20
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
10
$\begingroup$

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 $A\not\cong B$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.