Let $A$ be a Noetherian ring, $M$ a finite $A$-module and $I=(y\_1,\cdots,y\_n)$ an ideal of $A$ such that $M \neq IM$. Denote by $H\_i(y\_1,\cdots,y\_n;M)$ the homology at dimension $i$ of the augmented Koszul complex $K\_{\cdot}(y\_1,\cdots,y\_n) \otimes M$.



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> Theorem: If $y\_1,\cdots,y\_n$ is a regular sequence, then $H\_i(y\_1,\cdots,y\_n;M)=0, \, \forall i>0$. (see for example Matsumura, CRT, Theorem 16.5 (i)).
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Question: Is the converse true? I.e. if $H\_i(y\_1,\cdots,y\_n;M)=0, \, \forall i>0$, is it true that the generators of $I$, $y\_1,\cdots,y\_n$ are a regular sequence under some permutation?


Remark 1: The converse is true if $A$ is local or if $A$ and $M$ are $\mathbb{N}$-graded with all generators of $I$ having positive degree.


Remark 2: If $H\_i(y\_1,\cdots,y\_n;M)=0, \, \forall i>0$, then by the characterization of the $I$-depth of $M$ by means of the Koszul complex (e.g. Matsumura, CRT, Theorem 16.8), we have that $I$ contains an $M$-regular sequence of length $n$. 


Remark 3: In the proof of the Corollary to Theorem 16.8, Matsumura claims that $H\_i(y\_1,\cdots,y\_n;M)=0, \forall i>0 \Leftrightarrow y\_1,\cdots,y\_n$ is an $M$-sequence. I can't see why this is true, and this is the motivation for my question.


Edit: 


My question might become clearer to those who do not have Matsumura's book, if i state the Corollary to Theorem 16.8, to which i refer in my Remark 3:



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> Let $A$ be a Noetherian ring, $I=(y\_1,\cdots,y\_n)$ an ideal of $A$, $M$ a finite $A$-module such that $IM \neq M$. Then $y\_1,\cdots,y\_n$ is an $M$-sequence if and only if $\operatorname{depth}(I,M)=n$.
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Matsumura's proof goes as follows: "$\operatorname{depth}(I,M)=n \Leftrightarrow H\_i(y\_1,\cdots,y\_n;M)=0, \forall i>0 \Leftrightarrow y\_1,\cdots,y\_n$ is an $M$-sequence."


Remark 4: Certainly $\operatorname{depth}(I,M)=n \Leftrightarrow H\_i(y\_1,\cdots,y\_n;M)=0, \forall i>0$ and certainly $y\_1,\cdots,y\_n$ is $M$-sequence $\Rightarrow H\_i(y\_1,\cdots,y\_n;M)=0, \forall i>0$. But why $H\_i(y\_1,\cdots,y\_n;M)=0, \forall i>0 \Rightarrow y\_1,\cdots,y\_n$ is an $M$-sequence?



zero homology of augmented Koszul complex implies the sequence is regular?
I don't think the corollary is true without further assumptions. 


Take $R = k[x,y,z]$ and $I = (x(y-1), y,z(y-1)$. Since $x(y-1), y,z(y-1)$ is a regular sequence, hence depth $(I,R)$ = 3. But $x(y-1),z(y-1), y$ is not a regular sequence.


I believe that the statement is that $I$ can be generated by an $M$-sequence. If $M = R$, then it is Exercise 16.9 in the same section in Matsumura's book.


