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Suppose $A$ is a Noetherian ring, $I\subset A$ an ideal, and $M$ a finitely generated $A$-module. If $IM\neq M$, then the length of a maximal $M$-sequence inside $I$ is fixed by the number $$\mbox{depth}(I,M)=\min\{n\geq0|\mbox{Ext}^n(A/I,M)\neq0\}.$$ However, if $IM=M$, then it can be shown that $\mbox{Ext}^i(A/I,M)=0 \ \forall i\geq0.$ The usual convention is that $\mbox{depth}(I,M)=\infty$ in this case, but I was wondering if the connection with the maximal length of an $M$-sequence is altogether lost. Since the ring is Noetherian, a maximal $M$-sequence still exists. So presumably there may be two different $M$-sequences with different length? Or is the maximal length of an $M$-sequence still fixed but cannot be determined by the non-vanishing dimension of $\mbox{Ext}^n(A/I,M)$ ?

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When $M$ is a vector space and $I=A$, every $M$-sequence has length 0 (Every non-zero element of $A$ acts as a unit on $M$, so the condition $(a_1,\ldots,a_n)M\neq M$ is not satisfied). So the convention $\mbox{depth}(I,A)=\infty$ has nothing to do with the length of a maximal $M$-sequence, at least in this case.

$M=0$ is another trivial example in which this occurs.

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