Let $O_K$ be a complete discrete valuation ring and $\pi$ is its uniformizer.(This should have been a valuation ring associated to a completed discrete valued field $K$.) Consider $E/K$ finite separable extension. Then $O_E$(integral closure of $O_K$ in $E$) is a f.g. free module of rank $[E:K]$.
The book says "$O_E$ is a free module of rank $[E:K]$. Then $O_E\cong\varprojlim\frac{O_E}{\pi^nO_E}$"
$\textbf{Q:}$ If $O_E$ is free module over $O_K$, then I can pass the inverse limit through its components as tensor commutes with direct sum. I need finiteness condition as direct sum is direct limit which does not commute with inverse limit in genera. So it become completion against $O_K$ and completion of $O_K$ is $O_K$. So I have $O_E\cong\varprojlim\frac{O_E}{\pi^nO_E}$ isomorphic as $O_K$ module. How do I inherit algebraic structure? Note that I have provided a $O_K$ module map. I need a ring map now.
This is related to pg 106 of Taylor, Frohlich Algebraic Number Theory's algebraic proof of uniqueness of absolute value.
My current guess is that first $O_E\cong O_E\otimes_{O_K}O_K$. Since $O_K$ is complete, I know $O_E\otimes_{O_K}O_K\cong O_E\otimes_{O_K}\hat{O_K}$. So $O_E\otimes_{O_K}\hat{O_K}\cong\varprojlim\frac{O_E}{p^nO_E}$. Then inherit ring structre from $O_E\otimes_{O_K}\hat{O_K}$ side.
Turning my comment into a slightly longer answer:
You want to show that there is an isomorphism of rings $O_E\overset{(*)}\to \varprojlim\frac{O_E}{\pi^nO_E}$. While you have a good idea for how to attack this problem, your proposed solution contains a lot of mysterious unspecified isomorphisms, such as $O_E\cong O_K^{[E:K]}$. There are several problems with this:
The solution to both of these problems is to write down your isomorphisms explicitly, as functions. (Choose an integral basis of $E/K$, say $x_1, \dots, x_n$, where $n = [E:K]$...) Then, at the end, you can compose them all together and get an exact formula for what your eventual map $(*)$ does to an element of $O_E$. And if you've set it up correctly, and chosen $(\dagger)$ and $(\ddagger)$ correctly, then it will be obvious that $(*)$ is a ring map.