0
$\begingroup$

I am looking ring with nilpotent element such that $J(R)=0$ where $J(R)$ is Jacobson radical.

Any suggestion?

  • 1
    The example, if it exists, would have to be non-commutative, since for commutative rings, the Nilradical is contained in the Jacobson radical.2012-08-07
  • 0
    https://en.wikipedia.org/wiki/Semiprimitive_ring It seems Kevin's example below is von Neumann regular.2012-08-07

1 Answers 1

5

The best examples are the matrix rings over a field. These are simple, so they've got trivial Jacobson radical, and yet already the $2\times 2$ matrices have nilpotent elements $e_{12},e_{21}$.

There arguably is one commutative example: in the trivial ring, $1=0$ is nilpotent but the Jacobson radical is, naturally, $0.$

  • 0
    Aha, I overlooked the trivial ring. One more has fallen prey to its evil designs.2012-08-07
  • 0
    I often wonder whether having a terminal object in Rings is really worth all the trouble.2012-08-07