Show that the nilradical of a commutative ring \(R\) is equal to the intersection of all prime ideals of \(R\).
Solution
Let \(R\) be a commutative ring with identity, and put \(P=\bigcap_{\mathfrak p\in\operatorname{Spec}(R)}\mathfrak p\).
1. \(\operatorname{Nil}(R)\subseteq P\).
Let \(x^n=0\) with \(n\ge1\), and let \(\mathfrak p\) be prime. Then \(x^n=0\in\mathfrak p\). If \(x^k\in\mathfrak p\) with \(k\ge2\), then \(x\cdot x^{k-1}\in\mathfrak p\) gives \(x\in\mathfrak p\) or \(x^{k-1}\in\mathfrak p\); induction on \(k\) gives \(x\in\mathfrak p\).
2. For \(f\in R\) not nilpotent, the family \(\mathcal F\) of ideals \(I\) with \(I\cap\{f^n:n\ge0\}=\varnothing\) has a maximal element \(\mathfrak p\).
Put \(S=\{f^n:n\ge0\}\); since \(f\) is not nilpotent, \(0\notin S\), so \((0)\in\mathcal F\). The union of a nonempty chain in \(\mathcal F\) is an ideal disjoint from \(S\), hence an upper bound in \(\mathcal F\). Zorn’s lemma gives a maximal element.
3. The ideal \(\mathfrak p\) of step 2 is prime and \(f\notin\mathfrak p\).
Since \(1=f^0\in S\), \(\mathfrak p\neq R\), and \(f=f^1\in S\) gives \(f\notin\mathfrak p\). Let \(a,b\notin\mathfrak p\). By maximality, \(\mathfrak p+(a)\) and \(\mathfrak p+(b)\) meet \(S\): \(f^m\in\mathfrak p+(a)\) and \(f^n\in\mathfrak p+(b)\) for some \(m,n\ge0\). Then \(f^{m+n}\in(\mathfrak p+(a))(\mathfrak p+(b))\subseteq\mathfrak p+(ab)\). If \(ab\in\mathfrak p\), then \(f^{m+n}\in\mathfrak p\cap S=\varnothing\), a contradiction; so \(ab\notin\mathfrak p\).
4. Q.E.D.
Step 1 gives \(\operatorname{Nil}(R)\subseteq P\). Steps 2 and 3 show that every non-nilpotent \(f\) lies outside some prime ideal, so \(P\subseteq\operatorname{Nil}(R)\).