Can a prime in a Dedekind domain be contained in the union of the other prime ideals? Suppose $R$ is a Dedekind domain with a infinite number of prime ideals.  Let $P$ be one of the nonzero prime ideals, and let $U$ be the union of all the other prime ideals except $P$.  Is it possible for $P\subset U$?
As a remark, if there were only finitely many prime ideals in $R$, the above situation would not be possible by the "Prime Avoidance Lemma", since $P$ would have to then be contained in one of the other prime ideals, leading to a contradiction.
The discussion at the top of pg. 70 in Neukirch's "Algebraic Number Theory" motivates this question.
Many thanks,
John

If $R$ is the ring of integers $O_K$ of a finite extension $K$ of $\mathbf{Q}$, then I don't think this can happen. The class of the prime ideal $P$ is of finite order in the class group, say $n$. This means that the ideal $P^n$ is principal. Let $\alpha$ be a generator of $P^n$. Then $\alpha$ doesn't belong to any prime ideal other than $P$, because at the level of ideals inclusion implies (reverse) divisibility, and the factorization of ideals is unique.
This argument works for all the rings, where we have a finite class group, but I'm too ignorant to comment, how much ground this covers :-(
