# 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...

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Is $R$ Noetherian or non-Noetherian? – Pham Hung Quy Dec 2 '11 at 7:53
noetherian, and you can add more assumptions if you would... auch like $R$ is excellent, or $a_{\lambda}$ satisfies some properties... – Zhengyu Hu Dec 2 '11 at 9:55
I just want to know in which canses the linear topology induced by $a_{\lambda}$ is finer than that induced by $a+m^{k}$, where $a=\bigcap a_{\lambda}$. In this question, I assume $a=0$... – Zhengyu Hu Dec 2 '11 at 9:57
The nilradical is the set of nilpotent elements. I don't think this is what you mean. – Laurent Moret-Bailly Dec 2 '11 at 10:30
You can forget this condition... I just want to know in which cases $a_{\lambda}$ induces a finer topology. When $R$ is complete, this is by a Theorem of Chevalley (1946), and I am interested in the case $R$ is not complete... – Zhengyu Hu Dec 2 '11 at 10:49

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.

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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 $m^2$.

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I think "Thus the image of $a_{\lambda}$ is equal to the image of the intersection of all $a_{\mu}$, which is zero" is not a clear argument since in general this does not hold.. – Zhengyu Hu Dec 2 '11 at 15:25

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.

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I don't understand the notations and the Chevalley theorem... could you give a reference? – Zhengyu Hu Dec 2 '11 at 15:29
In fact, if we dot not require something on $a_{\lambda}$, this is clearly false. e.g. Take $f\in \widehat{R}$ be a transcendental function, then $(f)+m^{k}\bigcap R$ is not contained in $m^{k}$... So the crucial point is to see under which assumption the two topologies coincide... – Zhengyu Hu Dec 2 '11 at 15:35
I am sorry, I have an mistake. I need a bit time to correct it. – Pham Hung Quy Dec 2 '11 at 15:43