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 $S=k[X_{1},X_{2},\\ldots]$ 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 $x^{n}\eq 0$ , 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 $A$ .

单词	NilpotencyIsNotARadicalProperty
释义	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 $S=k[X_{1},X_{2},\\ldots]$ 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 $x^{n}\eq 0$ , 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 $A$ .
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