Intersection of Chain of Prime Ideals of Commutative Ring is Prime Ideal

Theorem
Let $R$ be a commutative ring.

Let $\operatorname{Spec} R$ be the spectrum of $R$, ordered by inclusion.

Let $\left\{{P_\alpha}\right\}_{\alpha \in A}$ be a non-empty chain of prime ideals of $\operatorname{Spec} R$.

Let $\displaystyle P = \bigcap_{\alpha \in A} P_\alpha$ be their intersection.

Then $P$ is a prime ideal of $R$.

Proof
By Intersection of Ideals, $P$ is an ideal of $R$.

We show that $P$ is prime.

Let $a,b\in R$ with $a,b\notin P$.

We show that $ab\notin P$.

Because $a\notin P$, there exists $\alpha \in A$ with $a\notin P_\alpha$.

Because $b\notin P$, there exists $\beta \in A$ with $b\notin P_\beta$.

Since $\left\{{P_\gamma}\right\}_{\gamma \in A}$ is totally ordered, $P_\alpha\subseteq P_\beta$ or $P_\beta \subseteq P_\alpha$.

, we can assume $P_\alpha \subseteq P_\beta$, which gives us $a,b \notin P_\alpha$.

Since $P_\alpha$ is an prime ideal, $ab \notin P_\alpha$.

Thus $ab \notin P$.

Also see

 * Union of Chain of Prime Ideals of Commutative Ring with Unity is Prime Ideal