Question: Let $R$ be a commutative ring with identity and $C$ be a chain of prime ideals of $R$. Show that $\cup_{P\in C}P$ and $\cap_{P\in C}P$ are prime ideals of $R$.
Thoughts: I am thinking I should use Zorn's Lemma here, but I am not quite sure. I was also thinking this is the Prime avoidance Lemma, but that only deals with a finite number of primes, and so I didn't know how to handle an infinite chain.
Simply use the definition of prime. Here's a sketch. Let $a, b \in R$ be arbitrary. (I'm going to assume that the chain $\mathcal{C}$ is non-empty.)
Intersection. Suppose $ab \in \bigcap_{P \in \mathcal{C}} P.$ Then, $ab \in P$ for every $P \in \mathcal{C}$. Thus, given any $P$, either $a \in P$ or $b \in P$. (Now, there's the issue that some primes could contain $a$ and not $b$ while others contain $b$ and not $a$, we would want to ensure that one of them is in every prime in $\mathcal{C}$.)
Now, if $a$ and $b$ are both in every prime $P \in \mathcal{C}$, we are done. Suppose that that is not the case.
Then, there is a prime $P_0 \in \mathcal{C}$ which contains one and not the other. WLOG, assume $a \in P_0$ and $b \notin P_0$.
We now claim that $a \in P$ for all $P \in \mathcal{C}$.
Indeed, given any such $P$, there are two cases:
(This is where we have used that $\mathcal{C}$ is a chain.)
Union. This is easier. Suppose $ab \in \bigcup_{P \in \mathcal{C}}P$. Then, $ab \in P$ for some $P \in \mathcal{C}$. But $P$ is prime and thus, either $a \in P$ or $b \in P$. In turn, $a \in \bigcup P$ or $b \in \bigcup P$. (We didn't use the chain condition but we need that to know that the union is an ideal.)
In both cases, you have to now check two more things: Both are ideals and are not all of $R$. The latter is easy since $1 \notin P$ for all $P$.
For the former, it is easy in the case of intersection (indeed, you don't need the chain condition). It is not too difficult for the union either but you need to use the chain condition now.