Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 6 / 2 / Solution

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.