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## Summary

In algebra, the nilradical of a commutative ring is the ideal consisting of the nilpotent elements:

${\displaystyle {\mathfrak {N}}_{R}=\lbrace f\in R\mid f^{m}=0{\text{ for some }}m\in \mathbb {Z} _{>0}\rbrace .}$

In the non-commutative ring case the same definition does not always work. This has resulted in several radicals generalizing the commutative case in distinct ways; see the article Radical of a ring for more on this.

The nilradical of a Lie algebra is similarly defined for Lie algebras.

## Commutative rings

The nilradical of a commutative ring is the set of all nilpotent elements in the ring, or equivalently the radical of the zero ideal. This is an ideal because the sum of any two nilpotent elements is nilpotent (by the binomial formula), and the product of any element with a nilpotent element is nilpotent (by commutativity). It can also be characterized as the intersection of all the prime ideals of the ring (in fact, it is the intersection of all minimal prime ideals).

Proposition[1] —  Let ${\displaystyle R}$  be a commutative ring. Then ${\displaystyle {\mathfrak {N}}_{R}=\bigcap _{{\mathfrak {p}}\varsubsetneq R{\text{ prime}}}{\mathfrak {p}}.}$

Proof

Let ${\displaystyle r\in {\mathfrak {N}}_{R}}$  and ${\displaystyle {\mathfrak {p}}}$  be a prime ideal, then ${\displaystyle r^{n}=0}$  for some ${\displaystyle n\in \mathbb {Z} _{>0}}$ . Thus

${\displaystyle r^{n}=r\cdot r^{n-1}=0\in {\mathfrak {p}}}$ ,

since ${\displaystyle {\mathfrak {p}}}$  is an ideal, which implies ${\displaystyle r\in {\mathfrak {p}}}$  or ${\displaystyle r^{n-1}\in {\mathfrak {p}}}$ . In the second case, suppose ${\displaystyle r^{m}\in {\mathfrak {p}}}$  for some ${\displaystyle m\leq n-1}$ , then ${\displaystyle r^{m}=r\cdot r^{m-1}\in {\mathfrak {p}}}$  thus ${\displaystyle r\in {\mathfrak {p}}}$  or ${\displaystyle r^{m-1}\in {\mathfrak {p}}}$  and, by induction on ${\displaystyle m\geq 1}$ , we conclude ${\displaystyle r^{m}\in {\mathfrak {p}},\,\forall m:0 , in particular ${\displaystyle r\in {\mathfrak {p}}}$ . Therefore ${\displaystyle r}$  is contained in any prime ideal and ${\displaystyle {\mathfrak {N}}_{R}\subseteq \bigcap _{{\mathfrak {p}}\varsubsetneq R{\text{ prime}}}{\mathfrak {p}}}$ .

Conversely, we suppose ${\displaystyle f\notin {\mathfrak {N}}_{R}}$  and consider the set

${\displaystyle \Sigma :=\lbrace J\subseteq R\mid J{\text{ is an ideal and }}f^{m}\notin J{\text{ for all }}m\in \mathbb {Z} _{>0}\rbrace }$

which is non-empty, indeed ${\displaystyle (0)\in \Sigma }$ . ${\displaystyle \Sigma }$  is partially ordered by ${\displaystyle \subseteq }$  and any chain ${\displaystyle J_{1}\subseteq J_{2}\subseteq \dots }$  has an upper bound given by ${\displaystyle J=\bigcup _{i\geq 1}J_{i}\in \Sigma }$ , indeed: ${\displaystyle J}$  is an ideal[Note 1] and if ${\displaystyle f^{m}\in J}$  for some ${\displaystyle m}$  then ${\displaystyle f^{m}\in J_{l}}$  for some ${\displaystyle l}$ , which is impossible since ${\displaystyle J_{l}\in \Sigma }$ ; thus any chain in ${\displaystyle \Sigma \neq \emptyset }$  has an upper bound and we can apply Zorn's lemma: there exists a maximal element ${\displaystyle {\mathfrak {m}}\in \Sigma }$ . We need to prove that ${\displaystyle {\mathfrak {m}}}$  is a prime ideal: let ${\displaystyle g,h\notin {\mathfrak {m}},gh\in {\mathfrak {m}}}$ , then ${\displaystyle {\mathfrak {m}}\varsubsetneq {\mathfrak {m}}+(g),{\mathfrak {m}}+(h)\notin \Sigma }$  since ${\displaystyle {\mathfrak {m}}}$  is maximal in ${\displaystyle \Sigma }$ , which is to say, there exist ${\displaystyle r,s\in \mathbb {Z} _{>0}}$  such that ${\displaystyle f^{r}\in {\mathfrak {m}}+(g),}$  ${\displaystyle f^{s}\in {\mathfrak {m}}+(h)}$ , but then ${\displaystyle f^{r}f^{s}=f^{r+s}\in {\mathfrak {m}}+(gh)={\mathfrak {m}}\in \Sigma }$ , which is absurd. Therefore if ${\displaystyle f\notin {\mathfrak {N}}_{R}}$ , ${\displaystyle f}$  is not contained in some prime ideal or equivalently ${\displaystyle R\setminus {\mathfrak {N}}_{R}\subseteq \bigcup _{{\mathfrak {p}}\varsubsetneq R{\text{ prime}}}(R\setminus {\mathfrak {p}})}$  and finally ${\displaystyle {\mathfrak {N}}_{R}\supseteq \bigcap _{{\mathfrak {p}}\varsubsetneq R{\text{ prime}}}{\mathfrak {p}}}$ .

A ring is called reduced if it has no nonzero nilpotent. Thus, a ring is reduced if and only if its nilradical is zero. If R is an arbitrary commutative ring, then the quotient of it by the nilradical is a reduced ring and is denoted by ${\displaystyle R_{\text{red}}}$ .

Since every maximal ideal is a prime ideal, the Jacobson radical — which is the intersection of maximal ideals — must contain the nilradical. A ring R is called a Jacobson ring if the nilradical and Jacobson radical of R/P coincide for all prime ideals P of R. An Artinian ring is Jacobson, and its nilradical is the maximal nilpotent ideal of the ring. In general, if the nilradical is finitely generated (e.g., the ring is Noetherian), then it is nilpotent.