$M$ be a complete $R$-module and $N$ be a closed submodule of $M$ implies $M/N$ is complete.

Question

Let $M$ be a complete $R$-module with respect to the filteration $\{M_n\}.$ Also let $N$ be a closed submodule of $M.$ Then how can I show that $M/N$ is complete with respect to the induced filteration, i.e., $(M/N)_n=(N+M_n)/N$.

We should show that $M/N$ is Hausdorff and every Cauchy sequence in $M/N$ converges. Since $N= \cap_{n=1}^{\infty}(N+M_n)$ clearly $M/N$ is Hausdorff. But How can I show that every Cauchy sequence in $M/N$ is convergent ? I need some help. Thanks.

Answer 1

Recall every Cauchy sequence in $M$ converging is equivalent to the natural map $M\to\varprojlim M/M_n$ being surjective. So we need to show the same holds for $M/N$. You should check that the following diagram commutes

$$\require{AMScd} \begin{CD} M @>>> M/N @>>> \varprojlim (M/N)/(M/N)_n\\ @VVV && @VV{\text{id}}V \\ \varprojlim M/M_n @>{\phi}>> \varprojlim M/(M_n+N) @>{\simeq}>>\varprojlim(M/N)/((M_n+N)/N) \end{CD}$$

and also notice every map except $\phi$ and $(M/N)\to\varprojlim(M/N)/(M/N)_n$ is evidently surjective. If we can prove $\phi$ is surjective, then we can conclude the other is as well and we will have our result. The issue is that $\varprojlim$ does not necessarily preserve surjections in general; but we have the following lemma

Suppose $\{A_n\}$, $\{B_n\}$, $\{C_n\}$ are projective systems of $R$-modules, and we have maps of projective systems $$0\longrightarrow A_n\overset {f_n}\longrightarrow B_n\overset{g_n}\longrightarrow C_n\longrightarrow0$$ (this means that if we include the maps $A_{n+1}\to A_n$, $B_{n+1}\to B_n$ etc., then all squares commute). If in addition the maps $A_{n+1}\to A_n$ are surjective for all $n$, then the induced map of system is a short exact sequence $$0\to\varprojlim A_n\to\varprojlim B_n\to\varprojlim C_n\to0.$$ [Side note: without the hypoythesis on $\{A_n\}$ it is always left exact, and the surjectivity condition is what will ensure right exactness.]

For us, for each $n$ we will look at the short exact sequence

$$0\to (M_n+N)/M_n\to M/M_n\to M/(M_n+N)\to0,$$

and then using the above lemma, if you check that $(M_{n+1}+N)/M_{n+1}\to (M_n+N)/M_n$ is surjective for all $n$ (which should be simple), then you can conclude $\phi$ is surjective.