Find one-dimensional $P,Q$ such that $PQ = P \cap Q$ and $P,Q$ not coprime I am looking for an example of the following situation:
Let $R = k[x,y,z]$ be the polynomial ring in three indeterminates where $k$ is a field. I want to find two prime ideals $P$ and $Q$ of $R$ which satisfy


*

*$R/P$ and $R/Q$ are Dedekind domains

*at least one of $P$ and $Q$ is not principal ,

*$P+Q \neq R$,

*$P \cap Q = PQ$.


In geometrical terms: I would like to find two regular affine curves in $\mathbb{A}_k^3$ given by prime ideals from which at least one is not principal (Points 1. and 2.). Moreover, they should intersect in at least one point (This is point 3.) and the curve obtained from these two curves (the curve given by $R/PQ$) should be reduced (This is point 4.).
 A: This can never happen. If such an example existed, localizing at a maximal ideal containing $P+Q$, the same holds and so assume that $R$ is a regular local ring of dimension three, $P,Q\subset R$ prime ideals with all your properties, 3) being automatic now. Then $R/P$ is a dvr, so let $z$ be generator of its maximal ideal. Since, $R/Q$ is also a dvr, $Q=(x,y)$ where they form a regular system of parameters in $R$. 
First, assume that neither $x$ nor $y$ is zero in $R/P$. Then, $x=uz^m, y=vz^n$ in $R/P$ with $u,v$ units. We may assume without loss of generality $m\geq n\geq 1$. By abuse of notation, we call the lifts of $u,v,z$ to $R$ by the same letters. Notice that $u,v$ are units in $R$ and $z$ is a non-unit. Then, $x'=x-v^{-1}uz^{m-n}y$ goes to zero in $R/P$. Thus, $x'\in P\cap Q$. But, $x'$ is part of a regular system of parameter and thus can not be in $PQ$, the latter is contained in the square of the maximal ideal.
The case when $x$ or $y$ is zero in $R/P$ reduces to the above.
