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Does there exist a local domain of infinite dimension in which every chain of prime ideals is finite?

Of course, such a ring must be neither noetherian nor catenary.

(This question arose while trying to compare different definitions of catenarity, and more precisely while trying to understand what may cause codimensions in topological spaces to be infinite.)

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  • $\begingroup$ It's not completely obvious to me that it can't be noetherian. If there's a noetherian example $A$, then we can suppose that $A/P$ has finite dimension for every nonzero prime $P$. How does one get a contradiction? $\endgroup$
    – YCor
    Commented Oct 2, 2018 at 20:46
  • $\begingroup$ @YCor: Local noetherian rings have finite dimension, since primes in noetherian rings have finite height. $\endgroup$
    – Fred Rohrer
    Commented Oct 2, 2018 at 20:50
  • $\begingroup$ Oh sorry, I hadn't seen the local assumption. I was thinking of a nonlocal noetherian ring, for which it sounds not obvious to me. $\endgroup$
    – YCor
    Commented Oct 2, 2018 at 20:51
  • $\begingroup$ @YCor: Actually, my original problem was whether there is a spectral topological space with closed irreducible subsets $Y\subseteq Z$ such that ${\rm codim}(Y,Z)=\infty$ and that every chain of closed irreducible subsets with extremities $Y$ and $Z$ is finite. This can be reduced to the question given here. $\endgroup$
    – Fred Rohrer
    Commented Oct 2, 2018 at 21:05
  • $\begingroup$ Just an obvious comment ... the ring you are looking for can't be Valuation ring either ... $\endgroup$
    – user521337
    Commented Oct 2, 2018 at 22:49

1 Answer 1

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Choose a field $k$ and a Noetherian $k$-algebra $R$ of infinite dimension. (I know you know such a thing exists.) Let $R' \subset R[x]$ be the set of polynomials $f = \sum a_i x^i$ whose constant term is constant, i.e., $a_0 \in k \subset R$. Then we have $$ \text{Spec}(R') = \text{Spec}(R[x]) \amalg_{\text{Spec}(R)} \text{Spec}(k) $$ by Tag 0B7J. Observe that $R'[1/x] = R[x, 1/x]$ is Noetherian. Observe that for any prime $\mathfrak p \subset R$ the prime $\mathfrak p' = R' \cap \mathfrak p R[x]$ is contained in the maximal ideal $\mathfrak m = \text{Ker}(R' \to k) = \sqrt{xR'}$ (small detail omitted). Thus the local ring of $R'$ and $\mathfrak m$ is an example. Namely, its punctured spectrum is Noetherian of infinite dimension (as the localization of $R'$ at $\mathfrak p'$ is the same as the localization of $R[x]$ at $\mathfrak p R[x]$ which is flat over $R_\mathfrak p$ and hence has dimension $\geq \dim(R_\mathfrak p)$. Enjoy!

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