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

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.