A module is finite when its quotient is finite? Let $A$ be a ring, $I$ an ideal of $A$ and $M$ an $A$-module. Suppose $A$ is complete and separated in the $I$-adic topology, $M$ is separated in the $I$-adic topology and $M/IM$ is finite over $A/I$.Then it's said that $M$ is finite over $A$.
Here is my attempt. By assumption, $M/IM\cong \sum_{i=1}^{i=m}(A/I)x_i$ for some $x_i\in M$. Then we may deduce $M/I^nM\cong \sum_{i=1}^{i=m}(A/I^n)x_i$ for $n>0$. Since $A$ is complete, $A=\underleftarrow{Lim}(A/I^n)$. Thus, we only need to show that $M=\underleftarrow{Lim}(M/I^n)=\sum_{i=1}^{i=m}\underleftarrow{Lim}(A/I^n)x_i=\sum_{i=1}^{i=m}Ax_i$. The first equation is due to $M$ is separated, i.e. $\bigcap_n I^nM=(0)$.
I am not sure about the argument above. Hope someone could help. Thanks!
 A: In your proof, it will not be true in general that $M=\varprojlim M/I^nM$, this is the definition of being complete & separated, and we are missing the "complete" part here.
Take elements $x_1,\dots, x_s\in M$ such that $x_i+ IM$ generate $M/IM$ over $A/I$; as you've correctly noted, it follows that the $x_i+I^nM$ generate $M/I^nM$ over $A/I^n$ for all $n\geqslant 1$. We will show that the $x_i$ generate $M$. To this end, let $x\in M$; begin by choosing for each $i$ any element $a_{i1}\in A$ such that $$x\equiv a_{11}x_1+\cdots+a_{s1}x_s\:\mathrm{mod}\:IM.$$
Now for any $n\geqslant 1$ suppose we are given elements $a_{in}\in A$ such that $x+I^nM\equiv a_{1n}x_1+\cdots+a_{sn}x_s\:\mathrm{mod}\:I^nM$; we will inductively define elements $a_{i,n+1}\in A$ such that $a_{i,n+1}\equiv a_{in}\:\mathrm{mod}\:I^n$ and $x\equiv a_{1,n+1}x_1+\cdots+a_{s,n+1}x_s\:\mathrm{mod}\:I^{n+1}M$.
To begin this process, take arbitrary elements $\widetilde a_{i,n+1}$ such that $x\equiv\widetilde a_{1,n+1}x_1+\cdots+\widetilde a_{s,n+1}x_s\:\mathrm{mod}\:I^{n+1}M$. Then we must have $(\widetilde a_{1,n+1}-a_{1n})x_1+\cdots+(\widetilde a_{s,n+1}-a_{sn})x_s\equiv0\:\mathrm{mod}\:I^nM.$ Or in other words, there exist elements $b_i\in I^n$ such that $$(\widetilde a_{1,n+1}-a_{1n})x_1+\cdots+(\widetilde a_{s,n+1}-a_{sn})x_s=b_1x_1+\cdots+b_sx_s.$$
Now define $a_{i,n+1}:=a_{in}+b_i$. It clearly satisfies $a_{i,n+1}\equiv a_{in}\:\mathrm{mod}\:I^n$ for each $i$, and you can easily verify that we still have $x\equiv a_{1,n+1}x_1+\cdots+a_{s,n+1}x_s\:\mathrm{mod}\:I^{n+1}M$.
Now, using the fact that $A$ is separated complete, for each $i$ there exists a unique element $a_i$ such that $a_i\equiv a_{in}\:\mathrm{mod}\:I^n$ for all $n$. We claim that $x=a_1x_1+\cdots+ a_s x_s$: to see this, simply verify that the difference is zero modulo $I^nM$ for all $n\geqslant 1$, and use the fact that $M$ is separated. In other words, you will be showing that $$x-(a_1x_1+\cdots a_s x_s)\in\bigcap_{n\geqslant 1}I^nM=\{0\}.$$
Hope this helps!
