Linear compact module Can you tell me 

Why a finite module over a complete local ring is a linear compact module? 

I am study about linear compact module, can you tell me some concerned paper? Thanks you very much!
 A: 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. 
