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Is there an infinite family $\lbrace R_\alpha\rbrace_\alpha $ of rings (with identity $1\neq 0$) such that their direct product is a hereditary ring ?

I think the answer must be negative but i have no proof or counterexample yet.

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  • $\begingroup$ I have no time right now, but if I did have time the first place I'd look is T.Y. Lam's Lectures on Modules and Rings. I think there are a number of pages available through Google Books. The answer might be there $\endgroup$
    – David White
    Commented Dec 29, 2012 at 19:37
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    $\begingroup$ @David White: Exactly, its always true that there are a number of pages available through Google Books ! $\endgroup$
    – user30230
    Commented Dec 29, 2012 at 21:33

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As @Jeremy Rickard has mentioned, the impossibility was shown by Osofsky in 1968. But it is interesting that there is another article dating back to 1968 (again!) that proves the same thing. See Cateforis, Sandomerski, The singular submodule splits off, J. Algebra, 10 (1968), 149-165, Theorem 4.1. This article was received by J. Algebra on November 20, 1967 while Osofsky's paper was received by Proc. Amer. Math. Soc. on July 10, 1967. I hope this information meets the expectations.

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Barbara Osofsky proves this is not possible (assuming the Axiom of Choice, I think) in "Noninjective cyclic modules", Proc. Amer. Math. Soc. 19 (1968), 1383-1384.

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Products of fields are semihereditary. This follows from the facts that products of fields are von Neumann regular and that von Neumann regular rings are semihereditary. A proof can be found (as suggested by David White) in Lam's Lectures on modules and rings, Example 2.32 d).

(In the above, fields and rings are not necessarily commutative.)

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  • $\begingroup$ @Fred Rohrer: But an infinite direct product of non-zero rings can not be countable so there is a gap in your argument for the hereditary case. $\endgroup$
    – user30230
    Commented Dec 29, 2012 at 21:52
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    $\begingroup$ But that von Neumann regular rings are semi-hereditary seems not to be in doubt, so that answers one question. $\endgroup$
    – Todd Trimble
    Commented Dec 29, 2012 at 21:54
  • $\begingroup$ Oh dear, let me edit my silly mistake. $\endgroup$
    – Fred Rohrer
    Commented Dec 29, 2012 at 21:57

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