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 Jul 24 '16 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 Jul 24 '16 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 Jul 24 '16 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 Jul 24 '16 at 15:31
  • $\begingroup$ Relatad on MO: mathoverflow.net/questions/22899/… $\endgroup$ – Watson Aug 21 '16 at 12:56

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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