For any nonempty family like $\{ {M_\alpha}\}_{\alpha \in I}$ of $R$-modules we know that if $\prod_{\alpha \in I} M_\alpha$ is a projective $R$-module,then for all $\alpha \in I$, $\{ {M_\alpha}\}_{\alpha \in I}$ is projective too. Is the converse true? If not give me a counter-example.

Any comments are welcome.

  • 1
    $\begingroup$ That's not true if $I$ is infinite; for example, you can take $R={\mathbb Z}$, $I={\mathbb N}$ and $M_\alpha={\mathbb Z}$ for all $\alpha$. However, if the direct product is projective, then so are its factors, because each of them is also a summand. $\endgroup$ – Hanno May 14 '17 at 7:45
  • $\begingroup$ Is this preposition is true? Any $\mathbb Z$-module is projective iff free? $\endgroup$ – B.K-Theory May 14 '17 at 7:50
  • $\begingroup$ Yes, for $\mathbb Z$ that's true. $\endgroup$ – Hanno May 14 '17 at 7:51
  • $\begingroup$ Could you explain more about second part of your first comment? $\endgroup$ – B.K-Theory May 14 '17 at 7:57
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    $\begingroup$ For any $\alpha\in I$ you have $\prod_{\beta\in I} M_\beta \cong M_\alpha\times\prod_{\beta\neq\alpha} M_\beta\cong M_\alpha\oplus\prod_{\beta\neq\alpha} M_\beta$. $\endgroup$ – Hanno May 14 '17 at 8:48

A theorem by S. U. Chase states that for a ring $R$ the following conditions are equivalent:

  1. $R$ is left perfect and right coherent

  2. every product of projective left $R$-modules is projective

It's Theorem 3.3 in S. U. Chase, Direct products of modules, Transactions of the American Mathematical Society, 97 (1960), 457–473.

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(“Finitely related” is nowadays more commonly referred to as “finitely presented”.)

So you just need to take a non left perfect ring and you have a counterexample: a suitable direct power of the regular module won't be projective.


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