NilpotencyIsNotARadicalProperty 2004-02-28 14:19:39 2004-02-28 14:19:39 Proof radical theory 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here 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 $S = k[X_1, X_2, \dotsc]$ 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 \PMlinkid{here}{5650}). 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 $A$.