Generating the annihilator ideal up to finite index

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 $$(h_1, h_2)$$?

• What do you mean by "finite index"? – Hailong Dao Nov 23 '18 at 15:54
• I want the quotient $\text{Ann}(M)/(h_1,h_2)$ to be a finite group. – user447242 Nov 24 '18 at 17:21
• That quotient has Krull dimension $1$. May be you meant something else. Like some power of $ann(M)$ lies in $(h_1,h_2)$? – Hailong Dao Nov 24 '18 at 19:42
• 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? – user447242 Nov 26 '18 at 8:00
• 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. – Hailong Dao Nov 26 '18 at 13:39