Castelnuovo-Mumford regularity and degree of generator. - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T23:29:03Z http://mathoverflow.net/feeds/question/109139 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109139/castelnuovo-mumford-regularity-and-degree-of-generator Castelnuovo-Mumford regularity and degree of generator. Knot 2012-10-08T12:30:46Z 2012-10-09T04:57:23Z <p>Let \$R\$ be a polynomial ring over a field \$k\$,: \$k[x_{1},..x_{n}]\$, \$\mathfrak{m}=(x_0,...,x_{n})\$ and \$M\$ be a finitely generated \$R\$ module.</p> <p>In a paper of Kodiyahlam, he define the Castelnuovo-Mumford regularity of \$M\$ to be the least integer number \$m\$ such that for every \$j\$ the \$j\$-th syzygy of \$M\$ is generated in degree less or equal \$m+j\$.Then, he conclude that the Castelnuovo-Mumford regularity of \$M/\mathfrak{m}M\$ is equal to the maximal degree of generator of \$M\$.</p> <p>Could you please so me why the CM regularity of \$M/\mathfrak{m}M\$ is equal to the maximal degree of generator of \$M\$ ? </p> <p>If \$I\$ and \$J\$ are two finitely generated ideal of \$R\$ and \$J\subseteq I\$ then do we have the CM regularity of \$J\$ is less or equal the CM regularity of \$I\$ ?</p> http://mathoverflow.net/questions/109139/castelnuovo-mumford-regularity-and-degree-of-generator/109202#109202 Answer by Pham Hung Quy for Castelnuovo-Mumford regularity and degree of generator. Pham Hung Quy 2012-10-09T04:13:08Z 2012-10-09T04:57:23Z <p>Notice that \$M/\mathfrak{m}M\$ is an \$R\$-module of finite length, so \$H^i_{\mathfrak{m} }(M/\mathfrak{m}M) = 0\$ for all \$i>0\$ and \$H^0_{\mathfrak{m} }(M/\mathfrak{m}M) = M/\mathfrak{m}M\$. Recalling that CM regularity can be computed via local cohomology module (see Brodmann-Sharp: local cohomology), we have \$\$reg (M/\mathfrak{m}M) = \max { end (H^i_{\mathfrak{m} }(M/\mathfrak{m}M)) +i | i= 0,...,n }.\$\$ Here, consider a graded \$R\$-module \$N = \oplus_iN_i\$, we denote \$end(N) = \sup {i | N_i \neq 0}\$. Therefore \$reg (M/\mathfrak{m}M) = end ((M/\mathfrak{m}M))\$. \$end ((M/\mathfrak{m}M))\$ is equal to the maximal degree of generator of \$M\$ by graded Nakayama lemma.</p>