Intersection of prime ideals in a chain is prime

Question

I'm trying to prove that the intersection of prime ideals in a chain is again prime. It seems easy enough for any pair of prime ideals: let $P$ and $P'$ be two prime ideals in the chain. Assume without loss of generality that $P \subset P'$. If a product $ab$ is contained in their intersection, then $ab$ is in $P$, and therefore either $a$ or $b$ is in $P$, and therefore also in $P'$, and therefore also in the intersection.

Now my question is, how do I formally extend this to a statement about the intersection of infinitely many ideals? If the statement holds for the intersection of any two prime ideals, then surely it must hold for countably many of them, right? How can I make this more rigorous?

I guess this type of question isn't really specific to ring theory, it's more of a general question about how one takes a statement about finitely many... things and turns it into a statement about infinitely many things. But I couldn't figure out how to phrase this question more generally or tag it appropriately, so any help with that would be good too.

Answer 1

Let $S$ be a non-empty chain of prime ideals. Then $\cap S$ is an ideal, in fact a prime ideal: First, since $S$ is non-empty, there is some $P \in S$, so that $\cap S \subseteq P$ and we see that $\cap S$ is a proper ideal (this is needed for a prime ideal - notice that $\cap \emptyset$ equals the whole ring and therefore is no prime ideal). Next, if $a,b \notin \cap S$, there are $P,Q \in S$ such that $a \notin P$ and $b \notin Q$. We may assume $P \subseteq Q$. Then $b \notin P$ and hence $ab \notin P$. Hence, $ab \notin \cap S$.

Answer 2

Let $P_i$, $i \in I$ be a collection of ideals indexed by $i$, and such that the order on $I$ corresponds to the inclusion of ideals. Denote the intersection by $Q$

Suppose $ab \in Q$ for some $a,b \in R$. Thus $ab \in P_i$ for all $i$. Thus for every $i$ at least one of $a,b$ is in $P_i$. If we say

$$A = \{i \in I| a \in P_i\}$$ $$B = \{i \in I| b \in P_i\}$$

Then $A \cup B = I$. I claim (and leave it as fairly quick exercise to the reader) that this means at least one of $A,B$ has no strict lower bound in $I$ (that is, an $i$ such that $I \notin A$ (resp. $B$) but bounding $A$ below). In fancier language, one of them is cofinal. Suppose $A$ isn't strictly bounded below. Then for any $i$, there is a $j \in A$ with $j \leq i$, thus $a \in P_j \subset P_i$ so $a \in P_i$ for all $i$.

Note however that often it is good to look at a contrapositive form of the statement, which may be only a finite problem (see Martin Brandenburg's answer)