Show that the nilradical is given by \(\nilrad{R} = \rad(0)\).
Solution
1. Write \(\rad{I} = \ts{ x \in R \st x^n \in I \text{ for some } n \geq 1 }\) for the radical of an ideal \(I\), and \(\nilrad{R} = \ts{ x \in R \st x^n = 0 \text{ for some } n \geq 1 }\) for the nilradical.
2. \(\nilrad{R} \subseteq \rad{(0)}\).
If \(x^n = 0\) then \(x^n \in (0)\), since \((0) = \ts 0\).
3. \(\rad{(0)} \subseteq \nilrad{R}\).
If \(x^n \in (0)\) then \(x^n = 0\), since \((0)\) has \(0\) as its only element.
4. Q.E.D.
Steps 2 and 3 give the two inclusions, so \(\nilrad{R} = \rad{(0)}\).