Show that for an ideal \(I\trianglelefteq R\) in a commutative ring \(R\), its radical \(\sqrt{I}\) is the intersection of all prime ideals containing \(I\).
Solution
We prove \[ \sqrt I=\bigcap_{\substack{\mathfrak p\in\Spec R\\ I\subseteq\mathfrak p}}\mathfrak p. \]
If \(x\in\sqrt I\), then \(x^n\in I\) for some \(n\ge1\). Every prime ideal \(\mathfrak p\supseteq I\) contains \(x^n\), hence contains \(x\). Therefore \[ \sqrt I\subseteq\bigcap_{\mathfrak p\supseteq I}\mathfrak p. \]
Conversely, suppose \(x\notin\sqrt I\). Then \[ S=\{1,x,x^2,\dots\} \] is disjoint from \(I\). Consider the ideals \(J\supseteq I\) with \(J\cap S=\varnothing\). By Zorn’s lemma there is a maximal such ideal \(\mathfrak p\).
We claim \(\mathfrak p\) is prime. If \(a,b\notin\mathfrak p\), maximality gives \[ x^m\in\mathfrak p+(a),\qquad x^n\in\mathfrak p+(b) \] for some \(m,n\). Hence \[ x^{m+n}\in\mathfrak p+(ab). \] If \(ab\in\mathfrak p\), then \(x^{m+n}\in\mathfrak p\), contradicting \(\mathfrak p\cap S=\varnothing\). Thus \(ab\notin\mathfrak p\), so \(\mathfrak p\) is prime.
By construction \(I\subseteq\mathfrak p\) and \(x\notin\mathfrak p\). Hence \(x\) is not in the intersection of all primes containing \(I\). This proves the reverse inclusion.