# 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 |