most recent 30 from http://mathoverflow.net 2013-05-19T12:00:34Z http://mathoverflow.net/feeds/question/77536 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/77536/sbn-and-ibn-rings Miroslav Korbelar 2011-10-08T14:30:31Z 2011-10-08T20:10:55Z

<p>Hello, I can not figure out why a ring that is not IBN (invariant basis number) must be SBN (single basis number). More precisely: Let $R$ be a ring (with unit, generally non-commutative) such that the free $R$-module $R^n$ is isomorphic to another free $R$-module $R^m$, where $n, m$ are different natural numbers. How does this imply that $R$ is isomorphic to $R^2$ as an $R$-module? </p>

http://mathoverflow.net/questions/77536/sbn-and-ibn-rings/77555#77555 Miroslav Korbelar 2011-10-08T20:10:55Z 2011-10-08T20:10:55Z

<p>It seems that the implication does not hold. Thanks to Lukas Vokrinek for noticing this: According the example "Tom Leinster (mathoverflow.net/users/586), when is A isomorphic to A^3?" there is an abelian group $A$ such that $A$ is isomorphic to $A^3$ but not to $A^2$. Now, let $R=End(A)$. Then R is isomorphic to $R^3$ (as $R$-modules - that is easy), but not to $R^2$ (otherwise one could construct an isomorphism between $A$ and $A^2$ using the matrices with endomorphism entries, which would arise from the isomorphism $R$ and $R^2$). </p>