Prove that the nilradical of a commutative ring \(R\) is the intersection of all prime ideals: \[ \sqrt{(0)}=\bigcap_{\mathfrak p\in\operatorname{Spec}(R)}\mathfrak p. \]
Solution
1. Every nilpotent element belongs to every prime ideal.
Let \(x\in R\) be nilpotent, so \(x^n=0\) for some \(n\ge1\), and let \(\mathfrak p\) be prime. Since \[ x^n=0\in\mathfrak p, \] primality implies \(x\in\mathfrak p\). Hence \[ \sqrt{(0)}\subseteq\bigcap_{\mathfrak p\in\operatorname{Spec}(R)}\mathfrak p. \]
2. If \(x\) is not nilpotent, there exists a prime ideal not containing \(x\).
Assume \(x\) is not nilpotent and set \[ S=\{1,x,x^2,\ldots\}. \] Then \(0\notin S\). Consider the set of ideals of \(R\) disjoint from \(S\), ordered by inclusion. It is nonempty because \((0)\) is disjoint from \(S\). The union of a chain of such ideals is again an ideal disjoint from \(S\), so Zorn’s lemma gives a maximal ideal \(\mathfrak p\) among those disjoint from \(S\).
We show that \(\mathfrak p\) is prime. Suppose \(ab\in\mathfrak p\) but \(a,b\notin\mathfrak p\). By maximality, both larger ideals \(\mathfrak p+(a)\) and \(\mathfrak p+(b)\) meet \(S\). Thus there exist \[ s_1=u+ra\in S,\qquad s_2=v+tb\in S \] with \(u,v\in\mathfrak p\). Their product is \[ s_1s_2=uv+utb+vra+rtab\in\mathfrak p, \] because \(u,v,ab\in\mathfrak p\). But \(s_1s_2\in S\) since \(S\) is multiplicatively closed, contradicting \(\mathfrak p\cap S=\varnothing\). Hence \(\mathfrak p\) is prime.
Since \(x\in S\) and \(\mathfrak p\cap S=\varnothing\), we have \(x\notin\mathfrak p\).
3. Therefore\[ \sqrt{(0)}=\bigcap_{\mathfrak p\in\operatorname{Spec}(R)}\mathfrak p. \]
By 1, the nilradical is contained in every prime ideal. By 2, every nonnilpotent element is omitted by at least one prime ideal. Thus an element lies in every prime ideal exactly when it is nilpotent.