Let be a ring (not necessarily with identity element). Suppose that for all Prove that is commutative.
Last edited by NonCommAlg; June 7th 2009 at 03:12 PM.
Follow Math Help Forum on Facebook and Google+
(1) Using this in (1), we obtain (2) Using (2), we get ( ).
Originally Posted by AMI (1) Using this in (1), we obtain (2) Using (2), we get ( ). Hi AMI. I think the proof can be simplified just a little. Putting completes the proof.
Thanks, Abstractionist! Oh, and I think the reason why the problem said "not necessarily with identity element" is that if we had the identity , then we could take in the place of , obtain , substract the initial identity, and we're done.
good job! i just don't know why it took you guys so long to figure this out!
View Tag Cloud