The nilradical is contained in the Jacobson radical, i.e. \[ \nilrad{R} \subseteq J(R) .\]
Solution
Let \(x\in\operatorname{nil}(R)\), so \(x^n=0\) for some \(n\). For every prime ideal \(\mathfrak p\), primality gives \(x\in\mathfrak p\) from \(x^n\in\mathfrak p\). Every maximal ideal is prime in a commutative ring with identity, hence \(x\) lies in every maximal ideal. Therefore \[ \operatorname{nil}(R)\subseteq\bigcap_{\mathfrak m\in\operatorname{mSpec}R}\mathfrak m=J(R). \]