The computation of an inverse limit of modules. I am trying to read the book Euler System by Karl Rubin. But I found something confusing, right at the page $1,$ or the page $9$ in the book. There we find that:  

Let $\mathcal O$ be the ring of integers of a finite extension $\Phi$ of $\mathbb Q_p,$ where $p$ is a rational prime, let $G_K=\text{Gal}(\bar K,K),$ where $K$ is a field, and let $T$ be a free $\mathcal O-$module of finite rank with a continuous action of $G_K.$ Then we define $V=T\otimes_\mathcal O \Phi, W=V/T=T\otimes_\mathcal O(\Phi/\mathcal O),$ and $W_M=M^{-1}T/T\subset W$ for non-zero $M\in\mathcal O.$
  Then $$T=\varprojlim W_M.$$  

I cannot understand the last equation: How shall we make these $W_M$ into an inverse system, by which maps? The only relation that came into my mind is the division: if $M\mid N,$ then $\exists \rho_{N,M}:W_N\rightarrow W_M,$ satisfying the two defining conditions. I don't see how this done.
I think that, if I know about the system in question, I can compute the limit directly, or projectively, thus I ask for only the system. :)
Thanks for any hints in advance.  
P.S. The above paragragh is not from the book, but aapted from the book, which does not change the question, I surmise.
 A: The map $\rho_{N, M}$ is just multiplication by $N/M$, which takes $N^{-1} T$ surjectively to $M^{-1}T$ and which is well-defined mod $T$. It is easier if you only consider $M, N$ to be powers of a chosen uniformizer of $\Phi$ (you will compute the same limit, but with less of a hassle).
You might want to learn about the Tate module of an elliptic curve before reading that book, then every time you see $T, V, W$ just think of that example (here $\Phi = \mathbb{Q}_p$).
(It is also helpful to consider the one-dimensional free module $T$.) 
A: I think that the fundamental lemma is the following: 
Let $A$ be an integral ring with fractional field $K$. For $x \neq 0$, denote by $A_x$ the localized ring of $A$ at $x$. if $x | y$, we have an inclusion $A_x \hookrightarrow A_y$, by sending $\frac{1}{x}$ to $\frac{y}{x} \frac{1}{y}$ and for any $x$ we have inclusion $A_x \hookrightarrow K$; and all these inclusions are compatible. Prove that this gives an isomorphism $\varinjlim A_x \cong K$ (I think that the limit is an injective one).
Your particular problem is of the same flavour.
Edit : In fact, we want $A$ and not $K$. So you have to prove the following equality: $\varprojlim A_x \cong A$.
