Existence of minimal prime ideal contained in given prime ideal and containing a given subset

Solution 1:

Hint: The intersection of a chain of prime ideals is prime.

Full solution:

The idea is to apply Zorn's lemma downwards. Consider the set $S:=\{Q \ | \ Q\text{ is a prime ideal of } R, X\subseteq Q\subseteq P\}$ ordered by inclusion. Note that $P\in S$ implies $S\neq\emptyset$. Consider a chain $\{Q_i\}_{i\in I}$ in $S$. Then $\bigcap_{i\in I} Q_i$ is in $S$, so by Zorn's lemma there exists an element minimal in $S$, i.e., minimal with respect being prime inside $P$ and containing $X$.

Let us see that $Q:=\bigcap_{i\in I} Q_i$ is actually prime: let $ab\in Q$ with $a,b\in R\setminus Q$. Then there exist indices $j,k$ such that $a\not\in Q_j$, $b\not\in Q_k$, say with $Q_j\supseteq Q_k$. Therefore $a,b\not\in Q_k$, but $ab\in Q$ implies $ab\in Q_k$, contradicting the primeness of $Q_k$.

Regarding your last question: Just apply the above proposition with $X:=0$ to get a minimal prime ideal of $R$ inside $P$.