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Consider the following situation: let $(R,m)$ be a local Noetherian ring and $f: F_1 \rightarrow F_0$ a morphism of finite free $R$-modules with $\operatorname{rank}(F_i) = r_i, \, i=0,1$. Suppose that the image of $f$ contains a free direct summand $U$ of $F_0$ of rank $r_1$. How can we prove that $im(f) = U$?

The above situation arose in the proof of Proposition 1.4.12 (b) $\Rightarrow$ (a) in Bruns and Herzog Cohen Macaulay Rings.

Remark: What confuses me is that since $\operatorname{Hom}(M,L \oplus N) = \operatorname{Hom}(M,L) \times \operatorname{Hom}(M,N)$, then we can write $F_0 = U \oplus H$ and so $f=(f_U,f_H)$. Then i don't see why $f_H$ should be zero.

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I'm affraid you didn't show us the whole picture: in the book it is also assumed that $F_1/\ker f$ is free of rank $r_1$. –  user26857 Dec 27 '13 at 21:55
    
@YACP: As i understand, this is assumed to perform the induction step for complexes of length greater than 2. But the authors don't say how they prove that for the case where our complex is $0 \rightarrow F_1 \rightarrow F_0 \rightarrow 0$, i.e. for the first step of the induction. –  Manos Dec 27 '13 at 22:24
    
I don't undestand your comment: isn't trivially satisfied the first step? –  user26857 Dec 27 '13 at 22:33
    
@YACP: If our complex is $0 \rightarrow F_s \stackrel{f_s}{\rightarrow} F_{s-1} \rightarrow \cdots \rightarrow F_1 \rightarrow F_0 \rightarrow 0$, then why is it true that $f_s$ is injective and why is it true that $f_s(F_s)$ is a direct summand of $F_{s-1}$? Why is this trivially true? –  Manos Dec 27 '13 at 22:36
    
Ah, it is a complex, not a resolution! –  user26857 Dec 27 '13 at 22:37
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1 Answer 1

up vote 2 down vote accepted

Use Proposition 1.4.9(b) taking into account that $I_{r_1+1}(f)=0$.

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