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 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 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>