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Proposition 1.4.11 in Bruns and Herzog Cohen-Macaulay Rings reads:

Proposition: Let $R$ be a Noetherian ring and $\phi: F \rightarrow G$ a homomorphism of finite free $R$-modules. Then $rank (\phi) = r$ if and only if $grade(I_r(\phi)) \ge 1$ and $I_{r+1}(\phi)=0$.

Remarks on notation: by $rank(\phi)$ we mean the rank of $im(\phi) \otimes Q$, where $Q$ is the total ring of fractions of $R$ and $I_r(\phi)$ is a Fitting ideal with index $r$.

Question: In proving the direction $\Rightarrow$ i am stuck in showing that $I_{r+1}(\phi)=0$. What i have shown is that $I_{r+1}(\phi) \subset m$, for every maximal element $m$ of $Ass(R)$. How do we continue?

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If $M=\operatorname{Coker}\varphi$, then you have an exact sequence $F\stackrel{\varphi}\to G\to M\to 0$ and $\operatorname{rank}\varphi=r$ iff $\operatorname{rank}M=n-r$, where $n=\operatorname{rank} G$. Then $M\otimes Q$ is free of rank $n-r$, and therefore $F_{n-r-1}(M\otimes Q)=0$, where $F_i$ stands for the $i$th Fitting invariant. But $F_{n-r-1}(M\otimes Q)=F_{n-r-1}(M)Q$ and this shows that $F_{n-r-1}(M)=0$, that is, $I_{r+1}(\varphi)=0$.

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