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Let $R$ be a commutative unitary ring and $P_1,P_2,...,P_m$ a finite family of prime ideals in $R$ such that $$ \bigcap_{k=1}^m P_k \subseteq P $$ for a prime ideal $P$ in $R$. Then $P_k\subseteq P$ for some $k\in\{1,2,...,m\}$ because $P_1P_2...P_m\subseteq \bigcap_{k=1}^m P_k$ and $P$ is prime.

Is the same true for an arbitrary infinite intersection $\bigcap_M P_k$ of prime ideals?

More precisely, let $P_k$ with $k\in M$ and $J$ be prime ideals in $R$ with $\bigcap_M P_k\subseteq P$. Do we have $P_k\subseteq P$ for some $k\in M$?

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  • $\begingroup$ $(x,ay+bz)$ is prime ideal of $\mathbb{Q}[x,y,z]$ and $\displaystyle\bigcap_{a,b \in \mathbb{Q}^*} (x,ay+bz)= (x)$ $\endgroup$ – reuns Oct 10 '17 at 21:35
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No: let $R=\mathbb{Z}$, $P=(0)$, and let $P_1=2\mathbb{Z},P_2=3\mathbb{Z},P_3=5\mathbb{Z}$, and so on. Then $$ \bigcap_{k=1}^{\infty}P_k=(0)=P$$ but none of the $P_k$ are contained in $P$.

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    $\begingroup$ Ah, you beat me by a minute! :D $\endgroup$ – RKD Oct 8 '17 at 20:46
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No. Take the prime ideals $(p)$ for all the prime numbers in $\mathbb Z$.

It's clear that $$\bigcap_{p}(p)=(0)$$ and $(0)$ is prime, but $(p)$ is never a subset of $(0)$.

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