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3 added 19 characters in body

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.

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: $$\mathfrak q \text{ is \mathfrak p-primary} \Leftrightarrow {\rm Ass}(A/\mathfrak q)=\{\mathfrak p\}.$$ So, in order One way to ensure that $A/\mathfrak q$ is CM one would need 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.

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\}.So, in order to ensure that $A/\mathfrak q$ is CM one would need 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.

1

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.