Let $p>3$ a prime integer and let $A$ be a commutative $\mathbb{Z}_p$-algebra which as a $\mathbb{Z}_p$-module is free of finite rank. Let $x \in A$. The goal of this exercise is to show that $\lim_{n \to \infty}x^{n!}$ exists in $A$ for the $p$-adic topology and that the limit is an idempotent.
- Suppose first that $A = O_K$, with $K$ a finite extension of $\mathbb{Q}_p$. Show that: $\lim_{n \to \infty} x^{n!}=1$ if $x$ is invertible in $A$ and $\lim_{n \to \infty} x^{n!} =0$ if $x$ isn't invertible in $O_K$.
- Suppose that $A \otimes_{\mathbb{Z}_p} \mathbb{Q}_p$ is semi-simple, i.e. it has no nilpotent elements. Show that $A \otimes_{\mathbb{Z}_p} \mathbb{Q}_p$ is a finite product of finite extensions of $\mathbb{Q}_p$. Then show that for every $x \in A$ we have: $e := \lim_{n \to \infty} x^{n!}$ exists and it is an idempotent of $A$ ,i.e. $e^2 =e$.
- Decide what happens if $A \otimes_{\mathbb{Z}_p} \mathbb{Q}_p$ has nilpotent elements?
Can anyone help me?