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Rings will be unital here but I don't require that subrings share the identity elements with superrings.

I just accidentally came up with an example of a ring $R$ with a proper subring $S$ such that $R\cong S$ as rings. I may be mistaken, but I think this is the first such example I've ever come across. The example is this. $R$ is the ring of $\mathbb Z\times \mathbb Z$-matrices over $\mathbb R$ with finite-support columns, and $S$ is the subring of matrices with zeroes in all places indexed by at least one negative number. Then $S$ is isomorphic to the ring of $\mathbb N\times\mathbb N$-matrices over $\mathbb R$. But then $R$ is isomorphic to the endomorphism ring of the direct sum of $\aleph_0$ copies of $\mathbb R$ and $S$ is too.

I would like to know if there are other examples of such rings and if it has anything to do with the rings being Dedekind-infinite. I suspect it might because Dedekind-infinite sets are those which are equinumerous with some of their proper subsets.

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Any endomorphism of a field which is not surjective gives you an example, and also shows that Dedekind finiteness (of rings) does not come into play.

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  • $\begingroup$ Right! Thanks a lot! Eh, this was a pretty silly question. $\endgroup$ – user23211 May 4 '12 at 15:36
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    $\begingroup$ Dedekind-infiniteness (otherwise known, in the presence of choice, to just plain infiniteness) certainly does come into play though. $\endgroup$ – Chris Eagle May 4 '12 at 15:36
  • $\begingroup$ It wasn't that silly, I just was lucky with inspiration :) I had a similiar experience recently mathoverflow.net/questions/95779/bimodule-version-of-ibn $\endgroup$ – rschwieb May 4 '12 at 15:44
  • $\begingroup$ As @Chris implicitly notes, a finite ring certainly cannot contain a proper subring that is isomorphic to it, so infiniteness does play some role! Regards, $\endgroup$ – Matt E May 4 '12 at 15:54
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Even simpler $R[X]\simeq R[X^n] \subset R[X]$.

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More simply, let $R$ be any ring with more than one element and let $R^{\mathbb{N}}$ be the set of all sequences of elements of $R$, made into a ring by the obvious pointwise operations. Then the subset of $R^{\mathbb{N}}$ consisting of sequences with first term $0$ is a proper subring isomorphic to $R^{\mathbb{N}}$ itself.

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  • $\begingroup$ Very nice, thank you very much! $\endgroup$ – user23211 May 4 '12 at 14:32

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