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Is there an example of a ring $A$ (with unity) which is isomorphic as unital rings to $A\times A$?

Any such ring can't have invariant basis number so in particular can't be commutative.

Matt Samuel
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Avi Steiner
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    Maybe I'm missing a subtle requirement of the ring isomorphism, but it seems that $A$ can be commutative (though not a domain obviously). – hardmath Jun 26 '15 at 23:02
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    $A$ and $A\times A$ can be isomorphic as rings without being isomorphic as $A$-modules; the action of $A$ is not the same. So the ring can retain the invariant basis number property. – Matt Samuel Jun 26 '15 at 23:04
  • @hardmath Shoot, you're right. Oops. Thanks for pointing that out. Also. – Avi Steiner Jun 26 '15 at 23:14
  • It's a good question, because the distinction between the free-module structure and the ring structure is worth bringing out. – hardmath Jun 26 '15 at 23:27

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$R^{\omega}\cong R^{\omega}\times R^{\omega}$ for any ring $R$, commutative or not. This does not contradict the fact that a commutative ring always has the invariant basis number property because while we can construct an "artificial" isomorphism of rings, it is not an isomorphism of $A$-modules, as the action of $A$ is different. So $A\times A$ really is a free module of rank $2$, and if $A$ is commutative then it is not a free $A$-module of rank $1$.

Matt Samuel
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