Prove that if $A$ is compact $\Bbb{Z}_p$ - module such that $A/pA$ is finite, then $A$ is finitely generated over $\Bbb{Z}_p$.
I have to show that every element $ x \in A$ can be written in form $x= a_1p_1 + \dots + a_np_n, \ a_1,\dots,a_n \in A, \ p_1,\dots,p_n \in \Bbb{Z}_p$. I know that compact set is bounded and closed. $A/pA$ is finite, so it has finite number of elements. Maybe should I use the fact that $\alpha: A \to A/pA$ is homomorphism?
Let $B$ be a finitely generated submodule of $A$ such that it surjects onto $A/pA$ under the reduction map. By replacing $A$ by $A/B$, we can assume that our goal is to show that for compact $A$ such that if $A/pA = 0 \iff A = pA$, then $A=0$. Note that $A = pA$ implies that $A = p^nA$ for all $n$ by induction.
Now, let $U$ be an open subset of $A$ around $0$. Then, for any $a \in A$, we can find an open set $U_a$ around it and by continuity, a large enough $n$ such that $p^nU_a \subset U$. Since $A$ is compact, finitely many of these suffice to cover $A$ and therefore, there is some $n_0 \gg 0$ such that $p^{n_0}A \subset U \neq A$ which contradicts our assumption that $pA = A$.