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For a commutative ring $R$ and a multiplicative subset $S$ containing $1$, the universal property of localization specifies a pair $(R_S,\iota )$ with $\iota :R\to R_S$ a unital ring homomorphism, such that every $\iota (s)$, $s\in S$, is a unit, and any unital ring homomorphism $f:R\to T$ into a commutative ring that sends $S$ to units factors uniquely through $\iota$. Thus there is a unique $\tilde{f}:R_S\to T$ satisfying $\tilde{f}\iota =f$. The property determines the pair up to a unique compatible isomorphism.

For existence, take pairs $(r,s)\in R\times S$ and declareThe extra multiplier is essential when $R$ has zero divisors. This is an equivalence relation: for transitivity, multiply the two cross-multiplication equations by appropriate denominators and add them, using the product of their annihilating multipliers. Write the class as $r/s$ and defineMultiplying the defining equivalence equations verifies that these operations are well defined. They give a commutative ring, and $s/1$ has inverse $1/s$. The required map is necessarilyIt is well defined because applying $f$ to $u(s^{\prime }r-s{r}^{\prime })=0$ and cancelling the units $f(u),f(s),f(s^{\prime })$ equates the two images. Its addition and multiplication laws follow directly from the fraction formulas. This establishes existence and the universal property of localization. If $0\in S$, this construction gives the zero ring, which is allowed as a localization.

For a prime ideal $P$, its complement $S=R\setminus P$ is multiplicatively closed, and localization at a prime ideal means $R_P=S^{-1}R$. The mapis well defined and surjective, with kernel of a ring homomorphism $P{R}_P$: a numerator has zero image exactly when it belongs to $P$. Since $R/P$ is an integral domain, its field of fractions is a nonzero field, so $P{R}_P$ is a proper maximal ideal. Every fraction $r/s$ with $r\notin P$ is a unit, with inverse $s/r$. Every fraction with $r\in P$ belongs to this proper ideal and cannot be a unit. Thus the nonunits are exactly $P{R}_P$, and every proper ideal lies within it. ConsequentlyNow suppose each $R_P$ is reduced. If $x\in R$ is nilpotent, its image $x/1$ is nilpotent in every $R_P$, hence is zero. If $x\ne 0$, its annihilator ${\mathrm{Ann}}_R(x)$ is a proper ideal, so choose a maximal ideal $P$ containing it. Maximal ideals in a commutative ring are prime ideals. The vanishing criterion in a module localization would give $sx=0$ for some $s\notin P$. But then $s\in {\mathrm{Ann}}_R(x)\subseteq P$, a contradiction. Hence $R$ is reduced. This is reducedness is detected by prime localizations; no finiteness hypothesis is needed.

The stronger local-domain condition does not force $R$ to be an integral domain. Take $R=k\times k$ for a field $k$. Its only prime ideals are $P_1=0\times k$ and $P_2=k\times 0$: a prime ideal must contain one of the orthogonal idempotents $(1,0)$ and $(0,1)$, and the remaining quotient is the field $k$. At $P_1$, inverting $(1,0)$ kills the second component, so $R_{P_1}\cong k$; similarly $R_{P_2}\cong k$. Both localizations are integral domains, butis a product of nonzero elements in $R$. Therefore the answer is no, as illustrated by domain prime localizations do not imply a domain.

Let $S$ be a multiplicative subset, containing one and closed under products. On $A\times S$, useThe relation is reflexive and symmetric. For transitivity, if $u(ta-sb)=0$ and $v(rb-tc)=0$, then multiplication and addition give $uvt(ra-sc)=0$. Thus it is an equivalence relation. Write $a/s$ for the equivalence class. The localization of a ring $A_S=S^{-1}A$ has operationsThey are well defined. For example, replacing $a/s$ by $b/t$ in its sum with $c/r$ gives cross-numerator difference $r^2(ta-sb)$, and in its product with $c/r$ gives $rc(ta-sb)$. The same annihilating element $u$ kills both differences. Replacing the other input works identically. Associativity and distributivity then follow by putting all terms over the product of their denominators and using the ring identities in $A$. Addition and multiplication are commutative, with identities $0/1$ and $1/1$, and the displayed negative is an additive inverse. Hence $S^{-1}A$ is a commutative ring. If $0\in S$, the fraction relation identifies all elements, producing the zero ring.

The canonical ring homomorphism $\iota :A\to S^{-1}A$, $a\mapsto a/1$, makes each $s\in S$ a unit, with inverse $1/s$. The universal property of localization states that any ring homomorphism $h:A\to C$ sending $S$ into the units factors uniquely through $\iota$. Define the factor byIf $u(ta-sb)=0$, then $h(u)$ is invertible, so $h(t)h(a)=h(s)h(b)$, proving well-definedness. The fraction operations prove that $\tilde{h}$ is a ring homomorphism. Its value is forced by $a/s=\iota (a)\iota (s{)}^{-1}$, so it is unique. This property characterizes the localization up to a unique compatible isomorphism.

For a prime ideal $\mathfrak{p}$, its complement is a multiplicative subset, and the localization at a prime ideal is $A_{\mathfrak{p}}=(A\setminus \mathfrak{p}{)}^{-1}A$. Fractions with numerator outside $\mathfrak{p}$ are units; those with numerator in $\mathfrak{p}$ form the unique maximal ideal $\mathfrak{p}{A}_{\mathfrak{p}}$. Thus it is a local ring.

For the nilpotent question, suppose $a\ne 0$ and $a^n=0$. Its annihilator is a proper ideal, so it lies in some maximal ideal $\mathfrak{m}$. The fraction criterion says $a/1=0$ in $A_{\mathfrak{m}}$ only if $sa=0$ for some $s\notin \mathfrak{m}$, which is impossible by the choice of $\mathfrak{m}$. Thus $a/1$ is a nonzero nilpotent element in that localization. This proves reducedness is detected by prime localizations: if every prime localization is reduced, $A$ has no nonzero nilpotent elements.

For the domain question take $A=k\times k$ with $k$ a field. Its only prime ideals are ${\mathfrak{p}}_1=0\times k$ and ${\mathfrak{p}}_2=k\times 0$: a prime ideal must contain one of the orthogonal idempotents $(1,0),(0,1)$, and the corresponding quotient is $k$. Localizing at ${\mathfrak{p}}_1$ makes $(1,0)$ invertible and kills $(0,1)$, giving $A_{{\mathfrak{p}}_1}\cong k$; similarly $A_{{\mathfrak{p}}_2}\cong k$. But $(1,0)(0,1)=0$ with both factors nonzero. Therefore domain prime localizations do not force $A$ to be an integral domain, as expressed by domain prime localizations do not imply a domain.