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

  • $\begingroup$ Just out of curiosity: why do you care about this? Also, is it the finite generation part that is the toughie here? $\endgroup$ Nov 22, 2010 at 23:13
  • 1
    $\begingroup$ @Keerthi: I think the tough part is to assure that ${\rm Ass}(M)=\{p\}$ $\endgroup$ Nov 22, 2010 at 23:17

2 Answers 2


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: $$ \mathfrak q \text{ is $\mathfrak p$-primary} \Leftrightarrow {\rm Ass}(A/\mathfrak q)=\{\mathfrak p\}. $$ 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.

  • $\begingroup$ Yes, I absolutely agree with this implication. See mathoverflow.net/questions/46872/primary-regular-sequences. I take a look at the reversed implication. Thank you. $\endgroup$
    – David
    Nov 22, 2010 at 23:41
  • $\begingroup$ Just for sure. So complete intersection can be translated to this: for any prime $p \subset A$, there exists a set of elements $x_1, \dots, x_t \in p$ such that $t=ht(p)$ and the $x1, \dots, xt$ generate a $p$-primary ideal? Thank you. $\endgroup$
    – David
    Nov 22, 2010 at 23:53
  • $\begingroup$ David, this is set-theoretic complete intersection. A complete intersection would mean that the ideal itself is generated by ${\rm ht}(p)$ number of elements. $\endgroup$ Nov 23, 2010 at 0:12

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).

  • $\begingroup$ All of this is standard technique, of course. It was first used by Hochster to show that small Cohen-Macaulay conjecture implies Serre's Positivity. $\endgroup$ Nov 23, 2010 at 2:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.