Let $A$ be a commutative ring with identity and $\mathfrak{a}$ and ideal of $A$. Then $A$ has a topological structure which is defined by the following chain of ideals \begin{equation*} A \supseteq \mathfrak{a} \supseteq \mathfrak{a}^2 \supseteq \mathfrak{a}^3 \supseteq \cdots. \end{equation*} Hence, we have the ring of completion $\widehat{A}$. On the other hand, if $M$ is an $A$-module, then $M$ also has a topological structure defined by the $\mathfrak{a}$-filtration \begin{equation*} M \supseteq \mathfrak{a}M \supseteq \mathfrak{a}^2M \supseteq \mathfrak{a}^3M \supseteq \cdots. \end{equation*} We also have $\widehat{M}$ induced by the $\mathfrak{a}$-topology on $M$. We have shown that $\widehat{M}$ is an $\widehat{A}$-module with the scalar multiplication \begin{equation*} ((a_n)_{n=1}^{\infty},(x_n)_{n=1}^{\infty}) \longmapsto (a_nx_n)_{n=1}^{\infty}. \end{equation*} We want to show that the scalar multiplication is actually a continuous map. In other words, the map
\begin{equation*} \phi : \widehat{A} \times \widehat{M} \longrightarrow \widehat{M}, \ ((a_n)_{n=1}^{\infty},(x_n)_{n=1}^{\infty}) \longmapsto (a_nx_n)_{n=1}^{\infty} \end{equation*} is a continuous map.
My question is that what are the topologies on $\widehat{A}$ and $\widehat{M}$ respectively?
My thought there are three choices to construct a topology on $\widehat{A}$. We have ring homomorphism $i : A \longrightarrow \widehat{A}$ by $a \longmapsto (a)_{i=1}^{\infty}$. Then $i(\mathfrak{a}^n)$ is an abelian subgroup of $\widehat{A}$. But $i(\mathfrak{a}^n)$ is not an ideal in $\widehat{A}$. However, $(\mathfrak{a}^n)^e = (\mathfrak{a}^e)^n$ is an ideal in $\widehat{A}$. Therefore, we have the following two topological structures on $\widehat{A}$. \begin{equation*} \widehat{A} \supseteq i(\mathfrak{a}) \supseteq i(\mathfrak{a}^2) \supseteq \cdots. \end{equation*} or \begin{equation*} \widehat{A} \supseteq \mathfrak{a}^e \supseteq (\mathfrak{a}^e)^2 \supseteq \cdots. \end{equation*} Because we know that $A \times M \longrightarrow M$ by $(a,x) \longmapsto ax$ is a continuous map (I have checked this fact). With the paralleled argument, the best choice is choosing the second one and consider the $\mathfrak{a}^e$-adic topology on $\widehat{M}$. And hence, \begin{equation*} ((a_n)_{n=1}^{\infty},(x_n)_{n=1}^{\infty}) \longmapsto (a_nx_n)_{n=1}^{\infty}. \end{equation*} is a continuous map. Is my thought correct?
I think the third posibility is to consider $\mathfrak{a}$ as a $A$-module, and consider the completion $\widehat{\mathfrak{a}}$ .... By I am not ready to think this possibility...
We also want to show that if $M$ and $N$ are $A$-modules with $\mathfrak{a}$-adic topologies and $f : M \longrightarrow N$ is a continuous $A$-module homomorphism. Then $\widehat{f} : \widehat{M} \longrightarrow \widehat{N}$ is a continuous $\widehat{A}$-module homomorphism.
My Thought If we can show that if $f : G \longrightarrow H$ is a continuous group homomorphism of topological abelian groups. Then $\widehat{f} : \widehat{G} \longrightarrow \widehat{H}$ is a continuous group homomorphism. Because the continuity only involves the fundamental neighborhood of abelian subgroups. Then continuity of $\widehat{f}$ inherits the proof in abelian group ($\mathbb{Z}$-module) case. Is my thought correct?