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The Fitting ideal is defined either as the ideal generated by the $n-i$ minors of matrix of the kernel of the map $R^{n}\rightarrow M$, or more generally via a free resolution. But no matter which definition we use, I do not understand that if $M$ is generated by $k$ elements over $R$, then $Fitt_{k}(M)=R$.

The most trivial case is $M=R$. In this case we have all relations to be $0$. Therefore all the minors can only be $0$ as well. Now - why $Fitt_{1}=R$ instead? Where is this $R$ coming from? Is this just convention?

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I disagree. In the example above we are working with $Fitt_{1}$, and in general case $Fitt_{i}$. I need to know why for modules over a local ring with $i$ generators $Fitt_{i}=R$, not why $Fitt_{0}=R$. – Bombyx mori Jan 21 '13 at 16:48
I see. It seems a notational confusion. – Bombyx mori Jan 21 '13 at 17:37

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