Let $R$ be an arbitrary ring, $\{P_1,....,P_n\}$ be a set of prime ideals. Verify that $P_1 \cap ... \cap P_n$ is prime if and only if there exists $1 \leq i \leq n$ such that $P_i$ is contained in all other $P_j$'s.

I know the lemma that if $P=P_1 \cap ... \cap P_n$, then there exists i such that $P=P_i$ (the = can also be containment as well).

Beyond this, I am a little stumped by how to prove this.

  • $\begingroup$ You're almost there - the lemma gives the result, you just need to do a little set-theoretic work. Recall that an intersection of sets is always contained in each set $\endgroup$
    – zcn
    Mar 30 '14 at 20:55
  • $\begingroup$ Right to left: intersection(p1 to pn) = p = p_i implies a prime intersection. Left to right: p = pn = intersection(pi) implies intersection(pi) contain pn The above is easy to me so I wanted to share. I believe it's correct. Note pn = p_n for all variables $\endgroup$ Dec 10 '21 at 6:34

If $P_i \subset P_j$ for every $j$, then $P_1 \cap \cdots \cap P_n = P_i$ is prime. Conversely, if $P_1 \cap \cdots \cap P_n = Q$ is prime then $Q \subset P_j$ for every $j$ and also there exists some $i$ such that $P_i \subset Q$. Hence $Q=P_i$. Here i used the fact that if $a$ is an ideal such that $a \supset p_1 p_2 \cdots p_m$ where the $p_i$ are prime, then $a$ contains some $p_k$. (If i am not mistaken, this is proposition 1.11b in Atiyah-MacDonald.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.