We must show that the nil property, $\mathcal{N}$, is a radical property, that is that it satisfies the following conditions:

1. 1.

The class of $\mathcal{N}$-rings is closed under homomorphic images.

2. 2.

Every ring $R$ has a largest $\mathcal{N}$-ideal, which contains all other $\mathcal{N}$-ideals of $R$. This ideal is written $\mathcal{N}(R)$.

3. 3.

$\mathcal{N}(R/\mathcal{N}(R))=0$.

It is easy to see that the homomorphic image of a nil ring is nil, for if $f\colon R\to S$ is a homomorphism and $x^{n}=0$, then $f(x)^{n}=f(x^{n})=0$.

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 $N$ is the largest nil ideal in $R$, and $I$ is an ideal of $R$ containing $N$ such that $I/N$ is nil, then $I$ is also nil (see proof http://planetmath.org/node/5650here). So $I\subseteq N$ by definition of $N$. Thus $R/N$ contains no nil ideals.

Title nil is a radical property NilIsARadicalProperty 2013-03-22 14:12:58 2013-03-22 14:12:58 mclase (549) mclase (549) 5 mclase (549) Proof msc 16N40 PropertiesOfNilAndNilpotentIdeals