The Field of Fractions and Localisation

Reflexivity has witness $1$, and symmetry uses the same witness with the negative equation. If $u(r{s}^{\mathrm{\prime }}-r^{\mathrm{\prime }}s)=0$ and $v(r^{\mathrm{\prime }}{s}^{\mathrm{\prime }\mathrm{\prime }}-r^{\mathrm{\prime }\mathrm{\prime }}{s}^{\mathrm{\prime }})=0$, then $uv{s}^{\mathrm{\prime }}\in S$ and the identity $s^{\mathrm{\prime }}(r{s}^{\mathrm{\prime }\mathrm{\prime }}-r^{\mathrm{\prime }\mathrm{\prime }}s)=s^{\mathrm{\prime }\mathrm{\prime }}(r{s}^{\mathrm{\prime }}-r^{\mathrm{\prime }}s)+s(r^{\mathrm{\prime }}{s}^{\mathrm{\prime }\mathrm{\prime }}-r^{\mathrm{\prime }\mathrm{\prime }}{s}^{\mathrm{\prime }})$ shows $uv{s}^{\mathrm{\prime }}(r{s}^{\mathrm{\prime }\mathrm{\prime }}-r^{\mathrm{\prime }\mathrm{\prime }}s)=0$; hence the relation is transitive.

Suppose $(r,s)\sim (\rho ,\sigma )$ with witness $u$. For fixed $(a,t)$, multiplying the relation by $t^2$ shows $(rt+as,st)\sim (\rho t+a\sigma ,\sigma t)$, and multiplying it by $at$ shows $(ra,st)\sim (\rho a,\sigma t)$. Changing the second representative in the same way proves both operations are well defined.

Associativity, commutativity, and distributivity follow by expanding the displayed fraction formulas over common denominators. The classes $0\mathrm{/}1$ and $1\mathrm{/}1$ satisfy the zero and identity laws, and additive inverses are $(-r)\mathrm{/}s$.

The formulas give ${\lambda }_S(r+r^{\mathrm{\prime }})={\lambda }_S(r)+{\lambda }_S(r^{\mathrm{\prime }})$, ${\lambda }_S(r{r}^{\mathrm{\prime }})={\lambda }_S(r){\lambda }_S(r^{\mathrm{\prime }})$, and ${\lambda }_S(1)=1\mathrm{/}1$. Finally $(s\mathrm{/}1)(1\mathrm{/}s)=s\mathrm{/}s=1\mathrm{/}1$, since $(s,s)\sim (1,1)$ with witness $1$.

The equality criterion is the definition of equality of equivalence classes. Taking $(r^{\mathrm{\prime }},s^{\mathrm{\prime }})=(0,1)$ gives $r\mathrm{/}s=0\mathrm{/}1$ exactly when $u(r\cdot 1-0\cdot s)=ur=0$ for some $u\in S$.

If $0\in S$, step 1.1 makes every fraction zero by using $u=0$. Conversely, if $S^{-1}R$ is the zero ring, then $1\mathrm{/}1=0$, so step 1.1 gives $u\in S$ with $u\cdot 1=0$, whence $u=0$ and $0\in S$.

Since ${\lambda }_S(r)=r\mathrm{/}1$, step 1.1 gives the displayed kernel. Thus ${\lambda }_S$ is injective exactly when $ur=0$ with $u\in S$ always forces $r=0$, which says precisely that every member of $S$ has trivial annihilator.

Assume $R$ is nonzero. If every member of $S$ has trivial annihilator, then $0\mathrm{\notin }S$, because $0\cdot 1=0$ and $1\mathrm{\ne }0$, and no $u\in S$ is a zero divisor by [F2]. Conversely, if $0\mathrm{\notin }S$ and $S$ contains no zero divisor, then $u\in S$ is nonzero and $ur=0$ forces $r=0$ by [F2].

Define $\tilde{f}(r\mathrm{/}s)=f(r)f(s{)}^{-1}$. If $r\mathrm{/}s=r^{\mathrm{\prime }}\mathrm{/}{s}^{\mathrm{\prime }}$, [F1] gives $u(r{s}^{\mathrm{\prime }}-r^{\mathrm{\prime }}s)=0$ for some $u\in S$. Applying $f$ and cancelling the unit $f(u)$ yields $f(r)f(s^{\mathrm{\prime }})=f(r^{\mathrm{\prime }})f(s)$; multiplying by $f(s{)}^{-1}f(s^{\mathrm{\prime }}{)}^{-1}$ proves the definition is independent of representatives.

Using [F3] and the homomorphism laws for $f$, direct calculation shows that $\tilde{f}$ preserves addition, multiplication, zero, and one. Also $\tilde{f}(r\mathrm{/}1)=f(r)f(1{)}^{-1}=f(r)$, so $\tilde{f}\circ {\lambda }_S=f$.

If $g:S^{-1}R\to A$ is another such homomorphism, then $g(r\mathrm{/}1)=f(r)$ and $g(s\mathrm{/}1)=f(s)$. Since $(s\mathrm{/}1)(1\mathrm{/}s)=1$, uniqueness of inverses gives $g(1\mathrm{/}s)=f(s{)}^{-1}$; hence $g(r\mathrm{/}s)=g(r\mathrm{/}1)g(1\mathrm{/}s)=\tilde{f}(r\mathrm{/}s)$ for every fraction.

If $0\in S$, the hypothesis says that $f(0)=0$ is a unit of $A$, so $A$ is the zero ring. The construction and uniqueness above still apply, with the unique map between zero rings.

Apply the universal property of $L$ to ${\lambda }^{\mathrm{\prime }}$ and that of $L^{\mathrm{\prime }}$ to $\lambda$. This gives unique homomorphisms $\mathrm{\Phi }:L\to L^{\mathrm{\prime }}$ and $\mathrm{\Psi }:L^{\mathrm{\prime }}\to L$ with $\mathrm{\Phi }\lambda ={\lambda }^{\mathrm{\prime }}$ and $\mathrm{\Psi }{\lambda }^{\mathrm{\prime }}=\lambda$.

Both $\mathrm{\Psi }\mathrm{\Phi }$ and ${\mathrm{id}}_L$ compose with $\lambda$ to give $\lambda$. Uniqueness for $L$ gives $\mathrm{\Psi }\mathrm{\Phi }={\mathrm{id}}_L$; similarly $\mathrm{\Phi }\mathrm{\Psi }={\mathrm{id}}_{L^{\mathrm{\prime }}}$. Thus $\mathrm{\Phi }$ is an isomorphism.

Any isomorphism compatible with the maps from $R$ is, in particular, a factorisation of ${\lambda }^{\mathrm{\prime }}$ through $\lambda$, so it equals $\mathrm{\Phi }$ by uniqueness.

If $ar\in S$, then $as\mathrm{/}(ar)$ is a valid fraction and $(r\mathrm{/}s)(as\mathrm{/}(ar))=ars\mathrm{/}(sar)=1$ by [F1]. Hence $r\mathrm{/}s$ is a unit.

Conversely, suppose $(r\mathrm{/}s)(b\mathrm{/}t)=1$. Then $rb\mathrm{/}(st)=1$, so [F1] supplies $u\in S$ with $u(rb-st)=0$. Therefore $(ub)r=ust\in S$, because $u,s,t\in S$. Taking $a=ub$ proves the criterion.

A map $h:R\to A$ extends to ${\bar{T}}^{-1}(S^{-1}R)$ exactly when it sends $S$ to units and, after the first extension, sends every $t\mathrm{/}1$ with $t\in T$ to a unit.

The image of $t\mathrm{/}1$ is $h(t)$, so the condition in step 1.1 is exactly that $h$ send every member of $S\cup T$ to a unit, equivalently every element of the generated set $U$ to a unit.

Hence ${\bar{T}}^{-1}(S^{-1}R)$ and $U^{-1}R$ have the same universal property over $R$, so [F2] gives the unique displayed $R$-algebra isomorphism.

For $S=\{f^n:n\in \mathbb{N}\}$ and $T=\{g^n:n\in \mathbb{N}\}$, the set $U$ is generated by $f$ and $g$. A map inverts both precisely when it inverts $fg$: if $fg$ is a unit, then $f(g(fg{)}^{-1})=1$ and $g(f(fg{)}^{-1})=1$. Thus [F1] identifies this localisation with $R_{fg}$, including $f=0$ or $g=0$.

Let $\pi \in R$ be irreducible. By [L1], any factorization of $ab$ into irreducibles is obtained by concatenating factorizations of $a$ and $b$, so if $\pi \mid ab$ then $\pi$ is associate to one factor from $a$ or one factor from $b$. Therefore every irreducible element of $R$ is prime.

Let $f={\sum }_i{a}_i{x}^i$ and $g={\sum }_j{b}_j{x}^j$ be primitive. Suppose some irreducible $\pi \in R$ divides every coefficient of $fg$. Choose the least indices $i$ and $j$ with $\pi \nmid a_i$ and $\pi \nmid b_j$. Then the coefficient of $x^{i+j}$ in $fg$ is $a_i{b}_j+{\sum }_{u+v=i+j,(u,v)\mathrm{\ne }(i,j)}{a}_u{b}_v\mathrm{.}$ Every summand in the sum is divisible by $\pi$, while $a_i{b}_j$ is not by step 1.1. This contradicts the choice of $\pi$. So no irreducible divides all coefficients of $fg$, and hence $fg$ is primitive.

If $p$ is reducible in $R[x]$, then it is reducible in $K[x]$ because $R[x]\subseteq K[x]$. Conversely, suppose $p=qr$ in $K[x]$ with both factors of positive degree. Choose nonzero $a,b\in R$ with $aq,br\in R[x]$, and factor out the greatest common divisor of the coefficients to write $aq=c{q}_0,br=d{r}_0,$ where $q_0,r_0\in R[x]$ are primitive. Then $abp=cd{q}_0{r}_0\mathrm{.}$ By step 2.1 the product $q_0{r}_0$ is primitive, so the right-hand side has content associate to $cd$, while the left-hand side has content associate to $ab$ because $p$ is primitive. Thus $ab$ and $cd$ are associates. Absorbing the unit into one factor yields $p=q_1{r}_1$ with $q_1,r_1\in R[x]$ of positive degree, contradicting irreducibility in $R[x]$.

Therefore a primitive positive-degree polynomial is irreducible in $R[x]$ exactly when it is irreducible in $K[x]$.

Every nonzero element of the domain $D$ has trivial annihilator. Hence [F2], applied to $S=D\setminus \{0\}$, makes the canonical homomorphism injective. In particular $1\mathrm{/}1\mathrm{\ne }0\mathrm{/}1$, so the localisation is nonzero.

Let $a\mathrm{/}b$ be nonzero. By the vanishing criterion in [F2], $a\mathrm{\ne }0$, because otherwise $a\mathrm{/}b=0$. Thus $a\in D\setminus \{0\}$, and [F3] gives $(a\mathrm{/}b)(b\mathrm{/}a)=1$.

Every nonzero element is therefore a unit, and [F3] supplies the commutative-ring structure. By [F4], $\mathrm{Frac}(D)$ is a field.

If $b\mathrm{\ne }0$ in $D$, injectivity gives $f(b)\mathrm{\ne }0$, so [F1] makes $f(b)$ a unit. Since the denominators defining $\mathrm{Frac}(D)$ are exactly the nonzero elements, [F2] gives the unique extension and its formula.

If $\tilde{f}(a\mathrm{/}b)=0$, multiply $f(a)f(b{)}^{-1}=0$ by the unit $f(b)$ to obtain $f(a)=0$. Injectivity of $f$ gives $a=0$, hence $a\mathrm{/}b=0$. Thus $\tilde{f}$ is injective.

By [F1], the embeddings of $D$ produce injective homomorphisms $\varphi :\mathrm{Frac}(D)\to K$ and $\psi :\mathrm{Frac}(D)\to L$. Their images contain every quotient of embedded elements of $D$, so the quotient-generation hypothesis makes both maps surjective.

The composite $\psi {\varphi }^{-1}:K\to L$ is therefore a field isomorphism fixing $D$. If $\theta :K\to L$ also fixes $D$, then for every $a\mathrm{/}b\in K$ one has $\theta (a\mathrm{/}b)=a\mathrm{/}b$, so $\theta =\psi {\varphi }^{-1}$.

By [F1] and [F2], $F[t]$ is an integral domain. Applying [F3] gives the displayed set of fractions, its field structure, and the embedding of $F[t]$.

Taking $F=\mathbb{R}$ gives the final assertion.

Assume $R$ is local with maximal ideal $\mathfrak{m}$. No element of $\mathfrak{m}$ is a unit. Conversely, if $x$ is a nonunit, then $(x)$ is proper and [F2] places it in a maximal ideal, necessarily $\mathfrak{m}$. Thus the nonunits are exactly $\mathfrak{m}$, proving condition 2.

Assume the nonunits form an ideal $N$. If both $x$ and $1-x$ were nonunits, then [F4] would give $1=x+(1-x)\in N$, contrary to [F3]. Hence condition 3 holds.

Assume condition 3 and let $N$ be the nonunits. By [F3], $0\in N$. If $x\in N$ and $r\in R$, then $rx$ cannot be a unit, since an inverse for $rx$ would make $x$ a unit; also $-x$ cannot be a unit.

If $x,y\in N$ and $u=x+y$ were a unit, then $x{u}^{-1}$ and $y{u}^{-1}=1-x{u}^{-1}$ would both be nonunits, because a unit among either would make $x$ or $y$ a unit. This contradicts condition 3. Thus $x+y\in N$, and [F4] with step 1.3 shows that $N$ is an ideal.

The ideal $N$ is proper because $1\mathrm{\notin }N$. Every proper ideal consists entirely of nonunits, so it is contained in $N$. Hence $N$ is maximal and is the only maximal ideal; by [F1], $R$ is local.

The ring is nonzero: if $1\mathrm{/}1=0$, [F3] gives $u\mathrm{\notin }\mathfrak{p}$ with $u=0$, impossible because $0\in \mathfrak{p}$.

By [F2], if $r\mathrm{\notin }\mathfrak{p}$, then $r\mathrm{/}s$ is a unit by taking $a=1$. If $r\in \mathfrak{p}$ and $ar$ lay outside $\mathfrak{p}$, the ideal property would be contradicted; hence $r\mathrm{/}s$ is not a unit. Thus the displayed set is exactly the set of nonunits.

Membership in the displayed set is independent of the chosen fraction: if $r\mathrm{/}s=r^{\mathrm{\prime }}\mathrm{/}{s}^{\mathrm{\prime }}$ with $r\in \mathfrak{p}$, then [F3] gives $u\mathrm{\notin }\mathfrak{p}$ with $u(r{s}^{\mathrm{\prime }}-r^{\mathrm{\prime }}s)=0$. Hence $u{r}^{\mathrm{\prime }}s=ur{s}^{\mathrm{\prime }}\in \mathfrak{p}$; primality and $u,s\mathrm{\notin }\mathfrak{p}$ force $r^{\mathrm{\prime }}\in \mathfrak{p}$.

The displayed set is an ideal: common-denominator addition and multiplication by arbitrary fractions preserve numerator membership in $\mathfrak{p}$. It is proper because $1\mathrm{\notin }\mathfrak{p}$, so $1\mathrm{/}1$ is not in it. Every proper ideal contains only nonunits, so step 1.2 makes every proper ideal lie inside it. It is therefore the unique maximal ideal, and [F4] makes $R_{\mathfrak{p}}$ local.

By [F1] and [F2], every denominator is already a unit in $R$. Applying [F3] to ${\mathrm{id}}_R$ gives $g:R_{\mathfrak{m}}\to R$ with $g(r\mathrm{/}s)=r{s}^{-1}$ and $g\lambda ={\mathrm{id}}_R$.

Both $\lambda g$ and ${\mathrm{id}}_{R_{\mathfrak{m}}}$ compose with $\lambda$ to give $\lambda$. The uniqueness clause of [F3] gives $\lambda g={\mathrm{id}}_{R_{\mathfrak{m}}}$, so $\lambda$ is an isomorphism with inverse $g$.

By [F1], either $e$ is a unit or $1-e$ is a unit. If $e$ is a unit, multiplying $e^2=e$ by $e^{-1}$ gives $e=1$.

If $1-e$ is a unit, then $(1-e)e=e-e^2=0$; multiplying by $(1-e{)}^{-1}$ gives $e=0$. The two cases prove the claim, including the endpoint idempotents themselves.

The displayed $S^{-1}I$ is an ideal: common-denominator addition preserves numerator membership in $I$, and multiplication by an arbitrary fraction does likewise. It is also the ideal generated by the elements $i\mathrm{/}1$ with $i\in I$, since $i\mathrm{/}s=(i\mathrm{/}1)(1\mathrm{/}s)$.

For completeness, let $\mathfrak{p}$ be prime and disjoint from $S$. It is saturated, since $sr\in \mathfrak{p}$ with $s\mathrm{\notin }\mathfrak{p}$ forces $r\in \mathfrak{p}$. Its extension is proper: $1=i\mathrm{/}s$ with $i\in \mathfrak{p}$ would give $u(s-i)=0$, hence $us=ui\in \mathfrak{p}$, contradicting $u,s\in S$.

An element $r$ lies in the contraction of $S^{-1}I$ exactly when $r\mathrm{/}1=i\mathrm{/}s$ for some $i\in I$. Equality gives $u(rs-i)=0$, hence $(us)r=ui\in I$, so $r\in I^{\mathrm{s}\mathrm{a}\mathrm{t}}$. Conversely, if $tr\in I$ for $t\in S$, then $r\mathrm{/}1=tr\mathrm{/}t\in S^{-1}I$. Thus the contraction is exactly $I^{\mathrm{s}\mathrm{a}\mathrm{t}}$.

If $J\subseteq S^{-1}R$ is an ideal and $I={\lambda }_S^{-1}(J)$, then $r\mathrm{/}s\in J$ implies $r\mathrm{/}1=(s\mathrm{/}1)(r\mathrm{/}s)\in J$, while $r\mathrm{/}1\in J$ implies $r\mathrm{/}s=(1\mathrm{/}s)(r\mathrm{/}1)\in J$. Hence $J=S^{-1}I$, as in [L1]. Step 2.1 also makes $I$ saturated.