Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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...

share|improve this question
    
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
1  
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

3 Answers 3

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.

share|improve this answer

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 $m^2$.

share|improve this answer
    
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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.