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Let $k$ be a field. Consider an infinite direct product of rings $\prod k$. This is an example of Von-Neumann regular ring (name also absolutely flat), that is, every module is flat.

I think this ring is nice! I have the following question:

1. Is $\prod k$ self-injective?

2. What is the global dimension of $\prod k$ ? Finite or infinite? I have the feeling that it is infinite.

I don't know how to deal with this question. Any help will be appreciated.

user26857
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Jian
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  • I only know for sure it is $>1$ (otherwise it would be hereditary+self-injective, which implies Artinian.) . You can prove it has infinite global dimension if you show that the projective dimensions of its cyclic modules are unbounded. – rschwieb May 04 '18 at 13:37
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    Actually here I see it claimed that $\prod_{i=1}^\infty \mathbb C$ has global dimension $2$! – rschwieb May 04 '18 at 13:39
  • @rschwieb how to show if R is hereditary and self-injective,then R is Artinian?thanks – Jian May 04 '18 at 15:10
  • Osofsky, Barbara. "Rings all of whose finitely generated modules are injective." Pacific Journal of Mathematics 14.2 (1964): 645-650. Corollary to the main Theorem. It seems to be open access at Project Euclid – rschwieb May 04 '18 at 15:15

2 Answers2

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The answer to question $1$ is that any product of fields is self-injective.

Actually, any product of self-injective rings is self-injective. See Self-injective ring on the Encyclopedia of Mathematics.

Bernard
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  • I was wondering why the direct product of self-injective rings remains self-injective. If possible, could you recommend a reference with a detailed proof? – Liang Chen Nov 12 '24 at 12:44
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The global dimension of $\prod_\kappa F$ depends on the cardinalities $\kappa$ and $|F|$. It can be both finite or infinite. You will find very interesting material on that topic on this MO post.

(And yes as previously noted, any product of right self-injective rings is right self-injective.)

user26857
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rschwieb
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  • Since the direct product of self-injective rings is also self-injective, the direct product of fields is therefore self-injective. However, the global dimension of self-injective rings is either 0 or infinite. So why would the global dimension of a direct product of fields be 2 on this MO post? – Liang Chen Nov 10 '24 at 10:32
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    @LiangChen I believe you found your answer in the intervening 16 hours. For other readers, the problem is that self-injective rings do not necessarily have that property. – rschwieb Nov 11 '24 at 03:29
  • Yes, thank you very much for your attention. – Liang Chen Nov 11 '24 at 04:02