Given a set $I$ and an $R$-module $M$, define the complete module product $M^I$ to be the set of all functions $I\to M$, with addition and scalar multiplication done componentwise. By contrast, given a set $I$ we can also form the free module on $I$, which is usually also denoted $M^I$ but I will denote by $M^{\oplus I}$ here for disambiguation, which is the submodule of $M^I$ consisting of functions $x:I\to M$ such that $x_\alpha=0$ for all but finitely many $\alpha\in I$. The question is:
Is the complete module product free? (where free means isomorphic to a free module)
Obviously if $I$ is finite then $M^{\oplus I}=M^I$ so the answer is yes. But if $I$ is infinite $I$ is no longer a basis for $M^I$. There is a theorem in linear algebra to the effect that every vector space has a basis, or every vector space is free, so if $M$ is a vector space then $M^I$ is as well and hence the answer is again yes. Are there counterexamples, then, with infinite dimensional module products?
In general, the answer is no. See here for example. There it is shown that $\mathbb {Z}^{\mathbb {N}}$ is not free as $\mathbb {Z} $-module.