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Can you tell me why: Finite module on a complete local ring is linear compact module? I am sturdy about linear compact module, can you tell me some concerned paper. Thanks you very much!

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sorry, my english is not good! –  tlquyen Nov 2 '12 at 3:50
    
@tlquyen. There are some definitions about linear compact module that I have known. This is one: A module M is called linear compact if for each filtration $(x_i+U_i,i\in I)$ (here, $x_i\in M, U_i$ are closure submodule in M) we have –  tlquyen Nov 2 '12 at 16:27
    
$$\bigcap_{i\in I} (x_i+U_i)\ne\emptyset $$. I am sure that has some relations between this definition to ordinary definition (every sequence has sub-sequencs converge) but I can not explain. Can you tell me more? Thanks. –  tlquyen Nov 2 '12 at 16:40
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1 Answer

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Sorry, I misunderstood your settings. I thought that linear compact modules are just modules which are compact for a linear topology. In fact you are talking about linearly compact modules. They are in general not compact.

In an usual topological compact space, the intersection of any centered family of closed subsets (centered means any finite intersection in the family is non-empty) is non-empty. For linearly compact modules, we only require this property for centered family of closed cosets $x_i+U_i$. So the condition is weaker.

Now let me show $M$ is linearly compact under the condition it is finitely generated over the complete local ring $R$ and complete. (So $R$ will be linearly compact). As I said before, the completeness of $M$ is automatic if $R$ is noetherian, but I am not sure this holds in general.

Let $C_i=\{ x_i + U_i \}_i$ be a centered family of closed cosets in $M$. Replacing the family with the family consisting in finite intersections of $C_i$'s (such an intersection is still a closed coset), we can suppose the family in "filtered": two members of the family always contain a third one. Let us first extract a Cauchy sequence of elemnts in the $C_i$'s. Fix a positive integer $r$. Consider the family of the classes of $C_i+\mathfrak m^r M$ in $M/\mathfrak m^r M$. As $M/\mathfrak m^r M$ is Artinian, this family has a minimal element $(C_{i_r}+\mathfrak m^r M)/\mathfrak m^r M$. Then $$C_{i}+ \mathfrak m^r M\supseteq C_{i_r}+\mathfrak m^r M, \quad \forall i$$ (they contain both some $C_j+\mathfrak m^r M$ which is then equal to $C_{i_r}+\mathfrak m^r M$). We construct inductively $y_r\in C_{i_r}$ such that $y_{r+1}\in y_r+\mathfrak m^r M$. This induces a Cauchy sequence in $M$. Let $y\in M$ be its limit. Let us show $y\in C_i$ for all $i$. We have $y-y_r\in \mathfrak m^r M$.

Fix $i$. Let $r\ge 1$. There exists $z_{i,r}\in C_i$ such that $y_r-z_{i,r}\in \mathfrak m^r M$. Then $$ y=z_{i,r}+(y-y_r)+(y_r-z_{i,r})\in z_{i,r}+\mathfrak m^r M.$$ Therefore $y$ belongs to the closure of $C_i$. But $C_i$ is closed by hypothesis, so $y\in C_i$.


Former answer:

Let $M$ be a finitely generated module over a noetherian (not necessarily complete) local ring $R$. Let $\mathfrak m$ be the maximal ideal of $R$. Then the completion of $M$ for the $ \mathfrak m$-adic topology is $$ \hat{M}=\lim_{\leftarrow} (M/\mathfrak m^n M)\simeq M\otimes_R \hat{R}$$ (the last isomorphism needs the noetherian hypothesis). In particular, if $R$ is already complete, and so is $M$.

If the residue field of $R$ is finite, then $R/\mathfrak m^n$ is a finite ring: we have an exact sequence $$ 0\to \mathfrak m^n/\mathfrak m^{n+1} \to R/\mathfrak m^{n+1}\to R/\mathfrak m^n\to 0$$ and $m^n/\mathfrak m^{n+1}$ is a finite dimensional vector space over the finite field $R/\mathfrak m$. An induction on $n$ shows that $R/\mathfrak m^n$ is finite for all $n\ge 1$.

As $M/\mathfrak m^n M$ is finitely generated over $R/\mathfrak m^n$, it is also a finite set, hence compact. This implies that $M=\hat{M}$ is compact.

Note that you really need to suppose the residue field of $R$ is finite. Otherwise $R$ is not compact.

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@tlquyen Thanks. It means $M$ is not compact if the residue field of $R$ is not finite. But in the paper which I read, the result need not the hypothesis finite residue field of R. –  tlquyen Nov 4 '12 at 3:25
    
Can you tell me more. In the definition of linear compact module, why does the filtration $(x_i+U_i\mid i\in I)$ must require $U_i$ is close submodule in M. Thank you very much. –  tlquyen Nov 4 '12 at 3:41
    
I tried to show $R$ be linearly compact in case it is a complete local ring. We have each $R/\mathfrak{m}^n$ is artin, so is linearly compact. Now, $R$ be linearly compact because of $R={\varprojlim}R/\mathfrak{m}^n$? –  tlquyen Nov 6 '12 at 15:40
    
You wrote: As $M/\mathfrak{m}^r$ is Artinian, the family of the classes of $C_i+\mathfrak{m}^rM$ has a minimal element. But in general, $C_i+\mathfrak{m}^r$ is not a submodules of $M/\mathfrak{m}^r$, how can we apply the property of artinian of $M/\mathfrak{m}^r$? Or may be I understood something. –  tlquyen Nov 6 '12 at 15:55
    
Sorry, I must write $M/\mathfrak{m}^nM$ and $C_i+\mathfrak{m}^nM$. –  tlquyen Nov 6 '12 at 16:02
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