Let $R$ be a unital commutative ring, $P$ $\subseteq$ $R$ a prime ideal, $X\subseteq P$ a subset. Show there exists a minimal (inclusion minimal) prime ideal contained in $P$ which contains $X$.

My approach: Let $\langle X\rangle$ be the ideal generated by $X$.
$P^c$ is a multiplicative set, as $P$ is prime. $\langle X\rangle\subseteq P\subseteq R$ and $P^c\cap\langle X\rangle=\emptyset $ so by Zorn's lemma there exist prime ideal $I_1$ satisfying $\langle X\rangle\subseteq I_1\subseteq P$. As $I_1$ is prime, it's complement is a multiplicative subset which does not intersect $\langle X\rangle$, hence I can find prime ideal $I_2$ satisfying $\langle X\rangle\subseteq I_2 \subseteq I_1$ and so on... I've got an infinite descending chain of prime ideals and I can't be sure that the process will stop.

Hints will be appreciated, thank you.

Edit: Can I conclude that every prime ideal contains a minimal prime ideal?

  • $\begingroup$ It may be noted that we don't need commutativity or the presence of identity for this. $\endgroup$
    – Atom
    Jan 7, 2023 at 13:42

1 Answer 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 $P$.


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