We must show that the nil property, $\nilrad$ , is a radical property, that is that it satisfies the following conditions:

1. The class of $\nilrad$ -rings is closed under homomorphic images.
2. Every ring $R$ has a largest $\nilrad$ -ideal, which contains all other $\nilrad$ -ideals of $R$ . This ideal is written $\nilrad(R)$ .
3. $\nilrad(R/\nilrad(R)) = 0$ .

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 here), 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 here). So $I \subseteq N$ by definition of $N$ . Thus $R/N$ contains no nil ideals.