Prime Ideal Including Ideal Includes Radical

Theorem

Let $R$ be a commutative ring with unity.

Let $\mathfrak p$ be a prime ideal.

Let $\mathfrak a$ be an ideal of $R$ such that:

$\mathfrak a \subseteq \mathfrak p$

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 Definition 3 of Prime Ideal:

\(\ds \forall n \in \N_{>0} : \ \ \)

\(\ds x^n\)

\(\not\in\)

\(\ds \mathfrak p\)

\(\ds \leadsto \ \ \)

\(\ds x^n\)

\(\not\in\)

\(\ds \mathfrak a\)

since $\mathfrak a \subseteq \mathfrak p$

Therefore, by Definition 2 of Radical:

$x \not\in \map \Rad {\mathfrak a}$

Thus:

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

Therefore, by Relative Complement inverts Subsets:

$\map \Rad {\mathfrak a} \subseteq \mathfrak p$

$\blacksquare$