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

If $\{M_i\}_{i \in I}$ is a family of free $R$-modules, then $\bigoplus_{i \in I}M_i$ is free.

Is this true for the product $\prod_{i \in I}M_i$ too?

• $\mathbb{Z}^\mathbb{N}$ over $\mathbb{Z}$ is probably the simplest and well-known counterexample. Jun 29, 2014 at 12:33
• Feb 13, 2018 at 17:09
• Feb 24, 2018 at 10:29

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

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.

• does have some counterexample easier? Jun 29, 2014 at 12:17
• I don't think there is an easier counterexample. Abelian groups are the simplest modules, and if you work with modules over $Z_n$, say, or with modules over a field, then it IS true that the infinite product of free modules is free. So my bet is that there is no simpler counterexample, although I may be wrong on this Jun 29, 2014 at 12:21
• Are you sure about $\mathbb{Z}/n\mathbb{Z}$-modules? Jun 29, 2014 at 12:32
• Yes. This follows from the fact that every abelian group of bounded exponent is isomorphic to the direct sum of cyclic groups (not easy). Jun 29, 2014 at 12:57