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 $f\colon R\to S$ 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
Canonical name	NilIsARadicalProperty
Date of creation	2013-03-22 14:12:58
Last modified on	2013-03-22 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