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Theorem 16.7 in Matsumura's Commutative Ring Theory reads as follows: "Let $A$ be a Noetherian ring, $I$ an ideal of $A$ and $M$ a finite $A$-module such that $IM \neq M$; then the length of a maximal $M$-sequence in $I$ is a well determined integer $n$, and $n$ is determined by $\operatorname{Ext}_A^i(A/I,M)= 0, i<n$ and $\operatorname{Ext}_A^n(A/I,M) \neq 0$."

Is it true that the hypothesis of the theorem as given guarantee that there exists a finite-length maximal $M$-sequence in $I$?

This question is motivated by the fact that Bruns and Herzog in Cohen Macaulay Rings page 10, conclude from $\operatorname{Ext}_A^i(A/I,M)=0, \forall i$ that $I M =M$.

So it seems to me that the existence of a finite length maximal $M$-sequence in $I$ should be included in the hypothesis of the theorem for it to be true. After all, its proof starts by saying "let $x_1,\dots,x_n$ be a maximal $M$-sequence in $I$..."

Manos
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1 Answers1

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To answer the stated question: yes, the hypotheses are enough to guarantee existence of a finite-length maximal $M$-sequence contained in $I$ (although it may be empty, i.e. $I$ may consist of zerodivisors on $M$). To be precise:

Proposition: $A$ Noetherian, $I$ an $A$-ideal, $M$ finite $A$-module, $M \ne IM$. Then there exists a finite-length maximal $M$-sequence contained in $I$ (possibly empty).

To see this, consider the set $\Sigma$ of subideals of $I$ generated by (finite) $M$-sequences, which, if nonempty, has a maximal element by Noetherianness. Then notice that maximality in $\Sigma$ implies maximality of the (finite-length) generating $M$-sequence.

zcn
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  • The proposition seems to be true even when $M = IM$, is that right? – Manos Mar 11 '14 at 15:56
  • Yes, the proposition only uses Noetherianness of $A$, so the proof works quite generally (even if $M = IM$ or $M$ is not finitely generated). But the conditions $M \ne IM$, $M$ finitely generated should be thought of as standing assumptions when talking about depth – zcn Mar 11 '14 at 20:39
  • Then does that mean that we can never have the condition $Ext_A^i(R/I,M)=0, \forall i$? I am basically confused by Bruns and Herzog page 10, when they discuss the case of an infinite grade. – Manos Mar 11 '14 at 23:17
  • Oh, I think I understand your question now. If $M = IM$, then $\text{depth}_I(M)$ is defined to be $\infty$. This doesn't mean that there is an $M$-sequence of infinite length contained in $I$. The reason it's defined this way when $M = IM$ is that it's consistent with the characterization using $\text{Ext}$, in the case $M \ne IM$. Infinite grade modules do occur: e.g. if $I + J = R$, then $M := R/J$ is a finite $R$-module, and by definition, $\text{depth}_I(R/J) = \infty$. – zcn Mar 12 '14 at 03:14
  • I think i finally understand: The existence of a finite length maximal $M$-sequence in $I$ implies that there exists an $n$ such that $Ext^n(R/I,M) \neq 0$ only when $IM \neq M$. If $IM=M$ we still have a finite length maximal $M$-sequence in $I$, but we necessarily have $Ext^i(R/I,M)=0, \forall i$ since $Supp(M) \cap Supp(R/I) = \emptyset$. Correct? – Manos Mar 12 '14 at 15:26
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    Yes, that's exactly correct – zcn Mar 12 '14 at 17:57
  • Thank you very much for your instruction, very valuable indeed :) – Manos Mar 12 '14 at 18:37