Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 101 / 2

Let $T=R\setminus P$. Since $P$ is a prime ideal, $T$ is a multiplicative subset containing $1$ and excluding $0$. The localization at a prime ideal isConcretely, its elements are equivalence classes of pairs $(r,s)\in R\times T$, written $r/s$, withThis extra multiplier is essential when $R$ has zero divisors. Addition and multiplication areThe equivalence criterion is compatible with these operations: after multiplying by the multipliers witnessing changes of representatives, the corresponding numerators agree. The commutative ring axioms follow from those of $R$; the identity is $1/1$.

The canonical ring homomorphism $\iota :R\to R_P$, $r\mapsto r/1$, supplies the algebra over a commutative ring structure, with scalar action $a(r/s)=ar/s$. It also has the universal property of localization: whenever a ring homomorphism $\psi :R\to A$ sends $T$ to units, the unique extension is $r/s\mapsto \psi (r)\psi (s{)}^{-1}$.

For later use, $R_P$ is a local ring with maximal ideal $P{R}_P$. Reduction of numerators and denominators gives a surjective ring homomorphism $R_P\to \mathrm{Frac}(R/P)$ with kernel $P{R}_P$, so this ideal is proper and maximal. Every fraction outside it has numerator outside $P$ and is a unit, with inverse $s/r$. Thus it is the unique maximal ideal.

The equivalence relation defining the localization at a prime ideal givesThereforeEquivalently, all zero divisors of $R$ must lie in $P$. To prove necessity, if $s\notin P$ and $sr=0$ with $r\ne 0$, then $r/1=0$ in $R_P$, contradicting the hypothesis that the map is injective. Conversely, if every denominator annihilates no nonzero element, the displayed kernel of a ring homomorphism is zero, so the map is injective.

An integral domain automatically satisfies this criterion at every prime ideal, but the criterion itself does not require $R$ to be an integral domain.

No: $R$ cannot contain a nonzero nilpotent element. We prove the needed principle, localization detects zero elements, rather than assume it.

Let $a\ne 0$ in $R$. Its annihilator $\mathrm{Ann}(a)=\{r:ra=0\}$ is a proper ideal, because $1$ does not annihilate $a$. Every proper ideal is contained in a maximal ideal, by the Zorn lemma; choose such a maximal ideal $\mathfrak{m}$ containing $\mathrm{Ann}(a)$. If $a/1=0$ in $R_{\mathfrak{m}}$, then some $s\notin \mathfrak{m}$ satisfies $sa=0$. This would put $s$ in $\mathrm{Ann}(a)\subseteq \mathfrak{m}$, a contradiction. Thus every nonzero element survives in at least one localization at a prime ideal.

Now suppose $a^n=0$. For every prime ideal $P$, $(a/1{)}^n=0$ in $R_P$. The assumed absence of nonzero nilpotent elements forces $a/1=0$ in every $R_P$. The preceding argument then forces $a=0$. In the terminology of reduced rings, we have proved

Yes: $R$ can have nonzero zero divisors. Let $R=k\times k$, a direct product of rings, where $k$ is any field. The nonzero idempotents $e_1=(1,0)$ and $e_2=(0,1)$ satisfy $e_1{e}_2=0$, so both are zero divisors.

There are exactly two prime ideals:Indeed, any prime ideal must contain $e_1$ or $e_2$, since their product is zero. If it contains $e_1$, it contains $k\times 0$ and corresponds to a prime ideal of the quotient ring $k$, whose only prime ideal is zero; hence it equals $P_2$. The other case gives $P_1$.

For the localization at a prime ideal $P_1$, every denominator $(u,v)\notin P_1$ has $u\ne 0$, andis an isomorphism. It is surjective using constant first-coordinate fractions. It is injective because a numerator with first coordinate zero is annihilated by $e_1\notin P_1$, so its fraction is zero. Interchanging the coordinates proves $R_{P_2}\cong k$.

Thus the localization at a prime ideal is a field in every case, although $R$ has zero divisors. The obstruction to using the previous part's argument is that the two nonzero factors of a zero product can survive at different prime ideals.

Let $e_j$ be the sequence with $1$ in coordinate $j$ and zero elsewhere. In the direct product of rings $R=k^{\mathbb{N}}$, the idealsform a strict ascending chain: every element of $(e_1,\dots ,e_n)$ vanishes after coordinate $n$, whereas $e_{n+1}$ does not. Hence $R$ is not a Noetherian ring.

Nevertheless, each $a=(a_j)\in R$ has a coordinatewise generalized inverse $b=(b_j)$, defined by $b_j=a_j^{-1}$ when $a_j\ne 0$ and $b_j=0$ otherwise. It satisfiesThis is the defining property of a Von Neumann regular ring. Fix any prime ideal $P$, and take $a\in P$. Since $ab\in P$, $1-ab\notin P$; butConsequently $a/1=0$ in the localization at a prime ideal $R_P$. All elements of $P{R}_P$ therefore vanish. By the local ring description in part (a), $P{R}_P$ is the unique maximal ideal, so $R_P$ is a nonzero field. A field has only the ideals zero and itself, and thus is Noetherian.

We have proved the stronger conclusionThe argument applies to every prime ideal, without assuming that it comes from a coordinate projection.