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

1.
The class of $\mathcal{N}$rings is closed under^{} homomorphic images^{}.

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.
$\mathcal{N}(R/\mathcal{N}(R))=0$.
It is easy to see that the homomorphic image of a nil ring is nil, for if $f:R\to S$ 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 

Canonical name  NilIsARadicalProperty 
Date of creation  20130322 14:12:58 
Last modified on  20130322 14:12:58 
Owner  mclase (549) 
Last modified by  mclase (549) 
Numerical id  5 
Author  mclase (549) 
Entry type  Proof 
Classification  msc 16N40 
Related topic  PropertiesOfNilAndNilpotentIdeals 