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Let $(A,m)$ be a local Noetherian ring, $M$ a finitely generated non-zero $A$-module and $a_1,\cdots,a_r$ an $M$-sequence. If $M=A$, then by using the Hauptidealsatz we can prove that $\operatorname{ht}(a_1,\cdots,a_r)=r$. For a general $M$, a natural question is

Question: Is it true that $\operatorname{ht}(a_1,\cdots,a_r,\operatorname{ann}M) = \operatorname{ht}(\operatorname{ann}M) + r$?

Remark:

In my study of this question, i proved the following result.

Proposition: Define $M_i = M/(a_1,\cdots,a_i)M$. Then $\operatorname{ht}(a_{i+1},\operatorname{ann}M_i) = \operatorname{ht}(\operatorname{ann}M_i) + 1$.

Now $\operatorname{ann}M_r$ contains the ideal $(a_r,\operatorname{ann}M_{r-1})$, whose height has increased by $r$ with respect to $\operatorname{ann}M$, using the above proposition. This proves that $\operatorname{ht}(\operatorname{ann}M_r) \ge \operatorname{ht}(\operatorname{ann}M) + r$.

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  • $\begingroup$ In general you get the following: $\dim R/(a_1,\dots,a_r,\operatorname{ann}M)=\dim R/\operatorname{ann}M-r$. If $R$ is Cohen-Macaulay you get what you want. Otherwise, I don't think so. $\endgroup$ – user89712 Nov 21 '13 at 21:25
  • $\begingroup$ @user: proof of your first statement? I am aware that $\dim M_r = \dim M - r$, i could prove this using the intermediate proposition that i mention. But your statement is slightly different. $\endgroup$ – Manos Nov 21 '13 at 21:40
  • $\begingroup$ @user: Ok, let me try to prove that. $\endgroup$ – Manos Nov 21 '13 at 21:57
  • $\begingroup$ @user: could you please provide a sketch of the proof of the fact that $\dim A/(a_1,\cdots,a_r,annM) = \dim A/annM -r$? I tried proving it by induction; i got the step for $r=1$, but for $r>1$ i am getting stuck. $\endgroup$ – Manos Nov 21 '13 at 22:50
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    $\begingroup$ Yeah, in a less explicit way you proved that $P\in Supp M_r$, but why so complicated? If not, then $M_P=(a_1,\dots,a_r)M_P$ and by NAK $M_P=0$ and this is false since $P\supset Ann(M)$. $\endgroup$ – user89712 Nov 22 '13 at 21:40
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From $\dim M_r=\dim M-r$ we get $\dim R/(a_1,\dots,a_r,\operatorname{Ann}M)=\dim R/\operatorname{Ann}M−r$.

If $R$ is Cohen-Macaulay then $\operatorname{ht}(a_1,\dots,a_r,\operatorname{Ann}M) = \operatorname{ht}(\operatorname{Ann}M) + r$.

Otherwise, I don't know.

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