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). {{math theorem|name=Proposition|
Let R be a commutative ring. Then the nilradical \mathfrak{N}_R of R equals the intersection of all prime ideals of R.}} {{Math proof| Firstly, the nilradical is contained in every prime ideal. Indeed, if a\in \mathfrak{N}_R, one has a^n=0 for some positive integer n. Since every ideal contains 0 and every prime ideal that contains a product, here a^n=0, contains one of its factors, one deduces that every prime ideal contains a. Conversely, let f\notin\mathfrak{N}_R; we have to prove that there is a prime ideal that does not contain f. Consider the set \Sigma of all ideals that do not contain any power of f. One has (0) \in \Sigma, by definition of the nilradical. For every chain J_1\subseteq J_2 \subseteq \dots of ideals in \Sigma, it is easy to check that the union J=\bigcup_{i\geq1} J_i is an ideal. Moreover, J belongs to \Sigma, since otherwise it would contain a power of f, which must belong to some J_i, contradicting the definition of J_i. So, \Sigma is a (nonempty)
partially ordered set by set inclusion such that every chain in \Sigma has a
upper bound in \Sigma. Thus,
Zorn's lemma applies, and there exists a maximal element \mathfrak{m} \in \Sigma. We have to prove that \mathfrak{m} is a prime ideal. If it were not prime there would be two elements g\in R and h\in R such that g\notin\mathfrak{m}, h\notin\mathfrak{m}, and gh\in\mathfrak{m}. By maximality of \mathfrak{m}, one has \mathfrak{m}+(g)\notin\Sigma and \mathfrak{m}+(h)\notin\Sigma. So there exist positive integers r and s such that f^r \in \mathfrak{m}+(g) and f^s \in \mathfrak{m}+(h). It follows that f^rf^s=f^{r+s} \in \mathfrak{m}+(gh)=\mathfrak{m}, contradicting the fact that \mathfrak{m} is in \Sigma. This finishes the proof, since we have proved the existence of a prime ideal that does not contain f.}} 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 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 (i.e., the ring is
Noetherian), then it is
nilpotent. == Noncommutative rings ==