Prime Ideal Including Ideal Includes Radical
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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:
- $\map \Rad {\mathfrak a} \subseteq \mathfrak p$
$\blacksquare$