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？
$\begingroup$
$\endgroup$
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
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.
-
$\begingroup$ What does reduced mean? The 2-adics are a local domain, so have no nilpotent elements, but it is not Jacobson semisimple. $\endgroup$ Commented Dec 3, 2010 at 15:19
-
$\begingroup$ 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! $\endgroup$– EmertonCommented Dec 3, 2010 at 18:41
$\begingroup$
$\endgroup$
5
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.
answered Dec 3, 2010 at 11:08
-
$\begingroup$ 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? $\endgroup$ Commented Dec 6, 2010 at 10:44
-
$\begingroup$ Because it is not artinian, in general... $\endgroup$ Commented Dec 6, 2010 at 11:15
-
$\begingroup$ Von-Neumann regular ring is also not artinian in general,but it has this kind of property. $\endgroup$ Commented Dec 8, 2010 at 4:58
-
$\begingroup$ I am at a total loss what you mean... $\endgroup$ Commented Dec 9, 2010 at 14:03
-
$\begingroup$ “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. $\endgroup$ Commented Dec 11, 2010 at 5:23