Localization and the Fraction Field

Introduction

Localization is the operation of adjoining inverses for a chosen set of elements. It is the ring-theoretic analogue of passing from $\mathbb{Z}$ to $\mathbb{Q}$, and it is the basic tool that lets one study a ring near a prime ideal while forgetting the rest. Two constructions are its extreme cases: inverting every nonzero element of an integral domain produces the fraction field $\operatorname{Frac}(R)$, and inverting every element outside a prime ideal produces a local ring $R_\mathrm{P}$, one in which a single maximal ideal controls the whole structure.

The construction is carried out here in full, including the equivalence relation, because for rings with zero divisors the naive rule $r/s = r'/s'$ if and only if $rs' = r's$ is wrong and must be replaced by a rule that permits clearing annihilators. Throughout, $R$ is a commutative ring with $1 \neq 0$; it need not be a domain, and where a domain is required this is stated. Units and zero divisors are as in Rings, §§8–9, associates as in Integral Domains, and ideals and quotients are as in Rings, §§6–7.

Multiplicative Sets

Definition and Examples

Definition. A subset $S \subseteq R$ is multiplicative if

$$ 1 \in S \qquad \text{and} \qquad s, t \in S \implies st \in S. $$

Examples.

(a) $S = R^\times$, the group of units. Localizing at $R^\times$ changes nothing, since the elements to be inverted are already invertible.

(b) $S = \{1, f, f^2, f^3, \ldots\}$ for a fixed $f \in R$, the multiplicative set generated by $f$. The resulting ring inverts $f$ and nothing more.

(c) $S = R \setminus \mathrm{P}$ for a prime ideal $\mathrm{P}$. This is multiplicative exactly because $\mathrm{P}$ is prime: if $s, t \notin \mathrm{P}$ and $st \in \mathrm{P}$, then $s \in \mathrm{P}$ or $t \in \mathrm{P}$, a contradiction. This is the case that produces local rings.

(d) $S$, the set of nonzero elements of $R$ that are not zero divisors: the set used for the total ring of fractions. It contains $1$ and is multiplicatively closed by Rings, where the convention that $0$ is not a zero divisor is fixed. For a domain, $S = R \setminus \{0\}$.

(e) $S = \{1\}$. Localizing at $\{1\}$ changes nothing either.

Remark. The intersection of any family of multiplicative sets is multiplicative, and the multiplicative set generated by a subset $T \subseteq R$ is the set of finite products of elements of $T$, with the empty product equal to $1$.

Saturation

Definition. A multiplicative set $S$ is saturated if $s t \in S$ implies $s \in S$ and $t \in S$.

Proposition. For a multiplicative set $S$, the set

$$ \bar{S} = \{r \in R : r \text{ divides some element of } S\} $$

is the smallest saturated multiplicative set containing $S$.

Proof. If $r \mid s$ and $r' \mid s'$ with $s, s' \in S$, then $rr' \mid ss'$ with $ss' \in S$, so $\bar S$ is multiplicative; it contains $S$ since $s \mid s$; and it is saturated, since a divisor of an element that divides an element of $S$ again divides an element of $S$. Any saturated set containing $S$ contains the divisors of the elements of $S$.

The complement of a saturated multiplicative set need not be an ideal: in $\mathbb{Z}$ with $S = \{1\}$, the saturated set is $R^\times = \{\pm 1\}$, whose complement is not an ideal. A saturated multiplicative set whose complement is an ideal is the complement of a prime ideal, and conversely the complement of a prime ideal is saturated and multiplicative; the complement of an arbitrary union of prime ideals is again a saturated multiplicative set, although the union itself need not be an ideal.

Localization

The Construction

Let $S \subseteq R$ be multiplicative. On the product $R \times S$ define

$$ (r, s) \sim (r', s') \iff \text{there exists } t \in S \text{ with } t(rs' - r's) = 0. $$

Proposition. The relation $\sim$ is an equivalence relation, and it is the smallest equivalence relation on $R \times S$ compatible with the rules of fraction arithmetic.

Proof. Reflexivity and symmetry are immediate. For transitivity, suppose $t(rs' - r's) = 0$ and $u(r's'' - r''s') = 0$ with $t, u \in S$. Multiplying the first by $u s''$ and the second by $t s$ gives

$$ tu s'' r s' = tu s'' r' s, \qquad tu s r's'' = tu s r'' s', $$

so $tu s'(r s'' - r'' s) = 0$, and $tu s' \in S$ because $S$ is multiplicative.

Definition. The localization of $R$ at $S$ is the quotient

$$ S^{-1}R = (R \times S)/{\sim}, $$

with the class of $(r,s)$ written $r/s$. Addition and multiplication are defined by

$$ \frac{r}{s} + \frac{r'}{s'} = \frac{rs' + r's}{ss'}, \qquad \frac{r}{s} \cdot \frac{r'}{s'} = \frac{rr'}{ss'}. $$

Proposition. These operations are well defined and make $S^{-1}R$ a commutative ring with zero $0/1$ and identity $1/1$. If $0 \notin S$ then $1 \neq 0$ in $S^{-1}R$; if $0 \in S$ then $0/1 = 1/1$, so that $S^{-1}R$ is the zero ring. The map

$$ \iota = \iota_S : R \to S^{-1}R, \qquad \iota(r) = \frac{r}{1}, $$

is a ring homomorphism, and $\iota(s)$ is a unit of $S^{-1}R$ for every $s \in S$, with $\iota(s)^{-1} = 1/s$.

Proof sketch. If $r/s = r'/s'$ and $p/t = p'/t'$, then multiplying the two witnesses by elements of $S$ shows $rs' + r's$ over $ss'$ and $rp$ over $st$ to be independent of the representatives; this uses only the closure of $S$ under multiplication. The ring axioms follow from those of $R$ after clearing denominators. Finally $(s/1)(1/s) = s/s = 1/1$, since $1 \cdot (s \cdot 1 - 1 \cdot s) = 0$.

Proposition. The kernel of $\iota$ is

$$ \ker \iota = \{r \in R : sr = 0 \text{ for some } s \in S\}. $$

In particular, if $0 \notin S$ then $\iota$ is injective if and only if every element of $S$ is a non-zero-divisor; and if $0 \in S$ then $\iota$ is injective only when $R$ is the zero ring, since then $r/1 = 0/1$ for every $r \in R$.

Proof. $r/1 = 0/1$ means $t(r \cdot 1 - 0 \cdot 1) = 0$ for some $t \in S$, that is, $tr = 0$.

Remark. The witnessing element $t$ in the equivalence relation is exactly what makes localization behave well when $S$ contains zero divisors: it permits the annihilator of $rs' - r's$ to be cleared. If $R$ is a domain and $0 \notin S$, then $t$ may be cancelled and the relation reduces to $rs' = r's$.

The Universal Property

Theorem (universal property of localization). Let $S \subseteq R$ be multiplicative and let $\varphi : R \to T$ be a ring homomorphism such that $\varphi(s)$ is a unit of $T$ for every $s \in S$. Then there is a unique ring homomorphism $\psi : S^{-1}R \to T$ with $\psi \circ \iota = \varphi$, namely

$$ \psi\left(\frac{r}{s}\right) = \varphi(r)\varphi(s)^{-1}. $$

Proof. The formula is forced, so uniqueness is clear. It is well defined: if $t(rs' - r's) = 0$ with $t \in S$, applying $\varphi$ gives $\varphi(t)(\varphi(r)\varphi(s') - \varphi(r')\varphi(s)) = 0$, and $\varphi(t)$ is a unit, so $\varphi(r)\varphi(s') = \varphi(r')\varphi(s)$; multiplying by the inverses of $\varphi(s)\varphi(s')$ gives the same value for $r/s$ and $r'/s'$. Additivity and multiplicativity follow from the corresponding properties of $\varphi$ together with the commutativity of $T$.

Corollary. Localization is characterised up to unique isomorphism by the universal property: if a ring $A$ with a map $R \to A$ inverting all elements of $S$ has the same universal property, then $A \cong S^{-1}R$ by a unique isomorphism commuting with the maps from $R$.

Corollary (functoriality). A ring homomorphism $f : R \to R'$ carrying $S$ into a multiplicative set $S'$ induces a unique ring homomorphism $S^{-1}R \to (S')^{-1}R'$ commuting with the structure maps.

Example (an alternative presentation). For $S = \{1, f, f^2, \ldots\}$, the localization $S^{-1}R$, written $R_f$ or $R[1/f]$, is isomorphic to the quotient

$$ R[x]/(f x - 1), $$

with $x$ corresponding to $1/f$. The universal property identifies the two: in the quotient, the class of $f$ is inverted by the class of $x$, and any map inverting $f$ factors through $R[x]$ and then through the quotient.

Ideals in $S^{-1}R$

Let $\iota : R \to S^{-1}R$ be the structure map. For an ideal $I \subseteq R$, write

$$ S^{-1}I = \left\{\frac{i}{s} : i \in I,\ s \in S\right\} = I \cdot S^{-1}R $$

for the ideal it generates in $S^{-1}R$.

Definition. An ideal $I \subseteq R$ is saturated with respect to $S$ if

$$ \iota^{-1}(S^{-1}I) = I. $$

Every saturated ideal is disjoint from $S$: if $s \in I \cap S$ then $1/1 = s/s \in S^{-1}I$, so $1 \in \iota^{-1}(S^{-1}I) = I$ and $I = R$. Conversely an ideal disjoint from $S$ need not be saturated; the next theorem describes the ones that are.

Theorem (ideal correspondence). The maps

$$ \mathrm{A} \longmapsto \iota^{-1}(\mathrm{A}), \qquad I \longmapsto S^{-1}I, $$

are inverse bijections between the ideals $\mathrm{A}$ of $S^{-1}R$ and the saturated ideals $I$ of $R$. Under this bijection prime ideals correspond to prime ideals, and the maximal ideals of $S^{-1}R$ correspond to the primes of $R$ that are disjoint from $S$ and maximal among such primes. In particular a maximal ideal $\mathrm{M}$ of $R$ with $\mathrm{M} \cap S = \emptyset$ has $\mathrm{M} S^{-1}R$ maximal in $S^{-1}R$.

Proof. Let $\mathrm{A}$ be an ideal of $S^{-1}R$ and put $I = \iota^{-1}(\mathrm{A})$. Then $S^{-1}I \subseteq \mathrm{A}$, since $I \subseteq \mathrm{A}$ and $\mathrm{A}$ is an ideal of $S^{-1}R$; conversely, if $x = r/s \in \mathrm{A}$ then $r/1 = s x \in \mathrm{A}$, so $r \in I$ and $x \in S^{-1}I$. Hence $S^{-1}I = \mathrm{A}$, and then $\iota^{-1}(\mathrm{A}) = \iota^{-1}(S^{-1}I)$ shows that $I$ is saturated and that the composite $\mathrm{A} \mapsto I \mapsto \mathrm{A}$ is the identity.

For the other composite, let $I$ be saturated. If $S^{-1}I = S^{-1}R$ then $1/1 = i/s$ for some $i \in I$, $s \in S$, so $t(s - i) = 0$ for some $t \in S$, giving $ts = ti \in I$ with $ts \in S$; but $I$ saturated, hence disjoint from $S$, gives a contradiction unless $I = R$. So a saturated proper ideal has $S^{-1}I$ proper, and the composite $I \mapsto S^{-1}I \mapsto \iota^{-1}(S^{-1}I) = I$ is the identity by the definition of saturated.

The bijection preserves inclusions in both directions. For primes, let $\mathrm{P}$ be prime with $\mathrm{P} \cap S = \emptyset$; then $\mathrm{P}$ is saturated, because if $r/1 = p/s$ with $p \in \mathrm{P}$, then $t(rs - p) = 0$ gives $trs = tp \in \mathrm{P}$ with $t, s \notin \mathrm{P}$, so $r \in \mathrm{P}$. If $(r/s)(r'/s') \in S^{-1}\mathrm{P}$ then $rr' \in \mathrm{P}$ after clearing the witness, so $r \in \mathrm{P}$ or $r' \in \mathrm{P}$; hence $S^{-1}\mathrm{P}$ is prime. Conversely if $\mathrm{A}$ is prime in $S^{-1}R$ and $rr' \in \iota^{-1}(\mathrm{A})$, then $(r/1)(r'/1) \in \mathrm{A}$, so $r/1 \in \mathrm{A}$ or $r'/1 \in \mathrm{A}$; hence $\iota^{-1}(\mathrm{A})$ is prime. Since the bijection preserves inclusions, it carries the primes maximal under inclusion to the primes maximal under inclusion; on the side of $S^{-1}R$ these are exactly the maximal ideals, and on the side of $R$ they are the primes disjoint from $S$ that are maximal among such primes. Finally, a maximal ideal $\mathrm{M}$ of $R$ with $\mathrm{M} \cap S = \emptyset$ gives $S^{-1}R / S^{-1}\mathrm{M} \cong S^{-1}(R/\mathrm{M}) = R/\mathrm{M}$, a field: localization commutes with quotients, and the image of $S$ in the field $R/\mathrm{M}$ consists of nonzero elements, hence of units, so inverting them changes nothing. Hence $S^{-1}\mathrm{M}$ is maximal.

Corollary. The prime ideals of $S^{-1}R$ are exactly the ideals $\mathrm{P} S^{-1}R$ with $\mathrm{P}$ prime in $R$ and $\mathrm{P} \cap S = \emptyset$.

Corollary. Localization preserves domains: if $R$ is an integral domain and $0 \notin S$, then $S^{-1}R$ is an integral domain. Indeed $0$ is prime in $R$ and $0 \cap S = \emptyset$, so $S^{-1}(0) = 0$ is prime in $S^{-1}R$.

The Fraction Field

Construction

Let $R$ be an integral domain and let $S = R \setminus \{0\}$, which is multiplicative since $R$ has no zero divisors. The equivalence relation simplifies to

$$ \frac{r}{s} = \frac{r'}{s'} \iff r s' = r' s, $$

because the witnessing element $t \neq 0$ can be cancelled.

Definition. The fraction field of an integral domain $R$ is

$$ \operatorname{Frac}(R) = (R \setminus \{0\})^{-1} R. $$

Theorem. $\operatorname{Frac}(R)$ is a field, and $\iota : R \to \operatorname{Frac}(R)$ is injective.

Proof. Injectivity follows since $S$ contains no zero divisors. If $r/s \neq 0$ then $r \neq 0$, so $r \in S$ and $s/r$ is an inverse of $r/s$: $(r/s)(s/r) = rs/rs = 1/1$. Hence every nonzero element is a unit.

Theorem (universal property of the fraction field). Let $R$ be an integral domain and let $\varphi : R \to K$ be an injective ring homomorphism into a field $K$. Then there is a unique field homomorphism $\psi : \operatorname{Frac}(R) \to K$ with $\psi \circ \iota = \varphi$. Consequently $\operatorname{Frac}(R)$ is the smallest field in which $R$ embeds, in the sense that it is initial among such fields.

Proof. The universal property of localization gives $\psi(r/s) = \varphi(r)\varphi(s)^{-1}$, which is defined since $\varphi$ is injective and $s \neq 0$. If $r/s \neq 0$ then $\varphi(r) \neq 0$, so $\psi(r/s) \neq 0$ and $\psi$ is a field homomorphism.

Examples

Domain $R$	$\operatorname{Frac}(R)$
$\mathbb{Z}$	$\mathbb{Q}$
$F[x]$, $F$ a field	$F(x)$, the rational function field
$F[x_1, \ldots, x_n]$	$F(x_1, \ldots, x_n)$
$\mathbb{Z}[i]$	$\mathbb{Q}(i)$
$\mathbb{Z}[\sqrt{2}]$	$\mathbb{Q}(\sqrt{2})$
$\mathbb{Z}_{(p)} = \{a/b \in \mathbb{Q} : p \nmid b\}$	$\mathbb{Q}$
$F[[x]]$	$F((x))$, the Laurent series field
a field $K$	$K$

The fraction field of $\mathbb{Z}_{(p)}$ is $\mathbb{Q}$ because $\mathbb{Z}_{(p)}$ already contains every integer denominator prime to $p$, and inverting the remaining prime $p$ produces all of $\mathbb{Q}$. The fraction field of the power series ring is the field of formal Laurent series, whose elements are finite-tailed series $\sum_{n \geq -m} a_n x^n$.

The Total Ring of Fractions

For a general commutative ring the same construction with $S$ the set of nonzero non-zero-divisors is available, and it is written $\operatorname{Frac}(R)$ or $Q(R)$ and called the total ring of fractions. It is not a field unless every nonzero element of $R$ is a non-zero-divisor.

Proposition. Let $S$ be the set of nonzero non-zero-divisors of $R$. Then $\iota : R \to S^{-1}R$ is injective, and $S^{-1}R$ is a ring in which every non-zero-divisor is a unit.

Proof. Injectivity is the characterisation of $\ker \iota$ together with the definition of non-zero-divisor: $sr = 0$ with $s \in S$ forces $r = 0$ because $s \neq 0$ is not a zero divisor. If $r/s$ is a non-zero-divisor in $S^{-1}R$, then $r$ is a non-zero-divisor in $R$: if $rr_0 = 0$ with $r_0 \neq 0$, then

$$ \frac{r}{s} \cdot \frac{r_0}{1} = \frac{1}{s} \cdot \frac{rr_0}{1} = 0, $$

with $r_0/1 \neq 0$ because $\iota$ is injective, contradicting the hypothesis on $r/s$. Hence $r \in S$ and $r/s$ is a unit.

Example. If $R$ is finite, the nonzero non-zero-divisors are exactly the units, and the total ring of fractions of $R$ is $R$ itself. Thus $\mathbb{Z}/6\mathbb{Z}$ and $\mathbb{Z}/12\mathbb{Z}$ are their own total rings of fractions, even though they are not fields.

Example. For $R = R_1 \times R_2$ with both factors nonzero, an element $(r_1, r_2)$ annihilates a nonzero element of $R$ exactly when $r_1$ annihilates one in $R_1$ or $r_2$ annihilates one in $R_2$. Hence the nonzero non-zero-divisors are the pairs with $r_1 \neq 0$, $r_2 \neq 0$ and neither coordinate a zero divisor, and the total ring of fractions is $\operatorname{Frac}(R_1) \times \operatorname{Frac}(R_2)$. This is a product of fields exactly when both factors are integral domains, as for $R = \mathbb{Z} \times \mathbb{Z}$, whose total ring of fractions is $\mathbb{Q} \times \mathbb{Q}$.

Local Rings

Definition and Characterizations

Definition. A commutative ring $R$ with $1 \neq 0$ is local if it has exactly one maximal ideal.

Theorem. For a commutative ring $R$ with $1 \neq 0$ the following are equivalent.

(a) $R$ is local, with unique maximal ideal $\mathrm{M}$.

(b) The set $R \setminus R^\times$ of non-units is an ideal, necessarily the unique maximal ideal.

(c) There is a proper ideal $\mathrm{M}$ such that $r \in R^\times$ if and only if $r \notin \mathrm{M}$.

(d) For all $r \in R$, at least one of $r$ and $1 - r$ is a unit, and $R$ is not the zero ring.

Proof. (a) $\Rightarrow$ (b): every proper ideal is contained in a maximal ideal, so every non-unit lies in some maximal ideal, hence in the unique one, $\mathrm{M}$; and no element of $\mathrm{M}$ is a unit. Thus $R \setminus R^\times = \mathrm{M}$, an ideal. (b) $\Rightarrow$ (c): take $\mathrm{M} = R \setminus R^\times$; it is a proper ideal, and $r \notin \mathrm{M}$ means exactly that $r$ is a unit. (c) $\Rightarrow$ (d): if neither $r$ nor $1-r$ is a unit, then $r, 1 - r \in \mathrm{M}$, so $1 \in \mathrm{M}$, contradicting propriety. (d) $\Rightarrow$ (a): if $\mathrm{M}_1 \neq \mathrm{M}_2$ are maximal ideals, then $\mathrm{M}_1 + \mathrm{M}_2 = R$, because the sum is an ideal properly containing the maximal ideal $\mathrm{M}_1$; so $1 = a + b$ with $a \in \mathrm{M}_1$ and $b \in \mathrm{M}_2$, and then $a$ is a non-unit, being in $\mathrm{M}_1$, while $1 - a = b$ is a non-unit too, being in $\mathrm{M}_2$. This contradicts (d), so the maximal ideal is unique.

Examples. A field is local with maximal ideal $0$. The ring $\mathbb{Z}/p^n\mathbb{Z}$ is local with maximal ideal $(p)$. The power series ring $F[[x]]$ over a field is local with maximal ideal $(x)$, since a series is a unit exactly when its constant term is nonzero. For a field $F$, the dual numbers $\mathbb{D}'_F$ of Dual Numbers Algebra are local with maximal ideal $(\varepsilon)$, an ideal of square zero.

Localization at a Prime

Definition. Let $\mathrm{P}$ be a prime ideal of $R$ and let $S = R \setminus \mathrm{P}$. The localization of $R$ at $\mathrm{P}$ is

$$ R_\mathrm{P} = (R \setminus \mathrm{P})^{-1} R. $$

Theorem. $R_\mathrm{P}$ is a local ring with maximal ideal

$$ \mathrm{P} R_\mathrm{P} = \left\{\frac{p}{s} : p \in \mathrm{P},\ s \notin \mathrm{P}\right\}, $$

and residue field $R_\mathrm{P}/\mathrm{P}R_\mathrm{P} \cong \operatorname{Frac}(R/\mathrm{P})$.

Proof. By the ideal correspondence, the prime ideals of $R_\mathrm{P}$ correspond to the prime ideals of $R$ disjoint from $R \setminus \mathrm{P}$, that is, to the prime ideals contained in $\mathrm{P}$. Among these, $\mathrm{P}$ is the largest, so $\mathrm{P}R_\mathrm{P}$ is the unique maximal ideal. For the residue field, $R_\mathrm{P}/\mathrm{P}R_\mathrm{P} \cong (R/\mathrm{P})_{\bar{S}}$ where $\bar S$ is the image of $R \setminus \mathrm{P}$ in the domain $R/\mathrm{P}$, namely the nonzero elements; localizing a domain at its nonzero elements gives its fraction field.

Example. For $R = \mathbb{Z}$ and $\mathrm{P} = (p)$,

$$ \mathbb{Z}_{(p)} = \left\{\frac{a}{b} \in \mathbb{Q} : p \nmid b\right\}, $$

a local domain with maximal ideal $p\mathbb{Z}_{(p)}$ and residue field $\mathbb{F}_p$. It is the ring of rational numbers whose denominator is prime to $p$, and it is a subring of $\mathbb{Q}$; its fraction field is $\mathbb{Q}$.

Example (localization at an idempotent). Let $R = R_1 \times R_2$ with both factors nonzero and let $S = \{1, e\}$ where $e = (1, 0)$ is idempotent. Then $e$ is a unit of $S^{-1}R$, so $e = 1$ there and $(0,1) = 1 - e = 0$. Hence $S^{-1}R \cong R_1$. Localizing at one idempotent isolates the complementary factor of the product decomposition of Rings. Concretely, in $\mathbb{Z}/6\mathbb{Z}$ with $S = \{1, 4\}$ one gets $(\mathbb{Z}/6\mathbb{Z})_4 = \mathbb{Z}/6\mathbb{Z}[1/4] \cong \mathbb{Z}/3$, the factor on which $4$ acts as the identity.

Remark. Localization at a prime is compatible with the ideal theory in both directions: an ideal $I$ of $R$ is contained in $\mathrm{P}$ if and only if $I R_\mathrm{P}$ is proper, and for an integral domain

$$ R = \bigcap_{\mathrm{M}} R_\mathrm{M}, $$

the intersection taken over the maximal ideals and computed inside $\operatorname{Frac}(R)$. This identity is the reason a property can be checked locally at maximal ideals; the module-theoretic side of the same principle belongs to Modules over an Algebra.

Remark (localization preserves factorization). If $R$ is a UFD and $S$ is multiplicative, then $S^{-1}R$ is a UFD, since the prime factorization of an element of $S^{-1}R$ is obtained from that of a numerator in $R$ after discarding the factors that become units. In particular $\mathbb{Z}_{(p)}$ is a UFD, as is $\mathbb{Q}[x]_{(x)}$.

The Countability of $\mathbb{Q}$

Theorem. $\mathbb{Q}$ is countably infinite.

Proof. Every rational has a unique representation $a/b$ with $b > 0$ and $\gcd(a,b) = 1$. The map sending $a/b$ to the pair $(\operatorname{sgn}(a), (|a|, b))$, and then to a natural number by the bijection $\mathbb{N} \times \mathbb{N} \to \mathbb{N}$, is injective from $\mathbb{Q}$ into $\mathbb{N}$; and $\mathbb{N} \to \mathbb{Q}$, $n \mapsto n$ is injective. By the Schröder–Bernstein theorem of Cardinality and the Axiom of Choice, $\mathbb{Q}$ is countably infinite.

The statement is a cardinality statement about the fraction field of $\mathbb{Z}$, and it is proved here because this is where $\mathbb{Q}$ is introduced; the cardinality theory it uses — the countability of $\mathbb{N} \times \mathbb{N}$, the corollary on products and the Schröder–Bernstein theorem — is the subject of Cardinality and the Axiom of Choice, which states the result for $\mathbb{Q}$ and defers the reasoning to this article.

The Additive Group of $\mathbb{Q}$

The fraction field of $\mathbb{Z}$ carries, besides its ring structure, the structure of its additive group. As an abelian group, $\mathbb{Q}$ is torsion-free of rank one and divisible: for every $q \in \mathbb{Q}$ and every $n > 0$ the equation $nx = q$ has the solution $q/n$. These two properties determine it up to isomorphism, so the additive group of $\mathbb{Q}$ is the standard model of the abstract rank-one divisible torsion-free group that Infinite Abelian Groups, in the category Groups, writes $D_1$ and reasons with group-theoretically; the identification of $D_1$ with the additive group of $\mathbb{Q}$ is the content of this section.

Two group-theoretic facts about the additive group are worth recording here. It is not finitely generated: a finite set of rationals has denominators dividing a common positive integer $N$, so it generates only rationals whose denominators divide $N$, whereas $\mathbb{Q}$ contains $1/p$ for every prime $p$. And it is the injective hull of $\mathbb{Z}$ in the category of abelian groups, the minimal divisible group containing $\mathbb{Z}$; the corresponding torsion divisible group is $\mathbb{Q}/\mathbb{Z}$, the direct sum over the primes of the Prüfer groups, and the subgroups of the additive group are classified up to isomorphism by their types, the invariant of the group article.

Summary

A multiplicative set $S \subseteq R$ is a subset containing $1$ and closed under multiplication. The localization $S^{-1}R$ is the ring of formal fractions $r/s$ with $r \in R$, $s \in S$, with equality $r/s = r'/s'$ when $t(rs' - r's) = 0$ for some $t \in S$; the witnessing element is what makes the construction correct in the presence of zero divisors. The map $R \to S^{-1}R$ is a ring homomorphism inverting every element of $S$, and it is universal among such maps: any ring homomorphism from $R$ inverting $S$ factors uniquely through $S^{-1}R$. Its kernel consists of the elements annihilated by an element of $S$, so the map is injective exactly when $0 \notin S$ and every element of $S$ is a non-zero-divisor.

The ideals of $S^{-1}R$ are in bijection with the saturated ideals of $R$, those $I$ with $\iota^{-1}(S^{-1}I) = I$; in particular every such ideal is disjoint from $S$, and under the bijection prime ideals correspond to prime ideals. Consequently localization preserves the property of being a domain and turns the primes contained in $\mathrm{P}$ into the primes of $R_\mathrm{P}$, with $\mathrm{P}R_\mathrm{P}$ the unique maximal ideal. When $S$ is the set of nonzero elements of a domain, the localization is the fraction field $\operatorname{Frac}(R)$, the initial field in which $R$ embeds; when $S = R \setminus \mathrm{P}$ it is the local ring $R_\mathrm{P}$, whose residue field is $\operatorname{Frac}(R/\mathrm{P})$. A ring with a single maximal ideal is local, equivalently a ring in which the non-units form an ideal, equivalently a ring in which for every $r$ at least one of $r$ and $1-r$ is a unit.

Summary of Notation

Symbol	Meaning
$R$	Commutative ring with identity $1 \neq 0$
$S$	Multiplicative set: $1 \in S$, closed under multiplication
$R^\times$	Group of units
$\mathrm{P}$	Prime ideal
$\mathrm{A}$	Ideal of $S^{-1}R$ or of $R$
$\mathrm{M}$	Maximal ideal of a local ring
$S^{-1}R$	Localization of $R$ at $S$
$r/s$	Class of $(r,s)$ in $S^{-1}R$
$\iota = \iota_S$	Structure map $R \to S^{-1}R$, $r \mapsto r/1$
$S^{-1}I$	Ideal generated by $I$ in $S^{-1}R$
$R_f$, $R[1/f]$	Localization inverting $f$; isomorphic to $R[x]/(fx-1)$
$R_\mathrm{P}$	Localization at the prime $\mathrm{P}$, $(R \setminus \mathrm{P})^{-1}R$
$\mathrm{P}R_\mathrm{P}$	Maximal ideal of the local ring $R_\mathrm{P}$
$\mathbb{Z}_{(p)}$	$p$-local integers $\{a/b \in \mathbb{Q} : p \nmid b\}$, a localization of $\mathbb{Z}$
$\operatorname{Frac}(R)$	Fraction field of an integral domain $R$, and total ring of fractions of a general $R$
$Q(R)$	Alternative notation for the total ring of fractions; the multiplicative set inverted is the set of nonzero non-zero-divisors
$F(x)$	Rational function field $\operatorname{Frac}(F[x])$
$F((x))$	Laurent series field $\operatorname{Frac}(F[[x]])$
$R \setminus R^\times$	Set of non-units; an ideal exactly when $R$ is local
$\bar{S}$	Saturation of $S$