nil is a radical property

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

  1. 1.

    The class of 𝒩-rings is closed underPlanetmathPlanetmath homomorphic imagesPlanetmathPlanetmathPlanetmath.

  2. 2.

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

  3. 3.


It is easy to see that the homomorphic image of a nil ring is nil, for if f:RS is a homomorphismPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath and xn=0, then f(x)n=f(xn)=0.

The sum of all nil ideals is nil (see proof, 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 So IN 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