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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.

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  • $\begingroup$ 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
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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.)

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