Infinite direct product of rings free.

Question

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?

Answer 1

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.