If $\{M_i\}_{i \in I}$ is a family of $R$-modules free, then the product $\prod_{i \in I}M_i$ is free?

Solution 1:

By a theorem of Chase (see Anderson-Fuller, theorem 19.20), the following conditions are equivalent:

1. every direct product of flat right $R$-modules is flat,

2. the right $R$ module $R^X$ (direct product) is flat for every set $X$,

3. $R$ is left coherent.

As a corollary, one can deduce (see note after Corollary 28.9 in Anderson-Fuller) that every direct product of projective (or free) right $R$ modules is projective if and only if $R$ is right perfect and left coherent. In particular, products of projective (free) right $R$-modules is projective whenever $R$ is left artinian.

(Note that a direct product of projective modules is a direct summand of a direct product of free modules, so in the corollary it's immaterial whether we say “every direct product of projective modules is projective” or “every direct product of free modules is projective”.)

Since, for example, $\mathbb{Z}$ is not right perfect, there is a set $X$ such that $\mathbb{Z}^X$ (direct product) is not even projective. Of course, when $R^X$ is not projective, then $R^Y$ isn't projective for any set $Y$ such that $|Y|\ge |X|$, because $R^X$ is easily seen to be a direct summand of $R^Y$. Since (not easy, see the already linked answer on MathOverflow) a countable product of copies of $\mathbb{Z}$ is not free (projective and free is the same for abelian groups), no infinite direct product of free abelian groups is free.

Reference

Anderson and Fuller, Rings and Categories of Modules, second edition, Springer, 1992

Solution 2:

The product of free modules need not be free. Even the infinite product of free abelian groups may not be free, although this is not easy. See this answer on MO.