The Field of Fractions and Localisation: Examples

Facts [F4] and [F5] give the commutative-ring and no-zero-divisor clauses for $\mathbb{Z}$, while [F6] gives $1\mathrm{\ne }0$; hence [F7] makes $\mathbb{Z}$ an integral domain. The map $a\mapsto a\mathrm{/}1$ from $\mathbb{Z}$ to $\mathbb{Q}$ is injective by the defining equivalence relation in [F1]. Since $\mathbb{Q}$ is a field by [F2], [F3] extends it uniquely to an injective homomorphism $\mathrm{Frac}(\mathbb{Z})\to \mathbb{Q}$ with the displayed formula.

Every rational is $a\mathrm{/}b$ with $b\mathrm{\ne }0$ by [F1], so the homomorphism is surjective and hence an isomorphism. It fixes each integer, which makes it canonical over $\mathbb{Z}$.

The inclusion $\mathbb{Z}\hookrightarrow \mathbb{Q}$ sends $6$ to a unit, so [F4] gives a map ${\mathbb{Z}}_6\to \mathbb{Q}$ with $a\mathrm{/}{6}^n\mapsto a\mathrm{/}{6}^n$. If its image is zero, multiplication by the nonzero element $6^n$ in the field $\mathbb{Q}$ gives $a=0$, and then [F5] makes $a\mathrm{/}{6}^n=0$; hence the map is injective. Its image is exactly the set displayed in [F1].

The elements $2$ and $3$ are units because $2(3\mathrm{/}6)=1$ and $3(2\mathrm{/}6)=1$.

If a positive prime $q$ is a unit, then $q(a\mathrm{/}{6}^n)=1$ for some $a,n$, so $qa=6^n$. The case $n=0$ is impossible because $q>1$. For $n>0$, [F3] applied repeatedly to $6^n=2^n{3}^n$ gives $q\mid 2$ or $q\mid 3$, and primality with positivity forces $q=2$ or $q=3$.

Fact [F1] makes $(p)$ a prime ideal, and [F7] says that the denominators in ${\mathbb{Z}}_{(p)}$ are exactly the integers outside $(p)$, namely those not divisible by $p$. Such integers are nonzero and hence units in $\mathbb{Q}$, so [F5] gives a map ${\mathbb{Z}}_{(p)}\to \mathbb{Q}$ sending $a\mathrm{/}b$ to the same fraction. If its image is zero, multiplication by the nonzero $b$ in $\mathbb{Q}$ gives $a=0$, and [F6] makes $a\mathrm{/}b=0$ in the localisation; thus the map is injective. Its image is exactly the displayed set.

By [F2], the ring is local with maximal ideal $(p){\mathbb{Z}}_{(p)}=p{\mathbb{Z}}_{(p)}$. By [F3], its residue field is $\mathrm{Frac}(\mathbb{Z}\mathrm{/}p\mathbb{Z})$. Since the latter base ring is already a field by [F1], every fraction satisfies $a\mathrm{/}b=(a{b}^{-1})\mathrm{/}1$, so its fraction field is canonically itself.

By [F1] and [F2], evaluation at $0$ has kernel $(x)$ and is onto. Explicitly, $f+(x)\mapsto f(0)$ is well defined and has inverse $c\mapsto c+(x)$, so $F[x]\mathrm{/}(x)\cong F$. Thus [F3] makes $(x)$ maximal, and [F6] makes it prime.

By [F9], the denominators are the elements outside $(x)$, which are exactly the polynomials $g$ with $g(0)\mathrm{\ne }0$ by [F2]. They are nonzero and hence units in $F(x)$ by [F5], so [F7] gives a map from the localisation to $F(x)$ sending $f\mathrm{/}g$ to the same fraction. If its image is zero, then $f=0$ because [F5] embeds the domain $F[x]$ in its fraction field, and [F8] makes $f\mathrm{/}g=0$ in the localisation. The map is therefore injective, and its image is precisely the displayed fractions, which are exactly those admitting a representative with denominator nonzero at $0$.

By [F4], the maximal ideal is $(x)F[x{]}_{(x)}=xF[x{]}_{(x)}$, and the residue field is $\mathrm{Frac}(F[x]\mathrm{/}(x))\cong \mathrm{Frac}(F)$. Every nonzero denominator in the field $F$ is already invertible, so $a\mathrm{/}b=(a{b}^{-1})\mathrm{/}1$ and the last fraction field is canonically $F$.

Let $R$ be any nonzero commutative ring and let $S$ be multiplicative with $0\in S$, for example $S=R$. For every $r\mathrm{/}s\in S^{-1}R$, choose $u=0$. Then $ur=0$, so [F1] gives $r\mathrm{/}s=0$.

Thus $S^{-1}R$ has one element and is the zero ring, refuting the statement even though $R$ itself is nonzero.

In $R=\mathbb{Z}\mathrm{/}6\mathbb{Z}$, the classes of $2$ and $3$ are nonzero, but their product is the class of $6$, hence zero.

Therefore $R\setminus \{0\}$ is not closed under multiplication and is not multiplicative by [F1]. This shows why the domain hypothesis in [F3] cannot be removed.

In $R=\mathbb{Z}\mathrm{/}6\mathbb{Z}$, let $S=\{1,3\}$, the multiplicative set generated by the class of $3$ because $3^2=3$. The class of $2$ is nonzero, while $3\cdot 2=0$.

Since the denominator $3$ annihilates $2$, [F1] gives $2\mathrm{/}1=0$ in $S^{-1}R$. Thus the localisation map kills a nonzero element and is not injective.

In $R=\mathbb{Z}\mathrm{/}6\mathbb{Z}$, the class $0$ is not regular because it annihilates $1$; the classes $2,3,4$ are nonzero zero divisors; and $1,5$ are units and hence regular. Thus the regular elements are exactly $S=\{1,5\}=R^{\times }$.

Since the identity map of $R$ already sends every element of $S$ to a unit, [F1] gives an inverse to the localisation map, so $Q(R)=S^{-1}R\cong R$.

The nonzero class of $2$ is not a unit because every product $2a$ is even modulo $6$ and cannot equal $1$. Hence $R$, and therefore $Q(R)$, is not a field by [F3].