Timeline for Purely noncommutative algebra-Morita equivalence

5 events

when toggle format	what		by	license	comment
Dec 24, 2014 at 14:46	vote	accept	truebaran
Dec 20, 2014 at 3:13	comment	added	Qiaochu Yuan		@truebaran: yes, that's what I meant by "the center is Morita invariant."
Dec 19, 2014 at 23:49	comment	added	truebaran		Thank you for your answer: I found that in fact: if $A$ and $B$ are Morita equivalent then their centers are isomorphic. Therefore it is enough to take a simple algebra $A$: if $A$ is Morita equivalent to commutative $B$ then $B=k$ (underlying field) and it is known that $A$ must be of the form $End_B(P)$ where $P$ is finitely generated $B$ module. In this case ($B=k$) this means that $A$ is a matrix algebra.
Dec 15, 2014 at 6:48	comment	added	Qiaochu Yuan		I should mention that nothing preceding the example is needed to justify the example, only to suggest where to look for it: it's clear that $\mathbb{H}$, and more generally any noncommutative division ring, is not Morita equivalent to a commutative algebra because the endomorphism algebra of every $\mathbb{H}$-module is noncommutative.
Dec 14, 2014 at 21:52	history	answered	Qiaochu Yuan	CC BY-SA 3.0