Radical of Prime Ideal

Theorem
Let $R$ be a commutative ring with unity.

Let $\mathfrak p$ be a prime ideal of $R$.

Let $\map \Rad {\mathfrak p}$ be the radical of $\mathfrak p$.

Then:
 * $\map \Rad {\mathfrak p} = \mathfrak p$

$\supseteq$
Let $x \in \mathfrak p$.

Since $x = x^1$, by :
 * $x \in \map \Rad {\mathfrak p}$

$\subseteq$
Let $x \in \map \Rad {\mathfrak p}$.

Then:
 * $\exists n \in \N_{>0} : x^n \in \mathfrak p$

$x \notin p$.

Then by :
 * $x^n \notin \mathfrak p$

which contradicts the assertion that $x$ is nilpotent.

Thus:
 * $x \in \mathfrak p$