Presentation of finite modules with null annihilator - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T05:33:12Z http://mathoverflow.net/feeds/question/68956 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/68956/presentation-of-finite-modules-with-null-annihilator Presentation of finite modules with null annihilator Mahdi Majidi-Zolbanin 2011-06-27T18:48:36Z 2011-06-27T20:07:54Z <p>Let $R$ be a noetherian local ring and let $M$ be a finite $R$-module. Assume that the annihilator of $M$ is zero. Consider a minimal presentation of M as follows: $R^n\stackrel{\varphi}{\longrightarrow}R^m\longrightarrow M\longrightarrow0$. Can we conclude that $m>n$, or is it also possible to have $m\leq n$ with all $m\times m$ minors of the presentation matrix $\varphi$ equal to zero?</p> http://mathoverflow.net/questions/68956/presentation-of-finite-modules-with-null-annihilator/68964#68964 Answer by Hailong Dao for Presentation of finite modules with null annihilator Hailong Dao 2011-06-27T20:07:54Z 2011-06-27T20:07:54Z <p>Graham's comment gave some simple counterexamples. I will show that even if $R$ is nice, say a Gorenstein domain, there will always be a lot of counter-examples. </p> <p>Let $M$ be a non-free maximal CM module over $R$. Consider a minimal presentation: $$0 \to N \to R^n \to R^m \to M \to 0$$ If $m\leq n$ we found our counter example. If $m>n$ then dualizing the sequence (note that since $R$ is Gorenstein dualizing preserve exactness), so one gets a sequence:</p> <p>$$0 \to M^* \to R^m \to R^n \to N^* \to 0$$</p> <p>hence $N^*$ is a counter-example! </p>