Does there exist a Valuation ring $(R, \mathfrak m)$ , with principal maximal ideal, of infinite global dimension ?

Corollary 2 of the following paper by Osofsky has an example of Valuation ring of infinite global dimension. But I'm not sure whether we can also construct such example with principal maximal ideal.

Please help


1 Answer 1


The example of a valuation domain of infinite global dimension Barbara Osofsky gives is a generalized power series ring $k[[\Gamma]]$ with a well-chosen ordered abelian group $\Gamma$. But as far as I see the residue field $k$ can be chosen arbitrarily. Hence one can take any field $k$, that carries a discrete valuation ring $\overline{O}\subset k$. Let $O\subset k[[\Gamma]]$ be the preimage of $\overline{O}$ under the natural homomorphism $k[[\Gamma]]\rightarrow k$. Then $O$ is a valuation ring of the fraction field of $k[[\Gamma]]$ and $k[[\Gamma]]$ is a localization of $O$. Consequently $O$ must have infinite global dimension, since global dimension decreases under localization. On the other side the maximal ideal of $O$ is generated by any preimage of a generator of the maximal ideal of $\overline{O}$.

  • $\begingroup$ I'm quite not sure why $O$ is a Valuation ring ... why are the ideals linearly ordered ? $\endgroup$
    – user521337
    Mar 24, 2019 at 0:40
  • $\begingroup$ To check that $O$ is a valuation ring one verifies that either $x\in O$ or $x^{-1}\in O$ for every $x\in k((\Gamma))$. So assume $x\not\in O$. Case 1: $x\not\in k[[\Gamma]]$. Then $x^{-1}$ is in the maximal ideal of $k[[\Gamma]]$, which is the preimage of $0$ under the natural map. Hence $x^{-1}\in O$. Case 2: $x\in k[[\Gamma]]$. Then the image $\overline{x}$ of $x$ is not in $\overline{O}$ and thus $\overline{x}^{-1}\in\overline{O}$. Consequently $x^{-1}\in O$. (Remark: in valuation theory $O$ is called the composite valuation ring of $\overline{O}$ and $k[[\Gamma]]$). $\endgroup$
    – Hagen Knaf
    Mar 24, 2019 at 9:41

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