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Let $R$ be a commutative ring (not necessarily Noetherian) of finite Krull dimension. Suppose that for any prime ideal $p$, $\dim(R) = \dim(R/p) + \mathrm{height}(p)$ (call this condition $DIM$) .

Does that mean that for any inclusion of prime ideals $p \subsetneq q$, there is a longest chain (a chain of length $\dim(R)$) that includes both $p$ and $q$ ?

My attempt is that this would be to try to prove that a quotient and a localization at a prime ideal of a ring that satisfies $DIM$ also satisfies $DIM$. But now I think this is false. The counterexample in https://math.stackexchange.com/a/49285/445002 might also be a counterexample here. I can't figure out how to write $R[t]$ (where $R$ is a DVR) as a localization at prime ideals and quotient of a ring that does satisfy $DIM$. I know that $k[x]_{(x)}$ is a DVR, and $k[x]_{(x)}[t]$ is a localization of $k[x,t]$, but I don't think this is a localization at a prime ideal.

user26857
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David Lui
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  • Every finite poset arises as $\operatorname{Spec}(R)$ for some commutative ring $R$, and you can easily build a poset that satisfies DIM but not your condition – math54321 Feb 10 '24 at 04:52
  • So, do you want an example of a ring satisfying DIM, but having a localization which doesn't have the property DIM? – user26857 Feb 18 '24 at 21:06
  • @user26857 The localization would have to be at a prime ideal. – David Lui Feb 19 '24 at 00:34

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