On the comparison of linear topologies on a local ring Let $R$ be a local ring, $a_{\lambda}$ be a decreasing net of ideals, indexed by a directed set, such that each $a_{\lambda}$ is contained in the nilradical ideal and $\bigcap a_{\lambda}=(0)$. Then for any integer $k$, there exists an index $\lambda$ such that $a_{\lambda}\subseteq m^{k}$. That is, the linear topology defined by $a_{\lambda}$ is finer than the topology defined by $m^{k}$.
Could anyone give a proof of this statement or a counter example? Feel free to add some more assumptions...
 A: Counterexample: Let $k$ be a field and let $(R,\mathfrak{m})$ be the localization of $k[x,y]$ at the origin. Its completion is $\widehat{R}=k[[x,y]]$. Denote by $j:R\to \widehat{R}$ the inclusion. 
Choose a series $f\in xk[[x]]$ which is transcendental over $k(x)$, and put $\varphi(x,y)=y-f(x)\in\widehat{\mathfrak{m}}$. For $n\in\mathbb{N}$, put $a_n=j^{-1}(\varphi\widehat{R}+\widehat{\mathfrak{m}}^n)$. This is a decreasing sequence of ideals in $R$. None of them is contained in $\mathfrak{m}^2$ since $a_n$ contains the obvious ''$n$th truncation'' of $\varphi$.
I claim that $\bigcap_na_n$ is zero. This is equal to $j^{-1}(\bigcap_n (\varphi\widehat{R}+\widehat{\mathfrak{m}}^n))=j^{-1}(\varphi\widehat{R})$. So let $h\in k[[x,y]]$ be such that $h\varphi=(y-f(x))h(x,y)$ is a rational function $R(x,y)$. Substituting $f(x)$ for $y$, we get $R(x,f(x))=0$, hence $R=0$ by assumption. 
A: If $R$ is not noetherian then it is not true, e.g  $R=k[[X_1,...]]$ modulo all the monomials
of degree $2$, so that $m^2=0$, $a_i= (X_i, X_{i+1}, ....)$. Then none of the $a_i$ is contained in $m^2$.
A: Edit: 25/12/2011.
I give here an example
First, we consider a local ring $(R, \frak{m})$ with a filtration ideals $a_{\lambda}$ such that $\cap a_{\lambda} = 0$ however the linear topology defined by $a_{\lambda}$ is not finer than the $\frak{m}$-adic topology. There are many example even in the case $R = k[X, Y]_{(X, Y)}$ (eg: [Matsumura 1986, exercise 8.10] is a nice example).
Using ideallization $S = R \ltimes R$, then $0 \ltimes R$ contained in the nilradical of $S$. Therefore the filtration ideals $0 \ltimes a_{\lambda}$ ia a sperated topology of $S$ contained in the nilradical.
We have $\frak{m}$$ \ltimes R$ is the maximal ideal of $S$, and the $\frak{m}$$S$-adic topology is
$(\frak{m}$ $\ltimes R)^n = \frak{m}^n \ltimes \frak{m}^{n-1}$
By the our assumption we have the linear topology $0 \ltimes a_{\lambda}$ is not finer than the $\frak{m}$$S$-adic topology.
