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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?

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Write $(0)$ in $R$ as an intersection of ideals $I_i$, and resolve the direct sum of the $R/I_i$. Counterexamples will appear quickly. – Graham Leuschke Jun 27 '11 at 19:34
Thank you Graham, nice example. And thank you Hailong for responding. – Mahdi Majidi-Zolbanin Jun 27 '11 at 20:11
I came up with another class of counterexamples. Consider any module $M$ with a finite presentation $R^n\rightarrow R^m\rightarrow M\rightarrow0$, where $m\leq n$ and annihilator of $M$ is not zero. Let $I$ be the annihilator of $M$, and tensor the given presentation with $R/I$. The result is an example of an $R/I$-module $M$ whose annihilator over $R/I$ is $(0)$ but has a minimal presentation over $R/I$ with $m\leq n$. – Mahdi Majidi-Zolbanin Jun 29 '11 at 17:44
Must assume $M$ is not free over $R/I$. – Mahdi Majidi-Zolbanin Jun 29 '11 at 18:19
up vote 2 down vote accepted

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.

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:

$$ 0 \to M^* \to R^m \to R^n \to N^* \to 0 $$

hence $N^*$ is a counter-example!

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$N$ is also maximal Cohen-Macaulay and that's why annihilator of $N^*$ is zero, since $R$ is a domain, right? – Mahdi Majidi-Zolbanin Jun 27 '11 at 20:59
Mahdi, that's right. – Hailong Dao Jun 28 '11 at 0:04

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