Radical of Power of Prime Ideal

Theorem

Let $A$ be a commutative ring with unity.

Let $\mathfrak p \subseteq A$ be a prime ideal.

Let $n > 0$ be a natural number.

Then the radical of the $n$th power of $\mathfrak p$ equals $\mathfrak p$:

$\map {\operatorname{Rad} } {\mathfrak p^n} = \mathfrak p$

Proof

$(\subseteq):$

Let $a \in \map {\operatorname{Rad} } {\mathfrak p^n}$.

Then by definition, $a^k \in \mathfrak p^n$ for some integer $k$.

By Power of Ideal is Subset, $\mathfrak p^n \subseteq \mathfrak p$.

Hence, $a^k \in \mathfrak p$.

By Power in Prime Ideal, $a \in \mathfrak p$.

$\Box$

$(\supseteq):$

Let $b \in \mathfrak p$.

Then by definition $b^n \in \mathfrak p^n$.

Hence $b \in \map {\operatorname{Rad} } {\mathfrak p^n}$.

$\blacksquare$