When two infinite direct products $\prod_I\Bbb{Z}$ and $ \prod_J\Bbb{Z}$ are isomorphic?

Question

It is known that two free $\Bbb{Z}$-modules $\bigoplus_{I}\Bbb{Z}$ and $\bigoplus_{J}\Bbb{Z}$ are isomorphic if and only if $|I|=|J|$. Moreover, it is true for any two free module over a commutative ring $R$. Now, a natural question in this context is "if we replace the direct sum $\bigoplus$ with the direct product $\prod$ is the ruling still true?"

Thus, I have the following quation on infinite direct product of $\Bbb{Z}$ as $\Bbb{Z}$-module.

Question. Suppose that $I$ and $J$ are two (non-measurable) infinite cardinals, such that $|\prod_I\Bbb{Z}|=|\prod_J\Bbb{Z}|$. Can we conclude that $\prod_I\Bbb{Z}\cong \prod_J\Bbb{Z}$ as $\Bbb{Z}$-modules?

Answer 1

The cardinality claim is just saying that $2^{|I|} = 2^{|J|}$. If we assume, for instance, the generalized continuum hypothesis, then the map $\kappa \mapsto 2^\kappa$ is injective, so we can conclude that $|I| = |J|$, from which the isomorphism of modules easily follows. I don’t know whether we can solve this problem without making such an assumption.

Edit: in Mahmood Behboodi’s answer below, he cites the literature saying that this isomorphism occurs if and only if $|I| = |J|$. With this piece of information, I can now say the problem is independent of ZFC. For suppose we have ZFC + Martin’s Axiom + $\neg$ the continuum hypothesis. Then we have $2^{\aleph_0} = 2^{\aleph_1}$, but $\aleph_0 \neq \aleph_1$ (and both cardinals, not being inaccessible, are also not measurable).

Answer 2

Professor Alberto Facchini sent me the following note.

Those two groups are isomorphic if and only if I and J have the same cardinality. This appears in the first lines of Section 5 of https://dept.math.lsa.umich.edu/~ablass/sets.pdf and in reference [1] of that paper. Best regards. Alberto Facchini

Answer 3

For $i\in I$, let $e_i$ denote the element of $\mathbb{Z}^I$ whose $i$th coordinate is $1$ and all other coordinates are $0$. By the Łoś–Eda Theorem, $\operatorname{Hom}(\mathbb{Z}^I,\mathbb{Z})$ is generated by the homomorphisms given by the countably complete ultrafilters on $I$ (if $U$ is a countably complete ultrafilter on $I$, it gives a homomorphism $\mathbb{Z}^I\to\mathbb{Z}$ that takes $f:I\to\mathbb{Z}$ to the unique $n$ such that $f^{-1}(\{n\})\in U$). In particular, then, any homomorphism $\mathbb{Z}^I\to\mathbb{Z}$ vanishes on all but finitely many of the $e_i$ (the nonprincipal ultrafilters vanish on every $e_i$ and each principal ultrafilter vanishes on all but one $e_i$). It follows that any homomorphism $\mathbb{Z}^I\to\mathbb{Z}^J$ (for $J$ infinite) vanishes on all but at most $|J|$ of the $e_i$. So in particular, such a homomorphism cannot be injective if $|J|<|I|$.