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Let $R$ be a commutative unit ring (not necessarily Noetherian). Is there an example such that weak global dimension of $R$ is finite but the global dimension of $R$ is infinite? Can we find such an example if $R$ is a local ring?

user26857
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Zac
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2 Answers2

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There exists a von Neumann regular ring ("absolutely flat ring", and so it has weak dimension $0$) which has infinite global dimension, so the answer to the first question is "yes." See this mathoverflow post which refers to an interesting article by Osofsky about $\prod_\kappa F$.

The ring is nowhere near local, of course. I do not have an example or proof handy for that case. A local ring with weak dimension $0$ is already a division ring, so it'll have to be something with positive weak dimension.

rschwieb
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The answer of rschweib refers to a result of Osofsky that gives a ring with infinite global dimension but finite weak global dimension.

Another result of Osofsky gives a local example.

In Corollary 2 of

Osofsky, B. L., Global dimension of valuation rings, Trans. Am. Math. Soc. 127, 136-149 (1967). ZBL0145.27602

she shows that there is a valuation domain $R$ with infinite global dimension.

But valuation domains are precisely the commutative local rings of weak global dimension at most one. See, for example, Corollary 4.2.6 of

Glaz, Sarah, Commutative coherent rings, Lecture Notes in Mathematics. 1371. Berlin etc.: Springer-Verlag. xi, 347 p. (1989). ZBL0745.13004.

  • valuation domains are precisely the commutative local rings of weak global dimension at most one really? Wow! That’s neat! – rschwieb May 14 '22 at 17:10
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    @rschwieb Yes, it is. It's not that hard to prove, but I struggled to find a source that gave a statement and proof without proving more. Maybe an easier sourse from which to distill a proof is https://stacks.math.columbia.edu/tag/092S. – Jeremy Rickard May 15 '22 at 19:00