Direct product of projective modules

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.

• 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. – Hanno May 14 '17 at 7:45
• Is this preposition is true? Any $\mathbb Z$-module is projective iff free? – B.K-Theory May 14 '17 at 7:50
• Yes, for $\mathbb Z$ that's true. – Hanno May 14 '17 at 7:51
• Could you explain more about second part of your first comment? – B.K-Theory May 14 '17 at 7:57
• For any $\alpha\in I$ you have $\prod_{\beta\in I} M_\beta \cong M_\alpha\times\prod_{\beta\neq\alpha} M_\beta\cong M_\alpha\oplus\prod_{\beta\neq\alpha} M_\beta$. – 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. (“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.