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

This answer refers to the contributions of Jyrki and Georges: assume that a maximal ideal $P$ of a Dedekind domain $R$ is NOT contained in the union of all other maximal ideals. Then there exists an element $f\in P$ such that $v_P(f)=n>0$ for the discrete valuation attached to $P$ and $v_Q(f)=0$ for all $Q\neq P$. Now $P^n$ consists of those elements $r\in R$ such that $v_P(r) \geq n$. Thus for every $r\in P$ we get $r=fs$ with $s\in R$. Hence $P^n =fR$. Jyrki has already shown that if $R$ has torsion class group, then no maximal ideal contained in the union of all others can exist.
So: a maximal ideal of $R$ contained in the union of all other maximal ideals exists if and only if the class group of $R$ is not torsion.
