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

## 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. $⇒$ 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 $F$. $⇐$ 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. $⇒$ 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. $⇐$ 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. $⇒$ 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. $⇐$ 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 $+:{F}_{R} × {F}_{R}\to {F}_{R}$ and $×:{F}_{R} × {F}_{R}\to {F}_{R}$ given by $a b + c d = ad+bc bd and a b ⋅ c d = ac bd$ are well defined. Proof. Assume $\frac{a}{b}=\frac{a\text{'}}{a\text{'}}$ and $\frac{c}{d}=\frac{c\text{'}}{d\text{'}}$ To show: $\frac{a}{b}+\frac{c}{d}=\frac{a\text{'}}{b\text{'}}+\frac{c\text{'}}{d\text{'}}.$ $\frac{a}{b}\cdot \frac{c}{d}=\frac{a\text{'}}{b\text{'}}\cdot \frac{c\text{'}}{d\text{'}}$ To show: $\frac{ad+bc}{bd}=\frac{a\text{'}d\text{'}+b\text{'}c\text{'}}{b\text{'}d\text{'}}$ 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 $adb' ⏟ d'+ b cb'd' ⏟ = a'bdd' +bdb'c' = a'd'+ b'c' bd.$ So $\frac{a}{b}+\frac{c}{d}=\frac{a\text{'}}{b\text{'}}+\frac{c\text{'}}{d\text{'}}.$ To show $\frac{ac}{bd}=\frac{a\text{'}c\text{'}}{b\text{'}d\text{'}}.$ To show: $acb\text{'}d\text{'}=a\text{'}c\text{'}bd.$ We know that $ab\text{'}=ba\text{'}$ and $cd\text{'}=dc\text{'}.$ So $acb'd'= ba'cd'= ba'dc'= a'c'bd.$ So $\frac{ac}{bd}=\frac{a\text{'}c\text{'}}{b\text{'}d\text{'}}.$ $\square$

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 $+:{F}_{R} × {F}_{R}\to {F}_{R}$ and $×:{F}_{R} × {F}_{R}\to {F}_{R}$ 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: $+:{F}_{R} × {F}_{R}\to {F}_{R}$ is well defined. $×:{F}_{R} × {F}_{R}\to {F}_{R}$ 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 $+:{F}_{R} × {F}_{R}\to {F}_{R},$ $p q + m n + r s = pn+mq qn + r s = pn+mqs +qnr qns = pns+mqs+qnr qns .$ By the definition of the operation $+:{F}_{R} × {F}_{R}\to {F}_{R},$ $p q + m n + r s = p q + ms+nr ns = pns+ qms+nr qns = pns+qms+qnr qns .$ Since $R$ is commutative ($R$ is an integral domain), $pns+mqs+qnr qns = pns+qms+qnr qns .$ So $p q + m n + r s = p q + m n + r s .$ Assume $p/q,m/n\in {F}_{R}$. To show: $pq+m/n=m/n+p/q.$ By the definition of $+:{F}_{R} × {F}_{R}\to {F}_{R},$ $p q + m n = pn+qm qn .$ By the definition of $+:{F}_{R} × {F}_{R}\to {F}_{R},$ $m n + p q = mq+np nq .$ Since $R$ is commutative, $pn+qm qn = mq+np nq .$ So $p q + m n = m n + p q .$ 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 $0 1 + m n = 0⋅n+m 1⋅n = 0+m n = m n .$ 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 $+:{F}_{R} × {F}_{R}\to {F}_{R}.$ Assume $m/n\in {F}_{R}.$ Then $m n + -m n = mn+-nn n2 = 0 n2.$ To show: $0/{n}^{2}=0/1.$ Since $0=0\cdot 1=0\cdot {n}^{2}=0,$ $0/{n}^{2}=0/1.$ So $m n + -m n = 0 1 .$ 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 $×:{F}_{R} × {F}_{R}\to {F}_{R},$ $p q m n ⋅ r s = p q ⋅ mr ns = pmr qns .$ By the definition of the operation $×:{F}_{R} × {F}_{R}\to {F}_{R},$ $p q ⋅ m n ⋅ r s = pm qn ⋅ r s = pmr qns .$ So $p q ⋅ m n ⋅ r s = p q ⋅ m n ⋅ r s .$ 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 $1 1 ⋅ m n = 1⋅m 1⋅n = m n .$ So $1/1$ is an identity element for $×:{F}_{R} × {F}_{R}\to {F}_{R}.$ 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 $m n ⋅ p q + r s = m n ⋅ ps+qr qs = mps+qr nqs = mps+mqr nqs$ and $m n ⋅ p q + m n ⋅ r s = mp nq + mr ns = mpns+nqmr nqns .$ To show: $\frac{mps+mqr}{nqs}=\frac{mpns+nqmr}{nqns}.$ To show: $\left(mps+mqr\right)nqns=nqsn\left(mpns+nqmr\right).$ By commutativity of $R$ and the distributative property in $R$, $mps+mqr nqns = nqsn mps+mqr = nqs mpns+nqmr .$ So $mps+mqr nqs = mpns+nqmr nqns .$ So $m n ⋅ p q + r s = m n ⋅ p q + m n ⋅ r s .$ By the definition of the operations $p q + r s ⋅ m n = ps+qr qs ⋅ m n = ps+qrm qsn = psm+qrm qsn$ and $p q ⋅ m n + r s ⋅ m n = pm qn + rm sn = pmsn+qnrm qnsn .$ To show: $\frac{psm+qrm}{qsn}=\frac{pmsn+qnrm}{qnsn}.$ To show $\left(psm+qrm\right)qnsn=qsn\left(pmsn+qnrm\right).$ By commutativity of $R$ and the distributive property of $R$, $psm+qrm qnsn = qsnn psm+qrm = qsn pmsn+qnrm .$ So $psm+qrm qsn = pmsn+qnrm qnsn .$ So $p q + r s ⋅ m n = p q ⋅ m n + r s ⋅ m n .$ 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 $×:{F}_{R} × {F}_{R}\to {F}_{R},$ $m n ⋅ p q = mp nq and p q ⋅ m n = pm pq .$ By commutativity in $R$, $mp nq = pm qn .$ So $m n ⋅ p q = p q ⋅ m n .$ 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 $×:{F}_{R} × {F}_{R}\to {F}_{R},$ $m n ⋅ n m = mn nm .$ To show: $\frac{mn}{nm}=\frac{1}{1}.$ But $mn=nm$, by commutativity in $R$. So, by the definition of equality in fractions, $mn nm = 1 1 .$ So, if $x=m/n$ and $m/n\ne 0$, then ${x}^{-1}=n/n\in {F}_{R}$ and $x{x}^{-1}=\frac{m}{n}\cdot \frac{n}{m}=\frac{1}{1}.$ 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 $ϕ: R → FR r ↦ r 1$ 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 $\varphi \left(rs\right)=\varphi \left(r\right)\varphi \left(s\right).$ $\varphi \left(1\right)=\frac{1}{1}.$ Assume $r,s\in R$. Then $ϕr+s= r+s 1 and ϕr+ ϕs= r 1 + s 1 .$ By the definition of $+:{F}_{R} × {F}_{R}\to {F}_{R},$ $r 1 + s 1 = r⋅1+1⋅s 1⋅1 = r+s 1 .$ So $ϕr+s= ϕr+ ϕs.$ Assume $r,s\in R$. Then, by the definition of $×:{F}_{R} × {F}_{R}\to {F}_{R},$ $ϕrs= rs 1 = r 1 ⋅ s 1 = ϕr ϕs.$ By the definition of $\varphi :R\to {F}_{R},$ $ϕ1= 1 1 .$ 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

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