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If $A$ and $B$ are two unital rings such that $A \times A \cong B \times B$, as rings, does it follows that $A$ and $B$ are isomorphic (as rings)?

I believe that the answer is no, but I can't come up with a counterexample. A similar question for groups has already been asked - the answer is not straightforward. Here is a possibly related question, but there are $R$-modules isomorphisms.

[If $A$ and $B$ are fields, then we can see $B^2$ as a $2$-dimensional $A$-vector space, so that $A \cong B$ as $A$-vector spaces, because they have the same dimension. I may be wrong about this, but anyway this is not sufficient to get a field isomorphism.]

Thank you for your comments!

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Watson
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    For an abelian group $G$, let $R(G)$ be the ring with underlying additive group $R(G)^+ = G$ and multiplication $g.h = 0$ for $g, h\in G$. Would that reduce this problem to the one for abelian groups you linked to? – anomaly Jul 24 '16 at 14:24
  • No because $A$ and be should be isomorphic as rings, not just as additive abelian groups. Edit: I misunderstood your idea, the structure you mentioned isn't even a ring – a25bedc5-3d09-41b8-82fb-ea6c353d75ae Jul 24 '16 at 14:26
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    What about something like taking the groups ring $\mathbb{Z}[G]$ and $\mathbb{Z}[H]$? What you would really want is a functor from groups to rings which is injective on objects and commutes with taking products. – Dan Rust Jul 24 '16 at 14:32
  • We could try repeating the $A\times A=A\times A\times A$ trick for rings – a25bedc5-3d09-41b8-82fb-ea6c353d75ae Jul 24 '16 at 15:31
  • Relatad on MO: http://mathoverflow.net/questions/22899/non-isomorphic-groups-with-isomorphic-nth-powers-and-similarly-in-other-categor?rq=1 – Watson Aug 21 '16 at 12:56

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In this answer (and with more detail in this answer) I gave an example, due to Sundaresan, of a compact Hausdorff space $X$ such that if $Y$ and $Z$ are the results of adding one and two isolated points, respectively, to $X$, then $X\sim Z\not\sim Y$, where $\sim$ denotes homeomorphism. Thus, $X\sqcup X\sim X\sqcup Z\sim Y\sqcup Y$, even though $X\not\sim Y$. Let $A=C(X)$ and $B=C(Y)$; then

$$A\times A\cong C(X\sqcup X)\cong C(Y\sqcup Y)\cong B\times B\;,$$

but $A\not\cong B$.

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Brian M. Scott
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    Why the isomorphism $X\sqcup Z\sim Y\sqcup Y$ holds? – user267839 Feb 18 '18 at 19:59
  • Related: https://math.stackexchange.com/questions/226736 – Watson Nov 26 '18 at 13:08
  • @KarlPeter: Follow the links in the answer: it depends very much on the particular space $X$. – Brian M. Scott Feb 03 '21 at 02:20
  • This also gives an example of a ring $B$ with $B\cong B^3$ but $B\not\cong B^2$, since in fact $X\sqcup X\cong X$ so $Y\sqcup Y\cong X\not\cong Y$ but $Y\sqcup Y\sqcup Y\cong X\sqcup Y\cong X\sqcup X\sqcup {}\cong X\sqcup{}\cong Y$. – Eric Wofsey Feb 24 '23 at 20:47
  • Also, since these compact Hausdorff spaces are totally disconnected, instead of $C(X)$ and $C(Y)$ you could take their Boolean algebras of clopen subsets. – Eric Wofsey Feb 24 '23 at 20:49
  • Also, to make your example more concrete, the ring $A$ can be described as the ring of pairs $((a_n),(b_n))$ of bounded sequences of real numbers such that $(a_n-b_n)$ converges to $0$, and $B=A\times\mathbb{R}$. (If you use the Boolean algebras of clopen subsets instead, this corresponds to using $\mathbb{F}_2$-valued sequences instead of real-valued sequences.) – Eric Wofsey Feb 24 '23 at 21:38