Definition of the Tate group associated with $p$-divisible groups Let $A$ be an abelian group, $p$ a prime number and $p_A:A \rightarrow A$ the multiplication by $p$. Let $A\left[p^n\right]$ be the kernel of $p_A$ composed $n$ times with itself. Then Lang, in his algebra p. 50, defines the Tate group $T_p(A)$ associated with the $p$-divisible group $A$ as the inverse limit of the inversely directed family $\left(A\left[p^{n+1}\right]\right)$. 
As i understand, $n=0,1,...$ By definition, for $n=0$, the corresponding component of any element of $T_p(A)$ is zero and so for $n=1$ the component of
any element of $T_p(A)$ must be inside $A\left[p\right]$. But substituting for $n=1$ at the defining formula for $T_p(A)$ yields $A\left[p^2\right]$.
Can anyone advise me about the correct interpretation of the definition?
Thanks :-)
 A: Lang defines the Tate group $T_p(A)$ as the subset of $V_p(A)$ consisting of infinite sequences $(x_0,x_1,\ldots)$ such that $x_0=0$. Here $V_p(A)$ is the inverse limit of the following diagram:
$$A\xleftarrow{p}A\xleftarrow{p}A\xleftarrow{p}\cdots.$$
Then he says that
$$T_p(A)=\lim A[p^{n+1}].$$
He doesn't make the map between $A[p^{n+1}]$ and $A[p^{n}]$ explicit, but a little thought reveals that the map $f_{n+1}:A[p^{n+1}]\to A[p^{n}]$ is nothing but the multiplication by $p$. Note that $f_{n+1}$ is surjective (hint: $A$ is $p$-divisible!). So the group on the right hand side of the equation is the inverse limit of the following diagram:
$$A[p]\xleftarrow{p}A[p^2]\xleftarrow{p}A[p^3]\xleftarrow{p}\cdots.$$

It is easy to see why $T_p(A)$ and $\lim A[p^{n+1}]$ are equal. Define $\varphi: T_p(A)\to \lim A[p^{n+1}]$ by $(0,x_1,x_2,\ldots)\mapsto (x_1,x_2,\ldots)$ and $\psi:\lim A[p^{n+1}]\to T_p(A)$ by $(x_1,x_2,\ldots)\mapsto (0,x_1,x_2,\ldots)$. I leave it to the reader to verify that these maps are inverses to each other.
