Consider a noetherian ring $R$ and a collection $m\_i$, $i\in I$ of maximal ideals of $R$. Let $P$ be a prime ideal of $R$. It is well-known that if the collection is finite (i.e. the index set $I$ is finite), then $P\subseteq \cup\_{i\in I}m\_i$ if and only if $P$ is contained in one of the $m\_i$. (This is true under weaker assumptions.)


Now my question is what about the case of an infinite collection. Does the same hold (under suitable assumptions, e.g. $R$ is normal integral, etc.)? Or can anyone give a counter-example? 



union of infinitely many prime ideals
I fixed your notation so that there wasn't any equivocation. I changed $P\_i$ to $m\_i$, changed the ideal $I$ to $P$, and kept the set $I$ as $I$.


There is indeed a counterexample. Take $R=\mathbb C[x,y]$, the prime ideal $P=(x,y)$ is not contained in any of the ideals $m\_{a,b}=(x-a,y-b)$ for $(a,b)$ not both $0$, since $P$ is also a maximal ideal. Yet every element of $P$ is contained in at least one of the $m\_i$, because there is no polynomial function on $\mathbb A^2$ that vanishes on one point but not any other point. Thus $P \subset \bigcup\_i m\_i$.


