The set of prime ideals of a commutative ring contains inclusion-minimal elements

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.4 Exercise 7.4.36
Solution:
Let $\mathcal{P}$ denote the set of prime ideals. Note that $\mathcal{P}$ is partially ordered by the superset relation, and is nonempty since every maximal ideal is prime. Let $\{P_k\}_K$ be a chain in $\mathcal{P}$; that is, $P_\ell \supseteq P_k$ for all $\ell \leq k$. We know that $\bigcap P_k$ is an ideal in $R$ by Exercise 7.3.18; we now show that this ideal is prime.
Suppose $ab \in \bigcap P_k$. Then $ab \in P_k$ for all $k$. Suppose that (without loss of generality) there exists $m \in K$ such that $a \notin P_m$. Then $b \in P_m$, and moreover, $b \in P_k$ for all $\ell \leq m$. Now let $\ell > m$; since $ab \in P_\ell$, either $a \in P_\ell$ or $b \in P_\ell$. If $a \in P_\ell$, then $a \in P_m$, a contradiction. So $b \in P_\ell$. Hence $b \in P_k$ for all $k$, and we have $b \in \bigcap P_k$. Thus $\bigcap P_k$ is prime. So $\bigcap P_k \in \mathcal{P}$, and thus the chain $\{P_k\}_K$ has an upper bound.
Since the chain $\{P_k\}_K$ is arbitrary, by Zorn’s Lemma, $\mathcal{P}$ contains an inclusion-minimal element.