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Let $S = \mathbb k[x_1, \ldots, x_n]$ be a polynomial ring over a field $\mathbb k$ and let $I \subseteq S$ be a monomial ideal. For a monomial ideal $J$, let $\#(J)$ be the smallest number of monomials that generate $J$. A monomial quotient $S/I$ is called Cohen-Macaulay if the projective dimension of $S/I$ is equal to $\min\{\#(J): J \text{ is an irreducible component of }I\}$.

Is there a name for the "opposite" property that the projective dimension of $S/I$ is equal to $\max\{\#(J): J \text{ is an irreducible component of }I\}$? Are there prior work on these rings?

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    $\begingroup$ What is the largest number of generators of any irreducible component of $I$? $\endgroup$
    – Sándor Kovács
    Commented Sep 11, 2016 at 21:48
  • $\begingroup$ This does not answer your question as written, but you might be looking for the concept of "sequentially Cohen-Macaulay". $\endgroup$
    – Alexander Woo
    Commented Sep 11, 2016 at 22:07
  • $\begingroup$ @SándorKovács, you are right my wording was quite ambiguous. I have edited it to hopefully make it more clear what I'm looking for. $\endgroup$
    – SorcererofDM
    Commented Sep 12, 2016 at 0:03
  • $\begingroup$ I'm not familiar with the intricacies of monomial ideals. Is it true that the minimal number of generators equals the height? If not, then this definition seems very strange. $\endgroup$
    – Sándor Kovács
    Commented Sep 12, 2016 at 20:10
  • $\begingroup$ @SándorKovács, yes, this can be proven using Alexander duality. $\endgroup$
    – SorcererofDM
    Commented Sep 12, 2016 at 23:54

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I don't think this "opposite CM" property would be very interesting. Here is why. Let's first localize at a maximal ideal, so we can work over a local ring. For simplicity let me denote this local ring still by $S$.

1) Using $n$ as in the question, for any module $M$ over $S$ we have an equality $$ {\rm proj.dim}\, M + {\rm depth}\, M = n $$

2) Furthermore, for any module $M$ over $S$ we have an inequality $$ {\rm depth}\, M \leq \dim M $$

3) For an ideal $I$ generated by $r$ elements we have an inequality $$ {\rm height}\, I \leq r. $$ Based on your comment in response to my question in this case this is actually an equality, which implies that $$ \dim (S/I) = n -r. $$

Putting 1)-3) together says that $$ {\rm proj.dim}\, (S/I) = n- {\rm depth}\, (S/I) \geq n-\dim(S/I) = r. $$

In other words, you could rephrase the CM condition that the projective dimension of $S.I$ takes the smallest possible value this inequality allows. It also implies that all the irreducible components of $I$ are generated by the same number of elements.

The notion you are suggesting has no property that would influence the other irreducible components. I think you could do the following to have an example: Take an arbitrary ideal $I_0\subseteq S$ and let $r={\rm proj.dim}\,(S/I_0)$. Now let $\mathfrak p\subseteq S$ be a prime ideal of height $r$ that is generated by $r$ elements and does not contain any minimal primes of $I_0$ and let $I=\mathfrak p\cap I_0$. Then $S/I$ has your "opposite CM" condition, but there seems to be very little chance for some interesting behaviour.

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  • $\begingroup$ Sorry, I misread your comment. I thought you were talking about the irreducible components of a monomial ideal, which was what I was referring to in my question when talking about generators. The irreducible components of a monomial ideal are prime ideals of the form $(x_{i_1}, \ldots, x_{i_k})$, and for these ideals the height is equal to the number of generators. In general, this may not be true. For example, look at the ideal in the polynomial ring S generated by all degree 2 monomials. There are $n(n-2)/2$ of them, but their height obviously can't be more than $n$. $\endgroup$
    – SorcererofDM
    Commented Sep 15, 2016 at 16:22

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