# Does there exist a non-trival set $W$ such that $W\times W \cong W\times W\times W$?

1. Does there exist a non-trival set $W$ such that $W\times W \cong W\times W\times W$ where $\times$ is the exterior product?

2. Consider $R^2$ and $R^3$, I don't think they were isomorphism to each other, but I wasn't sure how to prove their order(i.e. suppose $|R|=\omega$)

3. Could you find a subspace $W$ of $R$ or $C$(fields) such that $W$ was the smallest (in order) non trival set such that $W\times W \cong W\times W\times W$?

• $\cong$ here meant for bijection, and was used to emphasize the another question: if $W$ was a field, could the bijection also be a ring homomorphism?
• What do yuo mean by $\cong$, exactly? – Saucy O'Path Aug 15 '18 at 15:12
• The product of two infinite numerable sets is numerable. – Tsemo Aristide Aug 15 '18 at 15:12
• If $W$ is a field, then $W \times W$ and $W \times W \times W$ are vector spaces. There would be no vector space isomorphism because they have a different dimension over $W$, hence no ring isomorphism. – Joe Johnson 126 Aug 15 '18 at 15:17
• Yes. If $X$ is an infinite set, then there is a set bijection between $X$ and $X^n$ for any $n \in \mathbb{N}$. – Joe Johnson 126 Aug 15 '18 at 15:22
• @MaliceVidrine thank you. Existence of the bijection first, and then the isomorphism. Joe Johnson 126 has answered the question about the isomorphism already. – user416486 Aug 15 '18 at 15:25

In the category of sets, any bijection $A \to A \times A$ is an isomorphism. In particular, take $A= \mathbb N$. Then define $f:\mathbb N \times \mathbb N \to \mathbb N$ by $(n,m) \mapsto 2^n \cdot 3^m$.

If $A$ is a field, there are some problems. For example, if you define $(a,b) \cdot (c,d)=(ac,bd)$, then $A \times A$ is most certainly not a field, since $(1,0) \cdot (0,1)=0$.

One way to "make this work" is that if you work in the category of $R$-modules, and consider $\otimes$ instead (which is reasonable, I think) then you can take $R \otimes_R R \cong R$ to be $(a \otimes b) \mapsto ab$ which is always an isomorphism of modules.

• Thank you. math.stackexchange.com/questions/2000739/… Patrick Stevens's answer provided a solution on $N$ by using Cantor pairing function, could it also be used on $R$? Further, what if $W$ is an non ordered field(A subspace in $C$)? – user416486 Aug 15 '18 at 15:50

The problem of cancelletion of the direct product have been deeply studied and in most of cases finds an answer in the Krull-Remak-Schmidt Theorem.

A list of relevant links is the following:

On cancellation in Groups.