The concept "opposite" of Cohen-Macaulayness

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?

• What is the largest number of generators of any irreducible component of $I$? Sep 11 '16 at 21:48
• This does not answer your question as written, but you might be looking for the concept of "sequentially Cohen-Macaulay". Sep 11 '16 at 22:07
• @SándorKovács, you are right my wording was quite ambiguous. I have edited it to hopefully make it more clear what I'm looking for. Sep 12 '16 at 0:03
• 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. Sep 12 '16 at 20:10
• @SándorKovács, yes, this can be proven using Alexander duality. Sep 12 '16 at 23:54

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.

• 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 $n$. Sep 15 '16 at 16:22