2

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 $\operatorname{ht}(\operatorname{ann}M_r) \ge \operatorname{ht}(\operatorname{ann}M) + r$.

Manos
  • 25,833
  • 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. –  Nov 21 '13 at 21:25
  • @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. – Manos Nov 21 '13 at 21:40
  • @user: Ok, let me try to prove that. – Manos Nov 21 '13 at 21:57
  • @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. – Manos Nov 21 '13 at 22:50
  • In your relation in comment #2 we have $\dim M=\dim A/\operatorname{ann}M$ and $\dim M_r=\dim A/(a_1,\dots,a_r,\operatorname{ann}M)$. The last one is simple if you look at the support of the two modules. –  Nov 22 '13 at 05:53
  • @user: ok, here is the proof: Clearly $supp(M_r) \subset supp(A/(a_1,\cdots,a_r,annM))$. For the converse inclusion, take $P \in supp(A/(a_1,\cdots,a_r,annM))$. Then $A_P \neq (a_1,\cdots,a_r,annM) A_P$. Suppose that $P \not \in supp(M)$. Hence if $M=A x_1+\cdots A x_n$, then for every $i$ there exists $c_i \in ann(x_i) - P$. Thus $c_1\cdots c_n \in annM - P$, contradiction. Correct? – Manos Nov 22 '13 at 18:48
  • @user: what i proved in my comment above, is that $Supp(M_r) = Supp(A/(a_1,\cdots,a_r,annM))$. Don't you agree? – Manos Nov 22 '13 at 21:21
  • 1
    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 $P\supset Ann(M)$. –  Nov 22 '13 at 21:40
  • @user: Nice! I am satisfied, thanks for your help. – Manos Nov 22 '13 at 22:15
  • @user: Yes, i can accept your first comment. – Manos Nov 23 '13 at 13:47

1 Answers1

1

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.