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.


1 Answer 1


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.


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