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.

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 Definition 3 of Prime Ideal of Commutative and Unitary Ring.


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