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
Since $P$ is a proper ideal by, we have:
 * $1_R \in R \setminus P$

where $1_R$ is the unity of $R$.

$R \setminus P$ 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 $R$.

Proof 2
This follows immediately from.

Also see

 * Definition:Localization of Ring at Prime Ideal