Artinian property of local cohomology module over graded local ring - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T17:23:44Z http://mathoverflow.net/feeds/question/115726 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115726/artinian-property-of-local-cohomology-module-over-graded-local-ring Artinian property of local cohomology module over graded local ring Axy 2012-12-07T16:42:01Z 2012-12-07T17:19:05Z <p>We know that if \$(R, m)\$ is a local ring, \$M\$ is a finitely generated \$R\$-module, then the local cohomology module \$H^{i}_{m}(M)\$ is an Artinian module for every \$j\$.</p> <p>My question is : if \$(R,m)\$ is a graded local ring, then is \$H^{i}_{m}(M)\$ is an Artinian module for every \$j\$ ?</p> http://mathoverflow.net/questions/115726/artinian-property-of-local-cohomology-module-over-graded-local-ring/115729#115729 Answer by Karl Schwede for Artinian property of local cohomology module over graded local ring Karl Schwede 2012-12-07T17:19:05Z 2012-12-07T17:19:05Z <p>Yes, this even holds for non-graded rings. Indeed, suppose that \$R\$ is a (Noetherian?) ring and \$m \subseteq R\$ is a maximal ideal. </p> <p>Then \$\$ H^i_m(M) = H^i_{mR_m}(M_m) \$\$ essentially by Chapter III, Exercise 2.3(f) (excision) of Hartshorne. </p> <p>Also note that \$N \subseteq H^i_{mR_m}(M_m)\$ is an \$R\$-submodule if and only if it is an \$R_m\$-module. </p>