Localization of a ring ($S^{-1}R$)

For a multiplicatively closed set $S$, the localization $S^{-1}R$ formally makes every element of $S$ invertible. Elements are represented by fractions $r/s$.

Here $S$ consists of the non-zero-divisors of a commutative ring $A$. The localization map $A\to Q(A)$ is injective: a fraction $a/1$ is zero only if some $s\in S$ annihilates $a$. For a reduced ring with finitely many minimal prime ideals, its zero divisors form the union of these primes. Indeed, if $ab=0$ and $a$ avoids every minimal prime, reduction to each domain quotient forces $b$ into their intersection, which is zero. Conversely, for $a$ in a minimal prime ${\mathfrak{p}}_i$, choose $b$ in every other minimal prime but outside ${\mathfrak{p}}_i$, using their incomparability and multiplying suitable elements. Then $b\ne 0$ and $ab=0$.

For a multiplicatively closed set $T$ in a commutative ring $A$, the map sends $(m\otimes n)/t$ to $(m/t)\otimes (n/1)$. An inverse sends $(m/s)\otimes (n/t)$ to $(m\otimes n)/(st)$. The fraction relations and the tensor product of modules balancing relation make these maps well-defined and mutually inverse.

An element $a$ of a commutative ring $R$ is zero if and only if its image in every localization at a prime ideal is zero. If $a\ne 0$, its proper annihilator lies in a maximal ideal $\mathfrak{m}$. No element outside $\mathfrak{m}$ can annihilate $a$, so $a/1\ne 0$ in $R_{\mathfrak{m}}$. This also applies to elements of an $R$-module using localization of a module.

A multiplicative subset $S$ of a ring contains $1$ and is closed under multiplication. It may contain zero, in which case its localization is the zero ring.

A multiplicative subset is saturated if every factor of a product in the subset also lies in it. We include $1$ in a multiplicative subset. A saturated subset containing $0$ is the whole ring. The units form a saturated subset in a commutative ring.

In $\mathbb{Z}$, the saturated multiplicative subset $\{-1,1\}$ has complement equal to the union of $(p)$ over all positive primes. No finite subcollection of prime ideals covers this complement: a prime outside the finite list is a nonunit not in any listed ideal.

The complement of a multiplicative subset is a union of prime ideals exactly when the subset is saturated. If $a\notin S$, saturation ensures $(a)\cap S=\varnothing$; extend this to a prime ideal avoiding a multiplicative subset. Conversely, a product in $S$ cannot have a factor in a prime ideal contained in the complement.

The localization map $\iota :R\to S^{-1}R$ sends every element of $S$ to a unit. If a ring homomorphism $f:R\to T$ also sends every element of $S$ to a unit, there is a unique homomorphism $\tilde{f}:S^{-1}R\to T$ with $\tilde{f}(r/s)=f(r)f(s{)}^{-1}$ and $\tilde{f}\iota =f$.

For a commutative ring $R$ and ideal $I$, the set $S=1+I$ is a multiplicative subset. For $i\in I$ and $s,t\in S$, the numerator $st-ri$ of $1-(r/t)(i/s)$ again lies in $S$. Thus this difference is a unit in the localization of a ring. The unit criterion for the Jacobson radical proves $S^{-1}I\subseteq J(S^{-1}R)$. This requires no Noetherian hypothesis.

For a multiplicative subset $S\subseteq R$ and an $R$-module $M$, the localization $S^{-1}M$ consists of fractions $m/s$, with $m/s=m^{\prime }/s^{\prime }$ when some $u\in S$ satisfies $u(s^{\prime }m-s{m}^{\prime })=0$. It is an $S^{-1}R$-module, and the canonical map is $m\mapsto m/1$. If $M$ is finitely generated over a Noetherian ring, then $S^{-1}M$ is a Noetherian module over $S^{-1}R$.

The fraction relation defining localization of a module gives this criterion. For localization at powers of a single element $f$, a fraction vanishes exactly when some power of $f$ annihilates its numerator. Finitely many such vanishing relations can be cleared with one common exponent. This is distinct from detecting global zero by localizing at every prime ideal.

For a multiplicative subset $S$ of a commutative ring $A$, localization preserves short exact sequences of $A$-modules. Surjectivity follows by lifting a numerator. If a fraction maps to zero, some element of $S$ annihilates its image, so multiplying its numerator by that element puts it into the original kernel. For an injective original map, the same criterion shows its localized kernel is zero. Thus $S^{-1}A$ is a flat module over $A$.

For a module homomorphism $f:M\to N$ over a commutative ring $A$, $f$ is injective if and only if every $f_{\mathfrak{m}}$ is injective, with $\mathfrak{m}$ ranging over maximal ideals. In a localization of a module, $x/s$ maps to zero exactly when some denominator $t$ satisfies $tf(x)=0$. If $f$ is injective this gives $tx=0$, so the original fraction is zero. In general it identifies $\ker (f_{\mathfrak{m}})$ with $(\ker f{)}_{\mathfrak{m}}$. A nonzero element of $\ker f$ has a proper annihilator, contained in some maximal ideal; it remains nonzero in that localization. Thus vanishing of all localized kernels forces $\ker f=0$. No finiteness or Noetherian assumption is required.

Let $R$ be Noetherian, $I\subseteq R$ an ideal, $S=1+I$, and $M$ a finitely generated $R$-module. ThenIf $(1+a)m=0$ with $a\in I$, then $m=(-a{)}^{j}m\in I^{j}M$ for every $j$. Conversely, apply the Artin-Rees lemma to $Rm\subseteq M$. For sufficiently large $c$ it givesso $m=am$ for some $a\in I$ and $(1-a)m=0$.

Letbe the semigroup algebra inside $k[t^{\mathbb{Q}}]$, and let $I$ be generated by the $t^q$ with $q>0$. This integral domain is not Noetherian, sinceMoreover $I^2=I$, because $t^q=(t^{q/2}{)}^2$. Thus ${\bigcap }_{j\ge 1}{I}^j=I\ne 0$, whereas the map $R\to (1+I{)}^{-1}R$ is injective because $R$ is a domain.

For a set $\pi$ of prime numbers, the integer localizationconsists of fractions whose denominator has prime factors only in $\pi$. As an additive group it is divisible by every prime in $\pi$.

If $A_{\neg p}=\mathbb{Z}[q^{-1}:q\ne p]$, its finite quotients are exactly the cyclic $p$-groups $\mathbb{Z}/p^n\mathbb{Z}$. Divisibility by every $q\ne p$ excludes other prime torsion, while a quotient of exponent $p^n$ factors through $A_{\neg p}/p^n{A}_{\neg p}\cong \mathbb{Z}/p^n\mathbb{Z}$.

Extension and contraction give inverse inclusion-preserving bijections between prime ideals of $S^{-1}R$ and prime ideals of $R$ disjoint from $S$:

For $S=1+I$, the prime ideal correspondence for localization identifies $\mathrm{Spec}{S}^{-1}R$ withIndeed, $\mathfrak{p}\cap (1+I)=\varnothing$ exactly when no element of $I$ is congruent to $-1$ modulo $\mathfrak{p}$, which is exactly $\mathfrak{p}+I\ne R$.

Every localization of a Noetherian ring is Noetherian. Each ideal $J\subseteq S^{-1}R$ is the extension of its contraction to $R$, whose finite generating set consequently generates $J$.

If $R$ is a Dedekind domain and $0\notin S$, then $S^{-1}R$ is an integrally closed Noetherian domain of dimension at most one. It is therefore either a Dedekind domain or, when its dimension is zero, a field.

Localization at a prime ideal $\mathfrak{p}$ means inverting $R\setminus \mathfrak{p}$. The resulting local ring $R_{\mathfrak{p}}$ has maximal ideal $\mathfrak{p}{R}_{\mathfrak{p}}$.

The ring $k\times k$, for a field $k$, has two prime ideals, $0\times k$ and $k\times 0$. Their localizations at a prime ideal are both isomorphic to $k$. Nevertheless $(1,0)(0,1)=0$ is a product of nonzero elements, so $A$ is not an integral domain. Nonzero orthogonal idempotents separate the two components of its spectrum of a commutative ring.