Let $A$ be a non-zero commutative ring with unit, $I$ a infinite set.



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> Can $\prod\_{i\in I}A$ be free as an $A$-module?
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I found when $A$ is a field or is isomophic to $\mathbb{Z}/m\mathbb{Z}$, then it is free.


But even when $A=\mathbb{Z}$, it is not free. (Baer Specker group)


It seems $\prod\_{i\in I}A$ is always not free when $A$ is a domain with dim$\geq1$?


But I've found it difficult to prove. So I want to prove that when $A$ is PID, $\prod\_{i\in I}A$ is always not free.
It is suffice to prove that when $I=\mathbb{N}$, $\prod\_{i\in \mathbb{N}}A$ is not free.


we have already checked that when $A$ is not DVR,then $\prod\_{i\in \mathbb{N}}A$ is not free. (similar to $\mathbb{Z}$)


we remain DVR need to check. 


My question is :


1. Is any other $A$ such that $\prod\_{i\in I}A$ is free as an $A$-module ?
2. Is $\prod\_{i\in I}A$ always NOT free when $A$ is an integral domain with dim$\geq1$?



Can the I-fold direct product be free?
If $A$ is a noetherian domain and not a field then the infinite product $M=A\times A\times \dots$ is not free. Suppose there is a basis. For $x\in M$ define its support to be the finite set of basis elements for which the coefficient is not zero. Note that if the supports of $x$ and $y$ are disjoint then their union is the support of $x+y$. Choose $\pi\in A$ neither zero nor invertible. Define the $n$-support of $x$ to consist of those basis elements for which the coefficient is not divisible by $\pi^n$. Note that $n$-support is contained in $(n+1)$-support is contained in support.


Choose an infinite sequence of nonzero elements $m\_1,m\_2,\dots $ of $M$ such that 


(1) $m\_n$ projects to zero in the first $n-1$ factors of the infinite product,


(2) the $m\_n$ have pairwise disjoint support.


To get $m\_n$ when all the previous $m\_k$ are given, you just have to know that the kernel of a certain map from $M$ to a finite product of copies of $A$ is nontrivial (project the product on the first $n-1$ factors and project the free module on the span of a finite subset of basis).


Then divide each $m\_n$ by as high a power of $\pi$ as possible; this preserves 1 and 2 while also arranging


(3) $m\_n$ is nonzero mod $\pi$.


Now let $s\_n=\pi m\_1+\pi^2 m\_2+\dots +\pi^nm\_n$ and let $s$ be the limit of $s\_n$ (defined because of 1). 


The contradiction is that the support of $s$ must contain arbitrarily large finite sets $S\_n$: Let $S\_n$ be the $(n+1)$-support of $s\_n$. Then the support of $s$ contains the $(n+1)$-support of $s$, which equals $S\_n$. And $S\_n$ properly contains $S\_{n-1}$ because it is the disjoint union of the $(n+1)$-support of $s\_{n-1}$ and the $(n+1)$-support of $\pi^nm\_n$, this last being the (by 3 nonempty) $1$-support of $m\_n$.


EDIT This implies that if $A$ is noetherian and has dimension $>0$ then the infinite product is not free, because $(A/P)\otimes \prod A=\prod (A/P)$ if $P$ is a finitely generated ideal -- choose $P$ to be a non-maximal prime. Also, the argument above proves more than I said: for a noetherian domain the infinite product is not even a submodule of a free module.


