# Complement of Prime Ideal of Ring is Multiplicatively Closed

## Theorem

Let $R$ be a commutative ring with unity.

Let $P \subset R$ be a prime ideal of $R$.

Then its complement $R \setminus P$ is multiplicatively closed.

## Proof

Aiming for a contradiction, suppose $R \setminus P$ is not multiplicatively closed.

That is:

- $\exists a, b \in R \setminus P: a b \notin R \setminus P$

This means:

- $a b \in P$

This contradicts the assertion that $P$ is a prime ideal of $R$.

$\blacksquare$