# Basic rings (e.g. non commutative) $A$ such $A^n \simeq A^m$ and $n\neq m$

EDIT : precision and broadening of my question.

Almost all is in the title : I am looking for various structures $A$ such $A^n$ and $A^m$ (products of $A$) are isomorphic (in the sense that it is compatible with at least one structure of $A$, e.g. group or ring if $A$ is a ring...) and moreover $n\neq m$

For example, I'm looking for groups, rings, modules, vector spaces (when the sets at stake exist), whichever commutative or not. Simple and basic examples are welcome, I am not researcher...

Extension possible of the question to $A^{(I)} \simeq A^{(J)}$ (external direct sums for structures for which this makes sense) with $card(I)\neq card(J)$.

This question may be a classic one but I can't find the answer neither in book nor here. Please redirect me if possible...

Thanks

Edit bis : references are welcome (it doesn't mean I dont trust answers, but often, books have very interesting comments before or after examples that can't always be put in the answers here... ;))

• By $A^n$ do you mean the usual product $\underbrace{A \times \cdots \times A}_n$? – Travis May 7 '15 at 13:34
• "commutative, non commutative, modules..."?! This does not make sense in that list. Please clarfiy the question. – quid May 7 '15 at 13:36
• Yes, please write complete sentences for your question. Stack Exchange has a lot of disk space, so refusing to type complete sentences just makes it harder for answerers to help you. – Thomas Andrews May 7 '15 at 13:38
• The question is answered for rings; it takes a little bit more work to find an example of a ring $A$ for which $A^n \simeq A^m$ as left $A$-modules, for $n \neq m$. You should clarify whether or not that's a part of your question. – Dustan Levenstein May 7 '15 at 13:39
• OK, sorry, I thought it was clear enough. I'm going to detail :). – Carré rond May 7 '15 at 13:52

The two standard examples (given in multiple places on the site, but perhaps not easy to find directly) are these:

If $F$ is a field, and $R=\prod_{i\in \Bbb N}F$ is the ring direct product, then $R\cong R^m$ as rings for whatever positive integer $m$ you wish.

On the other hand, it is impossible for $R^m\cong R^n$ with $n\neq m$ as modules for a nonzero commutative ring $R$. The standard example to find a ring $R$ to allow such an isomorphism is the ring of linear transformations $End(V_k)$ of an infinite dimensional $k$-vector space $V$. It is possible to show that $R\cong R^m$ as $R$ modules for any positive integer $m$ in this ring.

• OK, I'm looking at that also. Do you have hints for the demonstration of these examples, or references (books, articles...), please ? – Carré rond May 7 '15 at 14:48
• I found the second example well explained in Roman, Advanced Linear Algebra, example 5.1 p. 129. :-) – Carré rond May 14 '15 at 20:02

Well it may be very simple but if you take for $A_n:=\mathbb{Z}_2$ (the field with two elements) then :

$$B:=\Pi_{n\in\mathbb{N}}A_n$$

The laws on this ring $B$ are direct product laws.

$$A:=\oplus_{n\in\mathbb{N}}A_n$$

The ring $A$ is defined to be the subset of $B$ whose all components except a finite number are null. One can show that it is both stable by addition and multiplication (Actually it is an ideal of $B$) but it does not contain the unit element.

You can easily show that $A^n=A^m$ and $B^n=B^m$ for any $n,m$.

• Never "too" simple, I'm going to study that. Merci ;). – Carré rond May 7 '15 at 14:08
• Perhaps you could clarify what structures (as rings or as modules etc) you want things things to be considered as. Right now it's a little ambiguous what you intend your examples to do. – rschwieb May 7 '15 at 14:11
• It's a little tricky... I keep on thinking about that ! – Carré rond May 14 '15 at 20:03