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Let $k$ be a field.

If $I$ is a finite index set, then the global dimension of the ring $\prod_I k$ is zero because $\prod_I k$ is a semisimple ring.

Now, the question arises: When $I$ is an infinite index set, what is the global dimension of the ring $\prod_I k$?

For instance, when $ I = $ $\aleph_1 $, what is $ \mathrm{gl. dim} \prod_{\aleph_1} k $?

Martin Brandenburg
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Liang Chen
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    I would not be surprised if this is not decidable in ZFC, because of this. – Martin Brandenburg Nov 10 '24 at 15:16
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    Actually, now I see that the wiki page already answers your question for countable $I$. It says that then $k^I$ has global dimension $2$ iff the continuum hypothesis holds. This of course is compatible with Jeremy's answer. – Martin Brandenburg Nov 10 '24 at 21:13

1 Answers1

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By Theorem 2.51 of

Osofsky, Barbara L., Homological dimensions of modules, Conference Board of the Mathematical Sciences. Regional Conference Series in Mathematics. No. 12. Providence, R.I.: American Mathematical Society (AMS). VIII,89 p. (1973). ZBL0254.13015,

the global dimension of $\prod_Ik$ is $d+1$ if $\vert2^I\vert=\aleph_d$ (and infinite if $\vert2^I\vert>\aleph_d$ for all $d \in \mathbb{N}$).

So if the generalized continuum hypothesis is true, and $\vert I\vert=\aleph_1$, then the global dimension of $\prod_Ik$ is $3$.

Martin Brandenburg
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Jeremy Rickard
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