace{0.17em}{F}_R\to F_R,$ $$\left(\frac{p}{q}\cdot \frac{m}{n}\right)\cdot \frac{r}{s}=\left(\frac{pm}{qn}\right)\cdot \frac{r}{s}=\frac{pmr}{qns}.$$ So $$\frac{p}{q}\cdot \left(\frac{m}{n}\cdot \frac{r}{s}\right)=\left(\frac{p}{q}\cdot \frac{m}{n}\right)\cdot \frac{r}{s}.$$ To show: There is an element $1\in F_R$ such that $1\cdot m/n=m/n$ for all $m/n\in F_R.$ Let $1=1/1\in F_R.$ To show: If $m/n\in F_R$ then $1/1\cdot m/n=m/n.$ Assume: $m/n\in F_R.$ Then $$\frac{1}{1}\cdot \frac{m}{n}=\frac{1\cdot m}{1\cdot n}=\frac{m}{n}.$$ So $1/1$ is an identity element for $\times :F_R\hspace{0.17em}\times \hspace{0.17em}{F}_R\to F_R.$ Assume $m/n,p/q,r/s\in F_R.$ To show: $m/n\left(p/q+r/s\right)=m/n\cdot p/q+m/n\cdot r/s.$ $\left(p/q+r/s\right)m/n=p/q\cdot m/n+r/s\cdot m/n.$ By the definition of the operations $$\begin{array}{rcl}\frac{m}{n}\cdot \left(\frac{p}{q}+\frac{r}{s}\right)& =& \frac{m}{n}\cdot \frac{ps+qr}{qs}\\ & =& \frac{m\left(ps+qr\right)}{nqs}\\ & =& \frac{mps+mqr}{nqs}\end{array}$$ and $$\begin{array}{rcl}\frac{m}{n}\cdot \frac{p}{q}+\frac{m}{n}\cdot \frac{r}{s}& =& \frac{mp}{nq}+\frac{mr}{ns}\\ & =& \frac{mpns+nqmr}{nqns}.\end{array}$$ To show: ${\displaystyle \frac{mps+mqr}{nqs}={\displaystyle \frac{mpns+nqmr}{nqns}.}}$ To show: $\left(mps+mqr\right)nqns=nqsn\left(mpns+nqmr\right).$ By commutativity of $R$ and the distributative property in $R$, $$\begin{array}{rcl}\left(mps+mqr\right)nqns& =& nqsn\left(mps+mqr\right)\\ & =& nqs\left(mpns+nqmr\right).\end{array}$$ So $$\frac{mps+mqr}{nqs}=\frac{mpns+nqmr}{nqns}.$$ So $$\frac{m}{n}\cdot \left(\frac{p}{q}+\frac{r}{s}\right)=\frac{m}{n}\cdot \frac{p}{q}+\frac{m}{n}\cdot \frac{r}{s}.$$ By the definition of the operations $$\begin{array}{rcl}\left(\frac{p}{q}+\frac{r}{s}\right)\cdot \frac{m}{n}& =& \frac{ps+qr}{qs}\cdot \frac{m}{n}\\ & =& \frac{\left(ps+qr\right)m}{qsn}\\ & =& \frac{psm+qrm}{qsn}\end{array}$$ and $$\begin{array}{rcl}\frac{p}{q}\cdot \frac{m}{n}+\frac{r}{s}\cdot \frac{m}{n}& =& \frac{pm}{qn}+\frac{rm}{sn}\\ & =& \frac{pmsn+qnrm}{qnsn}.\end{array}$$ To show: ${\displaystyle \frac{psm+qrm}{qsn}={\displaystyle \frac{pmsn+qnrm}{qnsn}.}}$ To show $\left(psm+qrm\right)qnsn=qsn\left(pmsn+qnrm\right).$ By commutativity of $R$ and the distributive property of $R$, $$\begin{array}{rcl}\left(psm+qrm\right)qnsn& =& qsnn\left(psm+qrm\right)\\ & =& qsn\left(pmsn+qnrm\right).\end{array}$$ So $$\frac{psm+qrm}{qsn}=\frac{pmsn+qnrm}{qnsn}.$$ So $$\left(\frac{p}{q}+\frac{r}{s}\right)\cdot \frac{m}{n}=\frac{p}{q}\cdot \frac{m}{n}+\frac{r}{s}\cdot \frac{m}{n}.$$ To show: $F_R$ is commutative. To show: If $m/n,p/q\in F_R$ then $m/n\cdot p/q=p/q\cdot m/n.$ Assume $m/n,p/q\in F_R.$ by the definition of $\times :F_R\hspace{0.17em}\times \hspace{0.17em}{F}_R\to F_R,$ $$\frac{m}{n}\cdot \frac{p}{q}=\frac{mp}{nq}\text{and}\frac{p}{q}\cdot \frac{m}{n}=\frac{pm}{pq}.$$ By commutativity in $R$, $$\frac{mp}{nq}=\frac{pm}{qn}.$$ So $$\frac{m}{n}\cdot \frac{p}{q}=\frac{p}{q}\cdot \frac{m}{n}.$$ So $F_R$ is commutative. To show: If $x\in F_R$ and $x\ne 0$ then there exists $x^{-1}\in F_R$ such that $x{x}^{-1}=1.$ Assume $x=m/n\in F_R$ and $m/n\ne 0/1$. Then, by equality of fractions, $m\cdot 1\ne 0\cdot n.$ So $m\ne 0$. Let $x^{-1}=n/n.$ Note: $n/m\in F_R$ since $m\ne 0$. To show: $m/n\cdot n/m=1/1.$ By the definition of $\times :F_R\hspace{0.17em}\times \hspace{0.17em}{F}_R\to F_R,$ $$\frac{m}{n}\cdot \frac{n}{m}=\frac{mn}{nm}.$$ To show: ${\displaystyle \frac{mn}{nm}}={\displaystyle \frac{1}{1}}.$ But $mn=nm$, by commutativity in $R$. So, by the definition of equality in fractions, $$\frac{mn}{nm}=\frac{1}{1}.$$ So, if $x=m/n$ and $m/n\ne 0$, then $x^{-1}=n/n\in F_R$ and $x{x}^{-1}={\displaystyle \frac{m}{n}}\cdot {\displaystyle \frac{n}{m}}={\displaystyle \frac{1}{1}}.$ So $F_R$ is a field. $\square$

Let $R$ be an integral domain with identity $1$ and $F_R$ be its field of fractions. Then the map $\varphi :R\to F_R$ given by $$\begin{array}{cccc}\varphi :& R& \to & F_R\\ & r& \mapsto & \frac{r}{1}\end{array}$$ is an injective ring homomorphism.

		Proof.
	To show: $\varphi$ is a ring homomorphism. $\varphi$ is injective. To show: If $r,s\in R$ then $\varphi \left(r+s\right)=\varphi \left(r\right)+\varphi \left(s\right).$ If $r,s\in R$ then $\varphi \left(rs\right)=\varphi \left(r\right)\varphi \left(s\right).$ $\varphi \left(1\right)={\displaystyle \frac{1}{1}}.$ Assume $r,s\in R$. Then $$\varphi \left(r+s\right)=\frac{r+s}{1}\text{and}\varphi \left(r\right)+\varphi \left(s\right)=\frac{r}{1}+\frac{s}{1}.$$ By the definition of $+:F_R\hspace{0.17em}\times \hspace{0.17em}{F}_R\to F_R,$ $$\frac{r}{1}+\frac{s}{1}=\frac{r\cdot 1+1\cdot s}{1\cdot 1}=\frac{r+s}{1}.$$ So $$\varphi \left(r+s\right)=\varphi \left(r\right)+\varphi \left(s\right).$$ Assume $r,s\in R$. Then, by the definition of $\times :F_R\hspace{0.17em}\times \hspace{0.17em}{F}_R\to F_R,$ $$\varphi \left(rs\right)=\frac{rs}{1}=\frac{r}{1}\cdot \frac{s}{1}=\varphi \left(r\right)\varphi \left(s\right).$$ By the definition of $\varphi :R\to F_R,$ $$\varphi \left(1\right)=\frac{1}{1}.$$ So $\varphi$ is a ring homomorphism. To show: If $r,s\in R$ and $\varphi \left(r\right)=\varphi \left(s\right)$ then $r=s$. Assume $r,s\in R$ and $\varphi \left(r\right)=\varphi \left(s\right).$ Then $r/1=s/1.$ Thus, by the definition of equality of fractions, $1\cdot r=1\cdot s.$ So $r=s$. So $\varphi$ is injective. So $\varphi$ is an injective ring homomorphism. $\square$

References

[CM] H. S. M. Coxeter and W. O. J. Moser, Generators and relations for discrete groups, Fourth edition. Ergebnisse der Mathematik und ihrer Grenzgebiete [Results in Mathematics and Related Areas], 14. Springer-Verlag, Berlin-New York, 1980. MR0562913 (81a:20001)

[GW1] F. Goodman and H. Wenzl, The Temperly-Lieb algebra at roots of unity, Pacific J. Math. 161 (1993), 307-334. MR1242201 (95c:16020)

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