Show that \(\sqrt I\) is the intersection of all prime ideals containing \(I\).
Solution
We prove \[ \sqrt I=\bigcap_{\substack{\mathfrak p\supseteq I\\ \mathfrak p\text{ prime}}}\mathfrak p. \]
If \(x\in\sqrt I\), then \(x^m\in I\) for some \(m\ge1\). For every prime \(\mathfrak p\supseteq I\), we have \(x^m\in\mathfrak p\), hence \(x\in\mathfrak p\). Thus \(x\) lies in the displayed intersection.
Conversely, suppose \(f\notin\sqrt I\) and let \[ S=\{1,f,f^2,\dots\}. \] Then \(S\cap I=\varnothing\). Consider the ideals \(J\supseteq I\) with \(J\cap S=\varnothing\), ordered by inclusion. A union of a chain is again such an ideal, so Zorn’s lemma gives a maximal one, say \(\mathfrak p\).
To see that \(\mathfrak p\) is prime, suppose \(a,b\notin\mathfrak p\). Maximality implies \[ (\mathfrak p+(a))\cap S\ne\varnothing, \qquad (\mathfrak p+(b))\cap S\ne\varnothing. \] Hence for some \(r,s\), \[ f^r\in\mathfrak p+(a),\qquad f^s\in\mathfrak p+(b). \] If \(ab\in\mathfrak p\), multiplying gives \[ f^{r+s}\in(\mathfrak p+(a))(\mathfrak p+(b))\subseteq\mathfrak p+(ab)=\mathfrak p, \] contrary to \(\mathfrak p\cap S=\varnothing\). Thus \(ab\notin\mathfrak p\), so \(\mathfrak p\) is prime. It contains \(I\) but not \(f\), proving that \(f\) is not in the intersection of the primes over \(I\).