5
$\begingroup$

By the fundamental work of De Concini, Eisenbud, and Procesi, an algebra with straightening law (ASL) must be Cohen-Macaulay if it is built on a Cohen-Macaulay poset. I would like to understand the state of the art regarding the converse question: If $A$ is a Cohen-Macaulay ASL over a field $k$ generated by the poset $P$, how far from Cohen-Macaulay can $P$ be (over $k$)? More precisely:

(a) Is there a known lower bound on the depth of the Stanley-Reisner ring $k[P]$?

(b) What is the "worst" known example?

(c) What are the known conditions on $P$ or $A$ that force $P$ to be Cohen-Macaulay?

I have been able to extract very little information about this from the literature I can find online. Here is what I know:

1) If $P$ is Buchsbaum and $A$ is Cohen-Macaulay, $P$ is Cohen-Macaulay (Miyazaki 2010). This is a partial answer to (c), but the Buchsbaum assumption is very strong.

2) Terai 1994 claims that $\operatorname{depth} k[P] \geq \operatorname{depth} A - 1$; however, Miyazaki 2010 claims Terai's proof is in error. This would have been a good answer to (a) if Terai had been correct, or to (b) if Miyazaki had given a counterexample.

3) In the more general setting of Hodge algebras, Hibi 1986 gives an example of a Cohen-Macaulay hodge algebra whose discretization is not Cohen-Macaulay. However, this example is not an ASL (or this would have been a partial answer to (b)).

This is all I have been able to find so far. I don't even know an example of a Cohen-Macaulay ASL whose discrete counterpart is not Cohen-Macaulay. I would appreciate any guidance you can offer about what is known about this.

Update 2/13/17: I reached Eisenbud and Terai by email. Neither is aware of an example of a C-M ASL whose discretization is not C-M, though Eisenbud says he would expect such a thing to exist. Terai says the problem is very difficult. I am willing to assume at this point that more or less nothing is currently known beyond the result of Miyazaki linked above. But if you do know something, I'm all ears.

$\endgroup$
4
$\begingroup$

Aldo Conca and Matteo Varbaro have posted a preprint that purports to answer this question, at least for graded ASLs:

Squarefree Gröbner degenerations

It appears that if $A$ is Cohen-Macaulay, then $k[P]$ must also be Cohen-Macaulay! See Corollary 3.9.

Addendum June 25, 2020: Conca and Varbaro's paper has now been published in Inventiones Mathematicae.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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