Prime Ideal Including Ideal Includes Radical

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

Let $\mathfrak p$ be a prime ideal.

Let $\mathfrak a \subseteq \mathfrak p$ be an ideal of $R$.

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

Then:
 * $\map \Rad {\mathfrak a} \subseteq \mathfrak p$

Proof
Let $x \in \relcomp R {\mathfrak p}$.

By :

Therefore, by :
 * $x \not\in \map \Rad {\mathfrak a}$

Thus:
 * $\relcomp R {\mathfrak p} \subseteq \relcomp R {\map \Rad {\mathfrak a} }$

Therefore, by Relative Complement inverts Subsets:
 * $\map \Rad {\mathfrak a} \subseteq \mathfrak p$