Homologically nice commutative rings

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).

My feeling is that this is really about set-theoretic complete intersections.

Let $X={\rm Spec} A$ be a noetherian affine scheme such that every irreducible subscheme of $X$ is a set-theoretic complete intersection. In other words, for any prime $\mathfrak p\subset A$, there exist a set of elements $x_1,\dots,x_t\in \mathfrak p$ such that $t={\rm ht} (\mathfrak p)$ and the $x_1,\dots,x_t$ generate a $\mathfrak p$-primary ideal with $\mathfrak p= \sqrt{( x_1,\dots, x_t)}$, or equivalently the zero set $Z(x_1,\dots,x_t)=\overline{\{\mathfrak p\}}$.

In this case, take $M=A/(x_1,\dots,x_t)$ has the property that ${\rm Ass} (M)=\{\mathfrak p\}$.

If in addition $A$ is CM, then so is $M$ and then its projective dimension satisfies that $$ {\rm pd} (M)=\dim A-\dim M= {\rm ht} (\mathfrak p). $$

I suppose the next step is to look at an affine scheme with an irreducible subscheme that is not a set-theoretic complete intersection and see what happens there.

(Addendum)

Regarding the case when there exists an irreducible subscheme that is not a set-theoretic complete intersection, one may mention, that in general, (still assuming that $A$ is CM, which follows if it is regular), $$ {\rm pd} (M)=\dim A-{\rm depth}_A M. $$ If furthermore ${\rm Ass} (M)=\{\mathfrak p\}$, then it follows that $$ {\rm depth}_A M \leq \dim M = \dim A - {\rm ht} (\mathfrak p), $$ So ${\rm pd} (M)\geq {\rm ht} (\mathfrak p)$ with equality iff $M$ is CM. In other words, your desired condition is to find a CM module whose only associated prime is $\mathfrak p$.

At least for modules generated by a single element this seems to be pretty close to $\overline{\{\mathfrak p\}}$ being a set-theoretic complete intersection as for an ideal $\mathfrak q\subseteq A$ in a noetherian ring the following holds: $$ \mathfrak q \text{ is $\mathfrak p$-primary} \Leftrightarrow {\rm Ass}(A/\mathfrak q)=\{\mathfrak p\}. $$ One way to ensure that $A/\mathfrak q$ is CM is to make sure that $\mathfrak q$ has the right number of generators and in order to have the condition on the associated primes one would need that $\mathfrak q$ is $\mathfrak p$-primary. Of course, I am not claiming that this is the only way to produce such modules, but this seems to be the obvious way.