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Nilradical

From AoPSWiki

Let $A$ be a commutative ring. The set of all nilpotent elements of $A$ (i.e., the set of all $x$ for which $x^n=0$ for some natural $n$ ) is called the nilradical of $A$ . It is sometimes denoted $\mathfrak{N}$ .

The nilradical is an ideal, for if $x^n = 0$ , then for any $a\in A$ , $(ax)^n = 0$ ; and if $x^n = y^m = 0$ , then $(x+y)^{m+n-1} = \sum_{k=0}^{m+n-1} \binom{n}{k} x^{m+n-1-k}y^{k} = 0 .$

Proposition. The nilradical $\mathfrak{N}$ of $A$ is the intersection of all prime ideals of $A$ .

Proof. Let $\mathfrak{p}$ be a prime ideal of $A$ , and let $x$ be a nilpotent element of $A$ . We claim that $x\in \mathfrak{p}$ . Indeed, let $n$ be the least positive integer for which $x^n \in \mathfrak{p}$ . (Such an integer exists, since $x$ is nilpotent.) Suppose that $n>1$ . Then $x \cdot x^{n-1} \in \mathfrak{p}$ ; since $\mathfrak{p}$ is a prime ideal, then $x \in \mathfrak{p}$ or $x^{n-1} \in \mathfrak{p}$ , a contradiction. This proves that $\mathfrak{N} \subset \bigcap_{\mathfrak{p} \text{ prime}} \mathfrak{p} .$ To show the converse, it suffices to show that for any non-nilpotent element $a$ , there is some prime ideal that does not contain $a$ .

So suppose that $a$ is an element of $A$ that is not nilpotent. Let $S$ be the set of ideals of $A$ that do not contain any element of the form $a^n$ . Since $(0) \in S$ , $S$ is not empty; then by Zorn's Lemma, $S$ has a maximal element $\mathfrak{m}$ .

It suffices to show that $\mathfrak{m}$ is a prime ideal. Indeed, suppose otherwise; then there exist elements $x,y \notin \mathfrak{m}$ for which $xy \in \mathfrak{m}$ . Then the set of elements $z$ for which $xz \in \mathfrak{m}$ is evidently an ideal of $A$ that properly contains $\mathfrak{m}$ ; it therefore contains $a^n$ , for some integer $n$ . By similar reasoning, the set of elements $z$ for which $a^n z \in \mathfrak{m}$ is an ideal that properly contains $\mathfrak{m}$ , so this set contains $a^m$ , for some integer $m$ . Then $a^{n+m} \in \mathfrak{m}$ , a contradiction.

Therefore $\mathfrak{m}$ is a prime ideal that does not contain $a$ . It follows from the generality of this argument that $\bigcap_{\mathfrak{p} \text{ prime}} \mathfrak{p} \subset \mathfrak{N} ,$ as desired. $\blacksquare$

See also

Jacobson radical

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Category: Commutative algebra

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