nilpotency is not a radical property

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

Let $k$ be a field, and let $S=k[X_1,X_2,\mathrm{\dots }]$ be the ring of polynomials over $k$ in infinitely many variables $X_1,X_2,\mathrm{\dots }$. Let $I$ be the ideal of $S$ generated by $\{X_n^{n+1}\mid n\in (N)\}$. Let $R=S/I$. Note that $R$ is commutative.

For each $n$, let $A_n={\sum }_{k=1}^{n}R{X}_n$. Let $A=\bigcup A_n={\sum }_{k=1}^{\mathrm{\infty }}R{X}_n$.

Then each $A_n$ is nilpotent, since it is the sum of finitely many nilpotent ideals (see proof http://planetmath.org/node/5650here). But $A$ is nil, but not nilpotent. Indeed, for any $n$, there is an element $x\in A$ such that $x^n\ne 0$, namely $x=X_n$, and so we cannot have $A^n=0$.

So $R$ cannot have a largest nilpotent ideal, for this ideal would have to contain all the ideals $A_n$ 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