
> 
> Let $R$ be a commutative regular local ring. Is it true that for every $\mathfrak p \in \mathrm{Spec}(R)$ there is a finitely generated $R$-module $M$ such that $\mathrm{projdim}(M)=\mathrm{ht}(\mathfrak p)$ and $\mathrm{Ass}(M)=\{\mathfrak p\}$?
> 
> 
> Or is there some family of commutative noetherian rings where is this true?
> 
> 
> 


I know that this holds if $R$ is commutative regular local of Krull dimension $\leq 4$ (up to dimension 3 it was easy, because factor rings $R/\mathfrak p$ where always CM, so we can take $M=R/\mathfrak p$. Dimension 4 was harder and in dimension 5 or more I don't know).


Is this an easy/hard/hopeless/already open/similar to something problem?


Thanks,
David



Homologically nice commutative 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.


