# nilpotency is not a radical property

Let $k$ be a field, and let $S=k[X_{1},X_{2},\ldots]$ be the ring of polynomials over $k$ in infinitely many variables $X_{1},X_{2},\dots$. Let $I$ be the ideal of $S$ generated by $\{X_{n}^{n+1}\mid n\in\mathbb{(}N)\}$. Let $R=S/I$. Note that $R$ is commutative   .

For each $n$, let $A_{n}=\sum_{k=1}^{n}RX_{n}$. Let $A=\bigcup A_{n}=\sum_{k=1}^{\infty}RX_{n}$.

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 $x^{n}\neq 0$, 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 NilpotencyIsNotARadicalProperty 2013-03-22 14:13:02 2013-03-22 14:13:02 mclase (549) mclase (549) 4 mclase (549) Proof msc 16N40