Can the I-fold direct product be free?

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

Can $\prod_{i\in I}A$ be free as an $A$-module?

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 you sketch the proof for $\mathbb{Z}/m$? We may reduce to a prime power. – Martin Brandenburg Dec 10 '10 at 9:49
• Every $\mathbb Z/p^k\mathbb Z$-module is a direct sum of cyclic modules: The fact that for this ring a free module is injective allows you to split any module as $F\oplus T$ where $F$ is free and $T$ is killed by $p^{k-1}$. Induct on $k$. – Tom Goodwillie Dec 10 '10 at 14:22

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.

A common generalisation covering the two examples of (1) is an Artinian self-injective (local) algebra $A$. Then the product is injective and any injective is a sum of indecomposable injectives and the only indecomposable injective is $A$ itself.

I believe it is also free for the p-adic integers $\mathbb{Z}_p$.

By restricting to the underlying additive group $G$ of the ring, you certainly require any Cartesian product of copies of $G$ to be isomorphic to some direct sum of copies of $G$, and this itself is an interesting question. For finitely generated abelian groups, I believe it holds if and only if the group has rank 0 (i.e. is torsion), this should follow easily from things you've mentioned already.

(Incidentally, in the world of non-abelian groups, I'm not sure if it holds for $Q_8$ or $D_8$.)

• @ndkrempel believes it to be true due to a result I have. Unfortunately I'm currently looking for a result I may have cited in the proof and have not yet found it. I believe my answer holds, but am not certain. – Jonathan Kiehlmann Dec 10 '10 at 11:35
• Dear ndkrempel: Here is a proof that it is not free when $A$ is a complete dvr. An infinite direct sum of copies of a complete dvr is not complete for the max-adic topology, whereas an arbitrary direct product of copies of a complete dvr is complete as a module (for the max-adic topology). Hence, an infinite direct product of copies of a complete dvr cannot be a free module. The same argument works (with the same conclusion) for any noetherian ring $A$ that is complete and separated with respect to a non-nilpotent ideal. – BCnrd Dec 10 '10 at 13:40
• @BCnrd: Thank you, very interesting! I've edited accordingly. This may be sad news for Jonathan Kiehlmann (from whom I got the result originally.) – ndkrempel Dec 10 '10 at 13:51
• @BCnrd: You should post this as an answer. – Martin Brandenburg Dec 10 '10 at 14:15
• Dear ndkrempel: Are you asking if an infinite direct product of copies of $\mathbf{Z}_p$ can be isomorphic to a direct sum of copies of $\mathbf{Z}_p$ as abelian groups? If that is the question, the answer is no, by the same argument. Indeed, the property of a $\mathbf{Z}_p$-module being $p$-adically separated and complete only depends on the underlying abelian group. So there is no such isomorphism as abelian groups either. As an aside, it seems more useful to consider flatness properties of huge direct products rather than freeness properties. – BCnrd Dec 10 '10 at 20:39