Are maximal Cohen-Macaulay modules supported everywhere? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T09:29:47Z http://mathoverflow.net/feeds/question/84685 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/84685/are-maximal-cohen-macaulay-modules-supported-everywhere Are maximal Cohen-Macaulay modules supported everywhere? ashpool 2012-01-01T15:43:37Z 2012-01-01T16:58:32Z <p>Let $A$ be a local CM ring, and $M$ a maximal CM $A$-module. Is it true that $\operatorname{Supp}M=\operatorname{Spec}A$ ? This suspicion stems from such statements as:</p> <ul> <li>If $\omega$ is a canonical module of $A$, then $\omega_{\mathfrak{p}}$ is a canonical module of $A_{\mathfrak{p}}$ for every $\mathfrak{p}\in\operatorname{Spec}A$.</li> <li>If $\omega$ is a canonical module of $A$, then $\mu_i(\mathfrak{p},\omega)=\delta_{i}^{\operatorname{ht}\mathfrak{p}}$ for every $\mathfrak{p}\in\operatorname{Spec}A$, where $\mu$ denotes the Bass number.</li> </ul> <p>And am I correct in understanding that maximal CM module is by definition nonzero?</p> http://mathoverflow.net/questions/84685/are-maximal-cohen-macaulay-modules-supported-everywhere/84688#84688 Answer by Graham Leuschke for Are maximal Cohen-Macaulay modules supported everywhere? Graham Leuschke 2012-01-01T16:58:32Z 2012-01-01T16:58:32Z <p>No. $R = k[x,y]/(xy)$, $M = R/(x)$.</p> <p>The zero module has infinite depth and support of dimension $-\infty$, so should not be considered MCM.</p>