OK, so this is weaker than your last question, which is false but related to some open problems and also weaker than the conjecture in your first question which is another famous open problem. All of these sound like a nice project (-:.

The short anwser is this one is also open in general. I will show that your statement, if true for all regular local ring, would imply one of the outstanding open problem, the so-called Serre's Positivity conjecture. I will sketch the proof below.

Suppose your statement is true for any regular local ring $(R,m)$. Let $p,q$ be such that $\sqrt{p+q} =m$ and $\dim R/p + \dim R/q = \dim R$. By your statement, we can choose Cohen-Macaulay module $M,N$ such that $Ass(M)= \{p\}$ and $Ass(M)= \{q\}$. Let:

$$\chi(M,N) = \sum_{i\geq 0} (-1)^i\text{length}(\text{Tor}_i^R(M,N))$$

the Serre's intersection multiplicity.

Then $M\otimes N$ has finite length and $M,N$ are Cohen-Macaulay, and a classical result by Serre (can be found in his book Local Algebra, V.6, Theorem 4, p 110 of the English version) says that $\text{Tor}_i^R(M,N)=0$ for $i>0$, so

$$\chi(M,N) =\text{length}(M\otimes N)>0$$

(This is a nice generalization of Bezout theorem, since curves are Cohen-Macaulay)

But since $Ass(M)= \{p\}$ one has a prime filtration of $M$ by $a>0$ copies of $p$ and primes of smaller dimensions. Similarly $N$ has a filtration with $b>0$ copies of $q$ and primes of smaller dimensions. As $\chi$ is additive on short exact sequences, one has:

$$\chi(M,N) = ab\chi(R/p,R/q)$$

(one needs to use the Vanishing part of Serre's conjectures, which is known, here)

So one can conclude that $\chi(R/p,R/q)>0$! But this has been open for about 50 years, so I doubt your statement is known (for all regular local rings).

`${\rm Ass}(M)=\{p\}$`

– Sándor Kovács Nov 22 '10 at 23:17