Radical of Prime Ideal
Jump to navigation
Jump to search
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:
- $x \in \mathfrak p$
$\blacksquare$