Question on Iwasawa Theory Let $F/\mathbb{Q}$ be a finite extension and $F_\infty/F$ be a $\mathbb{Z}_p$-extension. Let $\Gamma=\textrm{Gal}(F_{\infty}/F)\simeq \mathbb{Z}_p$ and $\gamma_0$ generate a dense subgroup of $\Gamma$. $\Gamma$ acts continuously on $A$, a $p$-primary, Abelian group with discrete topology. We know how to make $A$ into a $\mathbb{Z}_p[T]$-module (ie define $Ta=\gamma_0a-a$). 
How do we know that $T$ is topologically nilpotent and how does this make $A$ a $\mathbb{Z}_p[[T]]$-module? 
 A: Somewhat more concretely, you can show $T$ is topologically nilpotent as follows. We need to show that, given any $a\in A$, one has $T^na=0$ for $n\gg 0$. Fix some $a\in A$ and note that, since $\Gamma$ acts contintuously and $A$ is discrete, we have $A=\cup A^{\Gamma_n}$ where $\Gamma_n=\Gamma^{p^n}$ is a system of open neighborhoods of the identity. So there is some $n_0$ such that $a\in A^{\Gamma_{n_0}}$. In fact, $a\in A^{\Gamma_{n}}$ for all $n\geq n_0$. (This is because $\Gamma_n\supseteq \Gamma_{n+1}$, so $A^{\Gamma_n}\subseteq A^{\Gamma_{n+1}}$.) Thus, $a\in A^{\Gamma_n}$ for $n\gg 0$, so if $\gamma$ is a topological generator of $\Gamma$, we have $\gamma^{p^n}a=a$ for $n\gg0$. Since $\gamma$ acts by $1+T$ this gives
$$
\big((1+T)^{p^n}-1)a=0,\quad n\gg0.
$$
But $((1+T)^{p^n}-1)$ is a distinguished polynomial (i.e., the degree $p^n$ coefficient is 1 and all other coefficients are divisible by $p$). So $\big((1+T)^{p^n}-1\big)=pF(T)+T^{p^n}$ for some polynomial $F(T)\in \mathbb{Z}_p[T]$. Hence
\begin{equation}
F(T)(pa)+T^{p^n}a=0,\quad n\gg0.
\end{equation}
We can now induct on the order of $a$. If $a$ has order $p$ then we're done by the above equality. Suppose the result holds for elements of order $p^{N-1}$ and let $a$ have order $p^N$. Then $pa$ has order $p^{N-1}$ so $T^n(pa)=0$ for $n\gg0$. Applying $T^n$ to the above equality finishes the proof.
A: Keeping your notations, for any integer $n \ge 1$, $\Gamma$ admits exactly one quotient $ \Gamma_n$ of order $p^n$, and $\Gamma_n$ is cyclic; one way to recover $\Gamma$ from the $\Gamma_n$'s is to view it as the projective limit $\varprojlim \Gamma_n$ (see below) , so that $\Gamma$ is a compact topological group $\cong (\mathbf Z_p , +)$ and the completed group ring $\Lambda := \mathbf Z_p [[\Gamma]]$ is topologically isomorphic to the power series ring $\mathbf Z_p [[T]]$ (in which $T$ is naturally topologically nilpotent). The advantage of $\Lambda$ over the ordinary group rings $R_n:=\mathbf Z_p [\Gamma_n]$ ? Whereas $R_n$ admits very annoying divisors of zero, $\Lambda$ is a nice UFD which is "almost" principal, in the sense that the so called "structure theorem" describes the $\Lambda$- modules of finite type (= noetherian) up to pseudo-isomorphism. This (or a direct proof) implies that any such noetherian module $M$ is naturally a compact topological module on which $\Lambda$ acts continuously, and its Pontryagin dual $\hat M$ is a discrete module of finite co-type (not noetherian) for which one has no nice structure theorem (otherwise than re-dualizing, of course). Actually, one neat distinction lies in the description of $M$ and $A$ as limits : whereas $M = \varprojlim M_{\Gamma_n}$ (projective limit of co-invariant quotient-modules), $\hat M= lim_{\to} \hat M^{\Gamma_n}$ (inductive limit of invariant sub-modules).  A typical example is the sequence $(A_n)$ of the $p$-class groups along $F_{\infty}/F$ : the compact module $X=\varprojlim A_n$ and the discrete module $A=lim_{\to} A_n$ are related via the machinery of adjoints in Iwasawa theory.
This issue of duality is not specific to Iwasawa theory, it goes back at least to the foundations of Galois cohomology by Tate (see the famous notes by Serre - with numerous additional original results). There was a well established abstract cohomology of groups acting on modules, but the point in Galois theory is that infinite Galois groups are profinite (= proj. lim. of finite) groups. What makes Galois cohomology work is that one restricts to the category of profinite groups acting on discrete modules, because the only topological modules on which a profinite group acting continuously gives rise to a cohomological functor, must be discrete (op. cit.)
