Prime Ideal is Primary Ideal
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Theorem
Let $R$ be a commutative ring with unity.
Let $\mathfrak p$ be a prime ideal of $R$.
Then $\mathfrak p$ is a primary ideal of $R$.
Proof
Let $xy \in \mathfrak p$.
Let $x \not \in \mathfrak p$
By definition of prime ideal:
- $y^1 = y \in \mathfrak p$
Thus, by definition, $\mathfrak p$ is a primary ideal.
$\blacksquare$