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Bimodule version of IBN

Then $Z(R^n) \cong Z(R)^n$, so if $R^n \cong R^m$ as $(R, R)$-bimodules then $Z(R)^n \cong Z(R)^m$ as $Z(R)$-modules, and commutative rings satisfy IBN. …
Qiaochu Yuan's user avatar
15 votes

Dual of a bimodule

These duals come from thinking of an $(A, B)$-bimodule as a 1-morphism in the Morita 2-category whose objects are rings 1-morphisms are bimodules 2-morphisms are bimodule homomorphisms and applying …
Qiaochu Yuan's user avatar