Suppose $R$ is a regular local ring of dimension at least 3, and suppose $P_1, P_2$ are dimension 1 primes. Does there necessarily exist a dimension 2 prime $Q$ contained in both?
In other words, is every pair of "little curves" contained in a "little surface"?
If this is false, will it hold if the ring is complete?
I am particularly interested in the mixed characteristic case.
I prove below that the answer is positive in the following special case: $R$ is the localization of a finitely generated polynomial ring over a field $k$. By Cohen's structure theorem this will also cover the case that $R$ is finite dimensional, complete and equicharacteristic.
For the proof it suffices to consider the case that $k$ is algebraically closed and $R$ is the localization of $k[x_1, \ldots, x_n]$ at the maximal ideal generated by $x_1, \ldots, x_n$. let $C_1, C_2$ be two irreducible curves on $k^n$ passing through the origin. It suffices to show that there is an irreducible surface $S$ containing both $C_1$ and $C_2$.
Claim: there is a curve $C$ and non-constant (and therefore dominant) maps $\phi_j: C \to C_j$, $j = 1, 2$.
Proof of the claim: Take any polynomial $f$ which is non-constant on both $C_j$. The fields $k(C_j)$ of rational functions on $C_j$ are finite extensions of $k(f)$, so that there is a common finite extension $L$ of $k(C_1)$ and $k(C_2)$ (see this for a neat trick). Now define $C$ to be the (unique) nonsingular curve such that $k(C) = L$.
It is easy to construct $S$ with $C, \phi_1, \phi_2$ from the claim: take the closure of the image of $\phi: C \times k \to k^n$ given by $\phi(x, t) := t\phi_1(x) + (1-t)\phi_2(x)$.
