Integral domains with totally ordered spectra In my research I ended up trying to prove some properties of integral domains such that their spectrum is a totally ordered poset. Are there some nice (ubiqitous/natural) examples of such domains, which are not valuation domains? (Of course any local 1-dimensional domain is such, so non-noetherian almost perfect domains form a class of nice examples, so of higher Krull dimension?)
 A: A good source of such rings consist of the so-called "pseudo-valuation domains" of Hedstrom and Houston.  A link to their original (1978) article is: http://projecteuclid.org/download/pdf_1/euclid.pjm/1102810151.  Among other things in that article, they show the following facts that are relevant to your situation:


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*Given any pseudo-valuation domain $R$, there is a unique valuation domain $V$ with $R \subseteq V \subseteq$ the fraction field of $R$, such that the map of prime spectra is an order isomorphism.

*If you happen to start with a valuation domain of the form $V=K+M$ ($K$ a field, $M$ the maximal ideal of $V$) -- a form that many valuation rings found in nature take -- then take any proper subfield $F$ of $K$ (assuming of course that $K$ isn't a prime field). The subring $R=F+M$ will then be a pseudo-valuation ring whose spectrum is order-isomorphic to that of $V$.
A: If W is a totally ordered poset, k is a field, Xi: i∈W are indeterminates, and R=k[Xi, Xi/Xjn:i,j∈W, i<j, n≥0](Xi:i∈W), then R is a local domain with Spec(R)={(0)}∪{(Xi, Xi/Xjn:i∈V, i<j, n≥0)}:V=initial segment of W.  Thus any non-limit ordinal is the (poset of a) spectrum of a local domain.  Note R is a valuation domain with value group ⊕i∈Wℤ so it doesn't answer all requirements in the original question.
From W=ℚ we get the totally ordered poset: {(-∞,0)} ∪ (ℝ-ℚ)X{0} ∪ ℚX{0,1} ∪ {(∞,0)} (where 2nd component 0 corresponds to an open initial segment interval (-∞,i) and 1 corresponds to closed initial segment interval (-∞,i])
From W=ℝ we get the totally ordered poset: {(-∞,0)} ∪ ℝX{0,1} ∪ {(∞,0)}
