I know from this question that $R \times R$ can be isomorphic to $R$, as $R$-modules.
But can they ever be isomorphic as rings?
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Jean Valjean
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3Sure, make $R$ a direct sum of infinitely many copies of any ring. – anon Aug 12 '14 at 18:56
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2@blue: a direct sum of unital rings need not be unital, though. – Qiaochu Yuan Aug 12 '14 at 18:58
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3For preventing any confusion let me mention the following: for a commutative unitary ring $R$ it's not possible to have $R\times R\simeq R$ as $R$-modules. – user26857 Apr 21 '15 at 07:53
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Sure, if $R=\prod_{i\in\mathbb N} \mathbb Z$ then $R\times R\cong R$.
Thomas Andrews
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4The bijection from $R\times R\to R$ would be something like $(x_1,x_2,\ldots;y_1,y_2,\ldots)\mapsto(x_1,y_1,x_2,y_2,\ldots)$. – 2'5 9'2 Apr 20 '15 at 22:41
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Sure. Take $R = \Pi_{i=1}^\infty \mathbb{Z}$ with the product ring structure. Then clearly $R \times R$ is isomorphic to $R$. To find an isomorphism, just use some bijection $\mathbb{N} \amalg \mathbb{N} \rightarrow \mathbb{N}$.
user101036
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2Oh right, thanks! @DietrichBurde I wrote the answer at the same time as him and then saw his answer. I decided to post it anyways, but I can of course delete it. I thought I would add a bit of more detail. – user101036 Aug 12 '14 at 18:58