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Let $A$ be a commutative ring (viewed as an $A$-module over itself) that is not a field. Are there some conditions that guarantee that $\prod_{k=0}^\infty A$ is free? What if $A=\mathbf{Z}$ or more generally any pid?

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Well, if $A$ is a field, then $\prod_\mathbb{N} A$ is certainly free. I claim that the direct product $\prod_\mathbb{N} \mathbb{Z}_4$ is also free, where $\mathbb{Z}_4 = \mathbb{Z}/4\mathbb{Z}$.

To prove this let $\{c_i\}_{i\in\mathcal{I}}$ be a basis for $\prod_\mathbb{N}\mathbb{Z}_2$, where $\mathcal{I}$ is some indexing set. For each $i$ let $b_i$ be the corresponding element of $\prod_\mathbb{N}\mathbb{Z}_4$. That is, the entries of $b_i$ are all $0$'s and $1$'s, and agree with the entries of $c_i$. I claim that $\{b_i\}_{i\in\mathcal{I}}$ is a basis for $\prod_\mathbb{N}\mathbb{Z}_4$.

To see this, consider, the following commutative diagram of abelian groups $$ \begin{array}{ccccc} \textstyle\bigoplus_\mathcal{I} \mathbb{Z}_2 & \xrightarrow{\times2} & \textstyle\bigoplus_\mathcal{I} \mathbb{Z}_4 & \longrightarrow & \textstyle\bigoplus_\mathcal{I} \mathbb{Z}_2 \\ \downarrow & & \downarrow & & \downarrow \\ \textstyle\prod_\mathbb{N} \mathbb{Z}_2 & \xrightarrow{\times2} & \textstyle\prod_\mathbb{N} \mathbb{Z}_4 & \longrightarrow & \textstyle\prod_\mathbb{N} \mathbb{Z}_2 \end{array} $$ where the first and third vertical arrows are determined by the $c_i$'s, and the middle one is determined by the $b_i$'s. Both rows are short exact sequences and the vertical arrows on the left and right are known to be isomorphisms, so the middle arrow is also an isomorphism by the short five lemma.

The same argument shows that $\prod_{\mathbb{N}} \mathbb{Z}/p^2\mathbb{Z}$ is free for any prime $p$. Moreover, we can iterate the argument to prove that $\prod_{\mathbb{N}} \mathbb{Z}/p^{k}\mathbb{Z}$ is free for any prime $p$, where $k$ is a power of $2$.

I have no idea whether $\prod_{\mathbb{N}}\mathbb{Z}/8\mathbb{Z}$ is free.

Jim Belk
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  • I have a question: where does "take a basis ${c_i}I$ of $\mathbb{Z}$, then we have an isomorphism $\bigoplus_I\mathbb{Z}\to\prod\mathbb{N}\mathbb{Z}$" fail? What's the difference with the $\mathbb{Z}_2$ case? – Daniel Robert-Nicoud Jan 16 '14 at 11:24
  • $\mathbb{Z}_2$ is a field, so all of its modules have bases. The other rings don't but Jim uses the short five lemma to move it up. I believe his argument also works for all $\mathbb{Z}/p^k\mathbb{Z}$ (the short exact sequence could be p->p^k->p^(k-1) rather than p^i->p^2i->p^i). – Jack Schmidt Jan 16 '14 at 13:11
  • So I think this works for any artinian local ring. What about the p-adic integers, $\hat {\mathbb{Z}}_p$? – Jack Schmidt Jan 16 '14 at 14:12
  • @JackSchmidt I'm not convinced that the argument works for $p \to p^k \to p^{k-1}$. How would you make sure that the diagram commutes? – Jim Belk Jan 16 '14 at 20:37
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    @JimBelk: I haven't had time to write it down carefully, but I didn't see any difference. However, Ed Enochs mentioned a short proof for $\mathbb{Z}/n\mathbb{Z}$, $n\neq 0$: Over such a ring injective=projective, direct products of injectives are injective. Since we can assume $n$ is a prime power by euclid's algorithm, we can assume projective=free, which finishes it. This does not handle $k[x,y]/(x,y)^n$. – Jack Schmidt Jan 17 '14 at 16:19
  • @JackSchmidt Very nice! – Jim Belk Jan 17 '14 at 22:14
  • Ed also handled artinian local. In the non-local case, I believe he has an example with the direct product projective but not free. I have to brush up on my set theory to verify it. Let me know if we want another question opened for answers like this. I don't think it interests the OP. – Jack Schmidt Jan 21 '14 at 15:37