nil is a radical property
We must show that the nil property, , is a radical property, that is that it satisfies the following conditions:
- 1.
The class of -rings is closed under
homomorphic images
.
- 2.
Every ring has a largest -ideal, which contains all other -ideals of . This ideal is written .
- 3.
.
It is easy to see that the homomorphic image of a nil ring is nil, for if is a homomorphism and , then .
The sum of all nil ideals is nil (see proof http://planetmath.org/node/5650here), so this sum is the largest nil ideal in the ring.
Finally, if is the largest nil ideal in , and is an ideal of containing such that is nil, then is also nil (see proof http://planetmath.org/node/5650here). So by definition of . Thus contains no nil ideals.