Complement of Prime Ideal of Ring is Multiplicatively Closed
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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 $R$.
$\blacksquare$
