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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?

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2 Answers 2

An alternative phrasing to Bugs's answer is as follows:

A ring $R$ is Jacobson semisimple if and only if the only element which annihilates every simple $R$-module is the zero element.

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.

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What does reduced mean? The 2-adics are a local domain, so have no nilpotent elements, but it is not Jacobson semisimple. –  Jack Schmidt Dec 3 '10 at 15:19
    
Dear Jack, Yes, you are right, and I blundered. I should have said "in the case of finite type algebras over a field", and I will make this correction now! –  Emerton Dec 3 '10 at 18:41

Mighty Wikipedia to the rescue!! Seriously I dont think you can say anything better than a semisimple faithful module exists...

What is the cause of your curiosity?

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.

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I think this kind of property is interesting,so I want to find more examples.Why the Jacobson semisimple ring is not like the semisimple ring? –  Strongart Dec 6 '10 at 10:44
    
Because it is not artinian, in general... –  Bugs Bunny Dec 6 '10 at 11:15
    
Von-Neumann regular ring is also not artinian in general,but it has this kind of property. –  Strongart Dec 8 '10 at 4:58
    
I am at a total loss what you mean... –  Bugs Bunny Dec 9 '10 at 14:03
    
“a ring R is von-Neumann regular ring iff every module over R is flat”,but Von-Neumann regular ring is also not artinian in general,so nonartin is not the reason. –  Strongart Dec 11 '10 at 5:23

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