We know a ring R is semisimple ring iff every module over R is semisimple，a ring R is vonNeumann regular ring iff every module over R is flat，What about the Jacobson semisimple ring？
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 2adics are a local domain, so have no nilpotent elements, but it is not Jacobson semisimple. $\endgroup$ – Jack Schmidt Dec 3 '10 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! $\endgroup$ – 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 Moritainvariant: for any projective $P$, the homspace $Hom (P, \oplus_i S_i)$ is a faithful $End (P)$module where the sum is taken over nonisomorphic simples in the category.

$\begingroup$ 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? $\endgroup$ – Strongart Dec 6 '10 at 10:44


$\begingroup$ VonNeumann regular ring is also not artinian in general,but it has this kind of property. $\endgroup$ – Strongart Dec 8 '10 at 4:58


$\begingroup$ “a ring R is vonNeumann regular ring iff every module over R is flat”，but VonNeumann regular ring is also not artinian in general，so nonartin is not the reason. $\endgroup$ – Strongart Dec 11 '10 at 5:23