Intersection of family of prime ideals

Question

Here's the problem I am attempting...

Let $R$ be a commutative ring with identity. Let $\mathcal P=\{\mathfrak p_1, \mathfrak p_2, \ldots, \mathfrak p_m\}$ and $\mathcal Q=\{\mathfrak q_1, \mathfrak q_2, \ldots, \mathfrak q_n\}$ be two family of prime ideals of $R$ such that $\bigcap\limits_{i=1}^m \mathfrak p_i=\bigcap\limits_{i=1}^n \mathfrak q_i$. Assume that there are no non-trivial inclusion relations in either family. Prove that the two families coincide.

Here's how I'm interpreting

• "no non-trivial inclusion relations in either family": For each $\mathfrak p_i$, there is an $x_i\in\mathfrak p_i$ such that $x_i\notin \mathfrak p_j$ for all $j\neq i$. To put it simply, $\mathfrak p_i \not\subseteq \bigcup\limits_{j\neq i}\mathfrak p_j$. (Similar interpretation for the other family.)

• "two families coincide": Either $\mathcal P\subseteq \mathcal Q$ or, $\mathcal Q\subset \mathcal P$.

Let me know if you think the terms used in the question are ambiguous. It's possibly some university exam question (someone asked this question on a Telegram channel).

I know the avoidance lemma for prime ideals. However, I don't think it's relevant here. Please give me hints at solving this or provide counter example if the statement is false considering the way I interpreted it. I can see that it's true in the ring of integers. For instance, any family (with no non-trivial inclusion relation) of prime ideals whose intersection is $6\mathbb Z$ must contain $2\mathbb Z$ and $3\mathbb Z$.

Answer 1

Here is an answer in the affirmative.

Lemma: Suppose $\mathfrak{a}$ and $\mathfrak{b}$ are two ideals whose intersection is contained in some prime ideal $\mathfrak{p}$. Then at least one of the ideals is contained in $\mathfrak{p}$.

Proof: Suppose neither ideal is contained in our prime. Then there is some element $x$ of $\mathfrak{a}$ not in $\mathfrak{p}$ and some element $y$ in $\mathfrak{b}$ with the same property. But $xy$ is in $\mathfrak{p}$ by assumption, contradicting the fact that $\mathfrak{p}$ is prime.

One can iterate the above lemma to obtain a similar statement for an arbitrary number of ideals whose intersection is contained in a prime ideal.

To conclude the proof, we can deduce from the above that for each $j$ there is some $i$ such that $\mathfrak{p}_i \subseteq \mathfrak{q}_j$. Moreover, there is some $k$ such that $\mathfrak{q}_k \subseteq \mathfrak{p}_i$. Hence the prime ideals must be equal, else contradict the “no nontrivial inclusion relations” hypothesis.