Complement of Prime Ideal of Ring is Multiplicatively Closed

Theorem
Let $A$ be a commutative ring with unity.

Let $\mathfrak p \subset A$ be a prime ideal of $A$.

Then its complement $A \setminus \mathfrak p$ is multiplicatively closed.

Proof
$A \setminus \mathfrak p$ is not multiplicatively closed.

That is:
 * $\exists a, b \in A \setminus \mathfrak p: a b \notin A \setminus \mathfrak p$

This means:
 * $a b \in \mathfrak p$

This contradicts the assertion that $\mathfrak p$ is a prime ideal of $A$.