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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?

Mario Carneiro
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  • Free over what? Free over $I$, maybe? If that would be so then $M^{I}$ and $M^{\oplus I}$ would have to be isomorphic, wich is not the case if $I$ is not finite. – drhab Oct 03 '15 at 07:46
  • @drhab Free over anything. As I mentioned, $I$ is a basis for $M^I$ if and only if $I$ is finite, so clearly you need more, and if the solutions look anything like the Hamel bases in vector spaces, the basis will not be easily describable and will need recourse to the axiom of choice. You can probably choose the basis so that it contains $I$, but really I'm looking for any basis at all. There are various obstructions that cause some explicitly definable modules to be non-free, and the real question is whether those obstructions show up for this particular kind of module. – Mario Carneiro Oct 03 '15 at 07:48

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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.

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Ben
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