Radical of Primary Ideal is Smallest Prime Ideal

Theorem

Let $R$ be a commutative ring with unity.

Let $\mathfrak q$ be a primary ideal of $R$.

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

Then $\map \Rad {\mathfrak q}$ is the smallest prime ideal including $\mathfrak q$.

Proof

First, we shall show $\map \Rad {\mathfrak q}$ that $\map \Rad {\mathfrak q}$ is a prime ideal.

Let $x y \in \map \Rad {\mathfrak q}$.

Then by definition of radical of ideal:

$\exists n \in \N_{>0} : \paren {xy}^n \in \mathfrak q$

By Commutativity ($\text M 2$):

$x^n y^n \in \mathfrak q$

By definition of primary ideal:

$x^n \in \mathfrak q \lor \exists m \in \N_{>0} : \paren {y^n}^m = y^{n m} \in \mathfrak q$

Thus, by definition of radical of ideal:

$x \in \map \Rad {\mathfrak q} \lor y \in \map \Rad {\mathfrak q}$

So, $\map \Rad {\mathfrak q}$ is a prime ideal.

Finally, by Prime Ideal Including Ideal Includes Radical, every prime ideal $\mathfrak p$ satisfies:

$\mathfrak q \subseteq \mathfrak p \implies \map \Rad {\mathfrak q} \subseteq \mathfrak p$

That is, $\map \Rad {\mathfrak q}$ is the smallest prime ideal including $\mathfrak q$.

$\blacksquare$