Totally ordered sets and the prime spectra of rings

Abstract

Let $T$ be a totally ordered set and let $D(T)$ denote the set of all cuts of $T$. We prove the existence of a discrete valuation domain $O_v$ such that $T$ is order isomorphic to two special subsets of Spec$(O_v)$. We prove that if $A$ is a ring (not necessarily commutative) whose prime spectrum is totally ordered and satisfies (K2), then there exists a totally ordered set $U\subseteq \text{Spec}(A)$ such that the prime spectrum of $A$ is order isomorphic to $D(U)$. We also present equivalent conditions for a totally ordered set to be a Dedekind totally ordered set. At the end, we present an algebraic geometry point of view

arXiv comment13 pages