Let $\Lambda_2 = \mathbb{Z}_p[[T_1,T_2]]$ be the power series ring over $\mathbb{Z}_p$ in two variables, i.e., $\Lambda_2$ is a regular local ring of dimension 3. Let $M$ be the quotient of an elementary torsion $\Lambda_2$-module of the form $E = \bigoplus_i \Lambda_2/(f_i)$. Suppose that $M$ is a finitely generated $\Lambda_2$-module which is annihilated by two relatively prime elements (i.e., $M$ is pseudo-null). This means that one can choose elements $h_1, h_2$ in the annihilator ideal $\text{Ann}(M)$ of $M$ such that the Krull dimension of $\Lambda_2/(h_1,h_2)$ is 1.

Question: If the Krull dimension of $\Lambda_2/\text{Ann}(M)$ happens to be also 1, can one choose $h_1$ and $h_2$ contained in $\text{Ann}(M)$ such that some power of the annihilator ideal is contained in the ideal $(h_1, h_2)$?

  • $\begingroup$ What do you mean by "finite index"? $\endgroup$ – Hailong Dao Nov 23 '18 at 15:54
  • $\begingroup$ I want the quotient $\text{Ann}(M)/(h_1,h_2)$ to be a finite group. $\endgroup$ – user447242 Nov 24 '18 at 17:21
  • $\begingroup$ That quotient has Krull dimension $1$. May be you meant something else. Like some power of $ann(M)$ lies in $(h_1,h_2)$? $\endgroup$ – Hailong Dao Nov 24 '18 at 19:42
  • $\begingroup$ I admit that I am a bit confused now - why do you know that the Krull dimension of the quotient is 1? Couldn't it be that simply $\text{Ann}(M) = (h_1, h_2)$ holds? If some power of the annihilator ideal is contained in $(h_1, h_2)$, then I thought the quotient must have Krull dimension zero (and will therefore be finite). But I am not very familiar with this kind of notion (as you probably already noticed :-) ), therefore this is possibly too brave to expect. Can one choose $h_1, h_2$ such that at least the weaker statement holds? $\endgroup$ – user447242 Nov 26 '18 at 8:00
  • $\begingroup$ If the quotient is not zero, then it has dimension $1$. Because $R/(h1,h_2)$ has a nonzero divisor, and it will be a nzd on any non-zero submodule. The "weaker statement" is equivalent to $Ann(M)$ having the same support as $(h_1,h_2)$. In other words, it is a set-theoretic complete intersection. But it is a hard question unless we know more. $\endgroup$ – Hailong Dao Nov 26 '18 at 13:39

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