$T_p(E)/p^nT_p(E)$ is isomorphic to $E[p^n]$ In the case of an elliptic curve $E$ defined over a field $K$, I know that there is a good definition of the so-called Tate module, for every prime $p$, which is the $\mathbb{Z}_p$-module $T_p(E)=\underset{\underset{n}{\longleftarrow}}{\lim}E[p^n]$, where $E[p^n]$ is the $p$-torsion subgroup of $E$.
I would like to prove .
This seems analogy of $\mathbb{Z}_{p}/p^n\mathbb{Z}_{p}$ is isomorphic to $\mathbb{Z}
/p^n\mathbb{Z}$.
Is there any better generalization to module?
Could you tell me the proof of the title, and generalization of that proposition?
Thank you in advance.
 A: An element of the Tate module is an infinite vector $(P_n)$ of $p^n$ torsion points, such that $pP_n = P_{n-1}$.
Map $T_p(E) \to E[p^n]$ by projecting on the $n$th component. The kernel of this map is the set of vectors where the first $n$ coordinates are zero. I claim this is $p^n T_p(E)$.
Certainly $p^n T_p(E)$ has the property that the first $n$ coordinates are zero (the first $n$ coordinates of anything in $T_p(E)$ are killed by $p^n$).
Conversely, given an element of the form $(0, 0, 0, \ldots, 0, P_{n+1}, P_{n+2}, \ldots)$, we note that $P_{n+i}$ is really a $p^i$ torsion point (since the compatiability condition forces $pP^{n+1} = 0$. Therefore this element is $p^n(P_{n+1}, P_{n+2}, \ldots)$.
As a final note, underlying abelian group of the Tate module is (not naturally) isomorphic to the $p$-adics, hence the generalization you noted.
Note that it takes some care to write down the isomorphism, since the transition maps in the inverse limit defining the Tate module are multiplication by $p$ and the transition maps for the $p$-adics are projections. One way to get the isomorphism is to use the isomorphism of $\mathbb{Z}/p^n\mathbb{Z}$ with
$$
\frac{\frac{1}{p^n} \mathbb{Z}}{\mathbb{Z}},
$$
and now the transition maps become multiplication by $p$ on this side too.
