AS Cohen Macaulay algebras and dualizing complexes

2012-12-15T17:46:17Z

Let $A$ be an $\mathbb N$-graded algebra such that $A_0 = k$ is a field. This are usually called graded connected algebras.

One can define a torsion functor with respect to the ideal $\mathfrak m = \bigoplus_{i \geq 1} A_i$, setting for any graded $A$ module $M$

$$\Gamma_{\mathfrak m}(M) = \{m \in M | A_{\geq i}m = 0 \mbox{ for } i \gg 0 \} \cong \varinjlim_i Hom_A(A/A_{\geq i},M).$$

The derived functors of $\Gamma_{\mathfrak m}$ are the local cohomology functors with respect to $\mathfrak m$, and are denoted by $H^i_\mathfrak m$. As in the commutative case, there is a natural isomorphism $$H_\mathfrak m(-) \cong \varinjlim Ext_A^i(A/A_{\geq i}, -)$$

We say $A$ is

AS Cohen Macaulay if there is a natural number $n$ such that $H_\mathfrak m^i (A) = H_\mathfrak m(A^{op}) = 0$. Let us call $n$ the local dimension of the module $A$ (I'm not sure this is standard notation)
AS Gorenstein if it has finite injective dimension $n$ both as a right and left module, and furthermore $Ext_A^n(k,A) = k$, once again on both sides.

This are generalizations of ye olde condition of regularity for graded connected algebras introduced by Artin and Schelter, hence the AS. If $A$ is commutative and noetherian, then they are equivalent to their AS-less counterparts. (Maybe you can drop the noetherian hypothesis on this, but I'm not sure.)

We have the usual implication chain

AS regular $\Rightarrow$ AS Gorenstein $\Rightarrow$ AS Cohen Macaulay.

By Groethendick's vanishing theorem, if $A$ is a noetherian Cohen Macaulay algebra of local dimension $n$, then $H^i_\mathfrak m \equiv 0$ for $i > n$.

Question 1: Is this result still true for noncommutative noetherian AS Cohen Macaulay algebras?

The result is true for noetherian AS Gorenstein algebras, as explained in this paper by Yekutieli and Zhang, Corollary 4.3. The argument comes basically from the fact that AS Gorenstein algebras have balanced dualizing complexes, almost by definition. This brings me to my

Question 2: Are there AS Cohen Macaulay algebras without (balanced or unbalanced) dualizing complexes?

Thanks!

AS Cohen Macaulay algebras and dualizing complexes - MathOverflow

AS Cohen Macaulay algebras and dualizing complexes