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:

$\map \Rad {\mathfrak p} = \mathfrak p$

Proof

$\supseteq$

Let $x \in \mathfrak p$.

Since $x = x^1$, by Definition 1 of Radical of Ideal of Ring:

$x \in \map \Rad {\mathfrak p}$

$\Box$

$\subseteq$

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

Then:

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

Aiming for a contradiction, suppose $x \notin p$.

Then by Definition 3 of Prime Ideal:

$x^n \notin \mathfrak p$

which contradicts the assertion that $x^n \in \mathfrak p$.

Thus:

$x \in \mathfrak p$

$\blacksquare$