Every prime ideal is radical

Let \(\mathfrak p\) be a prime ideal of a commutative ring \(R\). The inclusion \[ \mathfrak p\subseteq\sqrt{\mathfrak p} \] is immediate.

Conversely, let \(x\in\sqrt{\mathfrak p}\). Then \(x^n\in\mathfrak p\) for some \(n\ge1\). Since \[ x^n=x\,x^{n-1}\in\mathfrak p \] and \(\mathfrak p\) is prime, either \(x\in\mathfrak p\) or \(x^{n-1}\in\mathfrak p\). Repeating this argument finitely many times gives \(x\in\mathfrak p\). Hence \[ \sqrt{\mathfrak p}\subseteq\mathfrak p. \] Therefore \[ \sqrt{\mathfrak p}=\mathfrak p, \] so every prime ideal is radical.