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I am looking to prove the above statement, if R is commutative then AB=1 in $M_n(R)$ implies BA=1, except without using brute force calculation. I was thinking about using some homomorphism, like left or right multiplication on a matrix by A. Can someone please provide guidance?

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  • Guidance would be to search for your question by putting your title into the search bar. This is a frequently asked question. I don’t think anyone tries to do it by brute force… – rschwieb Oct 04 '21 at 17:56
  • the supposed duplicate doesn't specify, but the answers assume there they're talking about linear algebra, ie assuming $R$ is a field. is there a duplicate of this actual question out there? – David C. Ullrich Oct 04 '21 at 18:12
  • @DavidC.Ullrich This post here shows it for unitary commutative semirings. – Dietrich Burde Oct 04 '21 at 18:19
  • @DavidC.Ullrich I've added the more accurate duplicate as well. Anyhow, as long as $R$ is commutative, this should be easily dispatched using the determinant, which I thought was the approach used by the linear algebra versions. e.g. https://math.stackexchange.com/a/3800863/29335. I guess I overestimated how well-known that argument was. – rschwieb Oct 05 '21 at 14:56
  • The goal of associating a duplicate, as I understand it, is that it has an answer that addresses the question at a reasonably similar (but not necessarily identical) question. – rschwieb Oct 05 '21 at 15:02

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