nilpotency is not a radical property

Nilpotency is not a radical property, because a ring does not, in general, contain a largest nilpotent idealPlanetmathPlanetmath.

Let k be a field, and let S=k[X1,X2,] be the ring of polynomials over k in infinitely many variables X1,X2,. Let I be the ideal of S generated by {Xnn+1n(N)}. Let R=S/I. Note that R is commutativePlanetmathPlanetmathPlanetmath.

For each n, let An=k=1nRXn. Let A=An=k=1RXn.

Then each An is nilpotent, since it is the sum of finitely many nilpotent ideals (see proof But A is nil, but not nilpotent. Indeed, for any n, there is an element xA such that xn0, namely x=Xn, and so we cannot have An=0.

So R cannot have a largest nilpotent ideal, for this ideal would have to contain all the ideals An and therefore A.

Title nilpotency is not a radical property
Canonical name NilpotencyIsNotARadicalProperty
Date of creation 2013-03-22 14:13:02
Last modified on 2013-03-22 14:13:02
Owner mclase (549)
Last modified by mclase (549)
Numerical id 4
Author mclase (549)
Entry type Proof
Classification msc 16N40