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 $K_{\cdot}(y_1,\cdots,y_n) \otimes M$.

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)).

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.


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:

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 $\operatorname{depth}(I,M)=n$.

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?

  • $\begingroup$ It is false for obvious reasons, just take $M=0$, but you only have to strengthen your hypothesis assuming the homology vanishes for all $M$ $\endgroup$ – Fernando Muro Oct 31 '13 at 15:34
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    $\begingroup$ @FernandoMuro: your "obvious reasons" are not in agreement with the hypothesis i am making: as i mention we must have $IM \neq M$. If i take $M=0$ this hypothesis is violated. $\endgroup$ – Manos Oct 31 '13 at 15:42
  • $\begingroup$ Sorry, I missed it $\endgroup$ – Fernando Muro Oct 31 '13 at 17:13
  • $\begingroup$ In remark 3 do you mean $H_i(y_1,\ldots,y_n;M)=0$? $\endgroup$ – Mahdi Majidi-Zolbanin Oct 31 '13 at 19:41
  • $\begingroup$ @MahdiMajidi-Zolbanin: Yes i believe Theorem 16.8. Could you please explain to me why based on that the claim in Remark 3 holds? (also: thanks for the typo) $\endgroup$ – Manos Oct 31 '13 at 19:46

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.

  • $\begingroup$ The point of debate is the Corollary to Theorem 16.8. The theorem itself is fine. $\endgroup$ – Manos Nov 1 '13 at 15:14
  • $\begingroup$ @Manos: I see I thought the statement highlinted was Theorem 16.8. Let me change theorem to corollary. $\endgroup$ – Youngsu Nov 1 '13 at 17:55
  • $\begingroup$ +1. Your example was the first that crossed my mind too (as a counterexample to Manos' question) since it is known as having a bad behaviour with respect to grade, that is, grade of $I$ is 1 while the grade of $I+yR$ is 3. $\endgroup$ – user26857 Nov 4 '13 at 16:20
  • $\begingroup$ @YACP: Since $y \in I$ we have $I+yR=I$. Is there a typo in your comment? $\endgroup$ – Manos Nov 4 '13 at 22:04
  • $\begingroup$ @Manos Sorry, I've thought that $I=(x(y-1),z(y-1))$. Please read my comment using this notation. $\endgroup$ – user26857 Nov 4 '13 at 22:28

In my second-hand copy of Matsumura's, that is pointed as a mistake, but actually I do not know the answer to your question. Maybe you are interested in http://arxiv.org/pdf/math/0406566 (since Koszul homology localizes). I'm sorry, this is not an answer, but I am not able to write comments (low reputation).

  • $\begingroup$ Interesting...You mean the previous owner has indicated that this is a mistake? $\endgroup$ – Manos Nov 1 '13 at 4:18
  • $\begingroup$ Yes, but only the obvious thing: that the proof was incomplete, not necessarily wrong. It is indeed known to be wrong as stated, but there is no comment on the validity with the required correction (as you state it), i.e, "up to some permutation". $\endgroup$ – Vinteuil Nov 1 '13 at 9:53
  • $\begingroup$ The original question by Manos, the most interesting one, remains unanswered: 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? Any hint? $\endgroup$ – Vinteuil Nov 5 '13 at 19:43

Matsumura's corollary is incorrect without additional assumptions.

I'd recommend to look into Bourbaki's volume on Homological algebra, where they prove that corollary is true with additional assumption that sequence is in the radical of the ring.

  • $\begingroup$ So given that $A$ is Noetherian, $M$ finite and $H_i(y_1,\cdots,y_n;M)=0, \forall i>0$, we must prove that i) $M$, $M/(a_1M), \cdots, M/(a_1 M + \cdots a_{n-1}M)$ are $I$-ally separated and ii) $y_1,\cdots, y_n$ is $M$-quasiregular. Then $y_1,\cdots, y_n$ will be $M$-regular. Is that what you are suggesting? $\endgroup$ – Manos Nov 1 '13 at 15:11
  • $\begingroup$ Matsumura's Corollary is wrong and the example given by @Youngsu shows this pretty clear. $\endgroup$ – user26857 Nov 4 '13 at 16:18
  • $\begingroup$ YACP, One condition is missing in Matsumura's book, that is used by Bourbaki: sequence should be in the radical of the ring. That looks like a very restrictive condition. But without this condition it is clearly wrong. $\endgroup$ – Sasha Pavlov Nov 4 '13 at 19:07
  • $\begingroup$ @SashaPavlov: Would you like to modify your answer to fit your comments? Right now your last comment essentially states that your answer is clearly wrong. $\endgroup$ – Manos Nov 4 '13 at 21:47
  • $\begingroup$ Actually Bourbaki proved that, in general, from $H_1(\underline{\bf x},M)=0$ one can obtain that the sequence $\underline{\bf x}$ is $M$-quasi-regular. Now, if we want to be $M$-regular need to impose extra-conditions, the most common being $R$ noetherian, $M$ finite and the sequence contained in the Jacobson radical. $\endgroup$ – user26857 Nov 4 '13 at 22:35

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