Let \(R\) be a commutative ring and \(I\trianglelefteq R\). Show that \[ \sqrt I=\bigcap_{\substack{\mathfrak p\supseteq I\\ \mathfrak p\text{ prime}}}\mathfrak p. \]
Solution
First suppose \(x\in\sqrt I\). Then \(x^n\in I\) for some \(n\ge1\). If \(\mathfrak p\) is any prime ideal containing \(I\), then \(x^n\in\mathfrak p\), and primality implies \(x\in\mathfrak p\). Thus \[ \sqrt I\subseteq\bigcap_{\mathfrak p\supseteq I}\mathfrak p. \]
For the reverse inclusion, suppose \(x\notin\sqrt I\). Then no positive power of \(x\) lies in \(I\). Let \[ S=\{1,x,x^2,\ldots\}. \] Thus \(I\cap S=\varnothing\).
Consider the set of ideals \(J\) such that \[ I\subseteq J, \qquad J\cap S=\varnothing, \] ordered by inclusion. Every chain has an upper bound given by its union, which is again an ideal disjoint from \(S\). By Zorn’s lemma there is a maximal such ideal \(\mathfrak p\).
We claim \(\mathfrak p\) is prime. Suppose \(ab\in\mathfrak p\) while \(a,b\notin\mathfrak p\). By maximality, both ideals \(\mathfrak p+(a)\) and \(\mathfrak p+(b)\) meet \(S\). Hence there exist \[ p_1+ra=x^m, \qquad p_2+sb=x^n, \] with \(p_1,p_2\in\mathfrak p\). Multiplying gives \[ x^{m+n}=(p_1+ra)(p_2+sb)\in\mathfrak p, \] because \(ab\in\mathfrak p\). This contradicts \(\mathfrak p\cap S=\varnothing\). Hence \(\mathfrak p\) is prime.
We have \(I\subseteq\mathfrak p\), but \(x\notin\mathfrak p\) since \(x\in S\). Therefore \(x\) is not in the intersection of all prime ideals containing \(I\).
Thus \[ \sqrt I=\bigcap_{\mathfrak p\supseteq I}\mathfrak p. \]