Complement of Prime Ideal of Ring is Multiplicatively Closed
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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.
Since $P$ is a proper ideal by Definition of Prime Ideal, we have:
- $1_R \in R \setminus P$
where $1_R$ is the unity of $R$.
Aiming for a contradiction, suppose $R \setminus P$ is not multiplicatively closed.
- $\exists a, b \in R \setminus P: a b \notin R \setminus P$
- $a b \in P$
This contradicts the assertion that $P$ is a prime ideal of $R$.
This follows immediately from Definition 3 of Prime Ideal of Commutative and Unitary Ring.