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

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.

## 1 Answer

A Cauchy sequence in $$M/N$$ has the form $$\{x_n+N\}$$, where the $$x_n\in M$$ satisfy relations $$x_{n+1}-x_n\in N+M_{s(n)}$$ with $$s(n)\to \infty$$. Though $$\{x_n\}$$ might not be a Cauchy sequence in $$M$$ we can a define a Cauchy sequence $$\{y_n\}$$ in $$M$$ such that $$x_n\equiv y_n\mod N$$. The procedure is as follows. We set $$y_1=x_1$$. Assume that we have defined $$y_n$$ so that $$y_n\equiv x_n\mod N$$. Note that $$x_{n+1}-x_n-u\in M_{s(n)}$$ for some $$u\in N$$. We thus define $$y_{n+1}=x_{n+1}-(x_n-y_n)-u$$. By construction we have $$y_{n+1}\equiv x_{n+1}\mod N$$ and $$y_{n+1}-y_n=x_{n+1}-x_n-u\in M_{s(n)}$$. It follows that $$\{y_n\}$$ is a Cauchy sequence in $$M$$ and has therefore a limit, say $$y\in M$$. It is now easy to see that $$y+N$$ is the limit of $$\{x_n+N\}=\{y_n+N\}$$.