Stable filtrations An $\mathfrak{a}$-stable filtration is (as many know) an $\mathfrak{a}$-filtration $\{M_n\}$ such that for large $n$; $\mathfrak{a}M_n=M_{n+1}$. This is saying in some sort of way (I think) that the filtration eventually behaves like the filtration $\{\mathfrak{a}^n\}$. Now I have an issue with the fact that I don't know any $\mathfrak{a}$-stable filtration which is not $\{\mathfrak{a}^n\}$ (or any trivial one, meaning that eventually $M_n=0$, or eventually they do become $\{\mathfrak{a}^n\}$ . Am I missing some easy example? Thanks in advance!
 A: $\newcommand{\ideal}[1]{{\mathfrak #1}}$
Let $\ideal{a} \subseteq A$ be an ideal of a noetherian ring. A typical case for considering $\ideal{a}$-stable filtrations is
$$0 \to M' \to M$$
an inclusion of finitely generated modules and $(M_n)$ an $\ideal{a}$-filtration of $M$ and $M' \cap M_n = M'_n$ the induced filtration of $M'$. 
Then if $M_n$ is $\ideal{a}$-stable $M'_n$ is $\ideal{a}$-stable too. So finally $\ideal{a} M'_k = M'_{k+1}$ for $k \gg 0$. But in the beginning this may fail.
Trying to give an example I started quite randomly with 
$A={\mathbb Q}[x,y,z,w]$ and $\ideal{a}=\ideal{m}=(x,y,z,w)$ together with 
$\ideal{b} = (x^2 y^2,z w)$. I let $\ideal{b} \subseteq A$ be the inclusion of $M' \subseteq M$. 
From the above general remark we know that $\ideal{b} \cap \ideal{m}^k = \ideal{b}_k$ becomes finally $\ideal{m}$-stable, but indeed we have
$$\ideal{b}_4/(\ideal{m} \ideal{b}_3) \neq 0$$
as can be easily calculated with Macaulay 2 (which is what I did). A further calculation shows that $\ideal{b}_4 \ideal{m} = \ideal{b}_5$ and it seems that from $4$ onwards strictness holds in the sequence $\ideal{b}_k$ although I have no proof for this.
Perhaps someone more knowledgeable could elucidate this contrived example by explaining how one could, based on general theorems, specifically construct an ideal $\ideal{b} \subseteq A$ so that $\ideal{b}_k$ becomes stable at a prechosen index $k_0$ (or how one could calculate $k_0$ from given $\ideal{m}$ and $\ideal{b}$).
