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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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    $\begingroup$ 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? $\endgroup$
    – anomaly
    Commented Jul 24, 2016 at 14:24
  • $\begingroup$ 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 $\endgroup$
    – a25bedc5-3d09-41b8-82fb-ea6c353d75ae
    Commented Jul 24, 2016 at 14:26
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    $\begingroup$ 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. $\endgroup$
    – Dan Rust
    Commented Jul 24, 2016 at 14:32
  • $\begingroup$ We could try repeating the $A\times A=A\times A\times A$ trick for rings $\endgroup$
    – a25bedc5-3d09-41b8-82fb-ea6c353d75ae
    Commented Jul 24, 2016 at 15:31
  • $\begingroup$ Relatad on MO: mathoverflow.net/questions/22899/… $\endgroup$
    – Watson
    Commented Aug 21, 2016 at 12:56

1 Answer 1

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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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    $\begingroup$ Why the isomorphism $X\sqcup Z\sim Y\sqcup Y$ holds? $\endgroup$
    – user267839
    Commented Feb 18, 2018 at 19:59
  • $\begingroup$ Related: math.stackexchange.com/questions/226736 $\endgroup$
    – Watson
    Commented Nov 26, 2018 at 13:08
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    $\begingroup$ 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 $Y\sqcup Y\sqcup Y\cong X\sqcup Y\cong X\sqcup X\sqcup \{*\}\cong X\sqcup\{*\}\cong Y$. $\endgroup$
    – Eric Wofsey
    Commented Feb 24, 2023 at 20:47
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    $\begingroup$ 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. $\endgroup$
    – Eric Wofsey
    Commented Feb 24, 2023 at 20:49
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    $\begingroup$ 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.) $\endgroup$
    – Eric Wofsey
    Commented Feb 24, 2023 at 21:38

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