most recent 30 from http://mathoverflow.net 2013-06-19T11:55:14Z http://mathoverflow.net/feeds/question/48158 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/48158/what-is-the-characteristic-of-the-module-over-jacobson-semisimple-ring Strongart 2010-12-03T11:00:05Z 2010-12-04T12:53:55Z

<p>We know a ring R is semisimple ring iff every module over R is semisimple，a ring R is von-Neumann regular ring iff every module over R is flat，What about the Jacobson semisimple ring？</p>

http://mathoverflow.net/questions/48158/what-is-the-characteristic-of-the-module-over-jacobson-semisimple-ring/48159#48159 Bugs Bunny 2010-12-03T11:08:51Z 2010-12-04T12:53:55Z

<p><a href="http://en.wikipedia.org/wiki/Semiprimitive_ring" rel="nofollow">Mighty Wikipedia</a> to the rescue!! Seriously I dont think you can say anything better than a semisimple faithful module exists...</p> <p>What is the cause of your curiosity?</p> <p>Having said that, I can think of a cute reformulation that makes it clear that the property is Morita-invariant: for any projective $P$, the hom-space $Hom (P, \oplus_i S_i)$ is a faithful $End (P)$-module where the sum is taken over non-isomorphic simples in the category.</p>

http://mathoverflow.net/questions/48158/what-is-the-characteristic-of-the-module-over-jacobson-semisimple-ring/48175#48175 Emerton 2010-12-03T14:21:54Z 2010-12-03T18:42:24Z

<p>An alternative phrasing to Bugs's answer is as follows:</p> <p>A ring $R$ is Jacobson semisimple if and only if the only element which annihilates every simple $R$-module is the zero element.</p> <p>In my experience (which is largely in the commutative case, or even finite type algebras over a field; in this latter case Jacobson semisimple coincides with reduced), this point of view on Jacobson semisimple rings is one that comes up frequently in arguments.</p>