SBN and IBN rings - MathOverflow 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 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 Answer by Miroslav Korbelar for SBN and IBN rings 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>