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I am viewing the sets above as $\mathbb{Z}$-modules. It is easy to show that they are not canonically isomorphic: we can view the elements on each side as $\mathbb{N} \times \mathbb{N}$ matrices with entries in $\mathbb{Z}$. The elements of $\prod_{\mathbb{N}} \bigoplus_{\mathbb{N}} \mathbb{Z}$ are such that each row is finitely supported. The elements of $\bigoplus_{\mathbb{N}} \prod_{\mathbb{N}} \mathbb{Z}$ are supported on finitely many rows. The transpose map embeds $\bigoplus_{\mathbb{N}} \prod_{\mathbb{N}} \mathbb{Z}$ into $\prod_{\mathbb{N}} \bigoplus_{\mathbb{N}} \mathbb{Z}$, but this map is clearly not surjective.

I am wondering if there is a non-canonical isomorphism between these two modules. Intuitively, I feel as though these groups are different, but I cannot show this. I think that if they are isomorphic, then the axiom of choice will be needed, but again I haven't found an argument. For example, if we replaced $\mathbb{Z}$ with $\mathbb{Q}$, then there are plenty of non-canonical isomorphisms, since we could just use the fact that each are infinite-dimensional $\mathbb{Q}$-vector spaces with the same cardinality and appeal to the axiom of choice. But over $\mathbb{Z}$ I don't know...

SFSH
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  • If you are trying to prove that they are not isomorphic, maybe you can look at their image from some functor. For example, $Hom(\oplus \prod \mathbb Z,\mathbb Z) = \oplus_{\mathbb Z \times \mathbb Z} \mathbb Z$ is countably generated – Aitor Iribar Lopez Oct 26 '21 at 23:25
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    @AitorIribarLopez Good idea, but isn't $\mathrm{Hom}(\oplus \prod \mathbb{Z},\mathbb{Z}) \simeq \prod \mathrm{Hom}(\prod \mathbb{Z},\mathbb{Z}) \simeq \prod \oplus \mathbb{Z}$? – Minseon Shin Oct 27 '21 at 01:11
  • I think you have the descriptions of the two groups as $\mathbb{N}\times\mathbb{N}$ matrices the wrong way round? – Jeremy Rickard Oct 27 '21 at 16:25
  • Yes thank you, fixed. – SFSH Oct 28 '21 at 23:42

1 Answers1

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They are not isomorphic.

In what follows, $\bigoplus$ and $\prod$ will denote direct sums or products of copies of groups indexed by $\mathbb{N}$, and $e_{n}$ will denote the element of $\prod\mathbb{Z}$ whose $n$th coordinate is $1$, with all other coordinates zero.

I will show that there is no surjective homomorphism $\bigoplus\prod\mathbb{Z}\to\prod\bigoplus\mathbb{Z}$.

An abelian group $G$ is called slender if, for every homomorphism $\alpha:\prod\mathbb{Z}\to G$, $\alpha(e_{n})=0$ for all but finitely many $n$. The following properties of slender groups are standard (see, for example, Section 13.2 of Fuchs's 2015 book Abelian Groups).

  • $\mathbb{Z}$ is slender (this is just the well-known theorem of Specker).

  • (Infinite) direct sums of slender groups are slender. So in particular $\bigoplus\mathbb{Z}$ is slender.

  • The image of any homomorphism from $\prod\mathbb{Z}$ to a slender group is a finite rank free abelian group.

So for any homomorphism $\alpha:\prod\mathbb{Z}\to\bigoplus\mathbb{Z}$, there is some $n$ such that the image of $\alpha$ contains only elements of $\bigoplus\mathbb{Z}$ whose $m$th coordinates are zero for $m\geq n$. Call the least such $n$ the type of $\alpha$.

A homomorphism $\beta:\prod\mathbb{Z}\to\prod\bigoplus\mathbb{Z}$ consists of a sequence of homomorphisms $\beta_{i}:\prod\mathbb{Z}\to\bigoplus\mathbb{Z}$. If $\beta_{i}$ has type $n_{i}$, we'll say that $\beta$ has type $(n_{0},n_{1},n_{2},\dots)$.

A homomorphism $\gamma:\bigoplus\prod\mathbb{Z}\to\prod\bigoplus\mathbb{Z}$ consists of a sequence of homomorphisms $\gamma_{i}:\prod\mathbb{Z}\to\prod\bigoplus\mathbb{Z}$.

Suppose $\gamma_{i}$ has type $(n_{i,0},n_{i,1},n_{i,2},\dots)$. Then every element of the image of $\gamma$, considered as an $\mathbb{N}\times\mathbb{N}$ matrix with finitely supported rows, has the following property:

  • There is some $k\in\mathbb{N}$ such that, for every $j$, the $m$th coordinates of the $j$th row are zero for $$m\geq\max(n_{0,j},n_{1,j},\dots,n_{k,j}).$$

But one can easily construct elements of $\prod\bigoplus\mathbb{Z}$ that do not have this property, and so $\gamma$ cannot be surjective, and in particular cannot be an isomorphism. For example, take an $\mathbb{N}\times\mathbb{N}$ matrix where the $j$th row has a single nonzero entry, in the $N_{j}$th column, where $N_{j}=\max(n_{0,j},n_{1,j},\dots,n_{j,j})$.

In fact, something more general is true. Contained in Corollary 3 of

Zimmermann-Huisgen, Birge, On Fuchs’ problem 76, J. Reine Angew. Math. 309, 86-91 (1979). ZBL0408.20038.

is the fact that the groups $\bigoplus\mathbb{Z},\;\prod\mathbb{Z},\;\bigoplus\prod\mathbb{Z},\;\prod\bigoplus\mathbb{Z},\;\bigoplus\prod\bigoplus\mathbb{Z},\;\prod\bigoplus\prod\mathbb{Z},\dots$ are pairwise nonisomorphic.

Jeremy Rickard
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