Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 3 / 2 / Solution

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.