AS Cohen Macaulay algebras and dualizing complexes - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T06:45:35Z http://mathoverflow.net/feeds/question/116469 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116469/as-cohen-macaulay-algebras-and-dualizing-complexes AS Cohen Macaulay algebras and dualizing complexes Pablo Zadunaisky 2012-12-15T17:46:17Z 2012-12-15T22:17:12Z <p>Let $A$ be an $\mathbb N$-graded algebra such that $A_0 = k$ is a field. This are usually called graded connected algebras. </p> <p>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 $M$</p> <p>$$\Gamma_{\mathfrak m}(M) = \{m \in M | A_{\geq i}m = 0 \mbox{ for } i \gg 0 \} \cong \varinjlim_i Hom_A(A/A_{\geq i},M).$$</p> <p>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 $$H_\mathfrak m(-) \cong \varinjlim Ext_A^i(A/A_{\geq i}, -)$$</p> <p>We say $A$ is</p> <ul> <li><p>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)</p></li> <li><p>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.</p></li> </ul> <p>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.)</p> <p>We have the usual implication chain </p> <blockquote> <p>AS regular $\Rightarrow$ AS Gorenstein $\Rightarrow$ AS Cohen Macaulay.</p> </blockquote> <p>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 $i > n$.</p> <p><strong>Question 1</strong>: Is this result still true for noncommutative noetherian AS Cohen Macaulay algebras? </p> <p>The result is true for noetherian AS Gorenstein algebras, as explained in <a href="http://www.ams.org/mathscinet/search/publdoc.html?arg3=&amp;co4=AND&amp;co5=AND&amp;co6=AND&amp;co7=AND&amp;dr=all&amp;pg4=AUCN&amp;pg5=TI&amp;pg6=PC&amp;pg7=ALLF&amp;pg8=ET&amp;r=1&amp;review_format=html&amp;s4=Yekutieli&amp;s5=Serre&amp;s6=&amp;s7=&amp;s8=All&amp;vfpref=html&amp;yearRangeFirst=&amp;yearRangeSecond=&amp;yrop=eq" rel="nofollow">this paper</a> 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</p> <p><strong>Question 2</strong>: Are there AS Cohen Macaulay algebras <em>without</em> (balanced or unbalanced) dualizing complexes?</p> <p>Thanks!</p>