# limit in $p$-adic numbers

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.

1. 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 $$O_K$$.
2. 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. $$e^2 =e$$.
3. Decide what happens if $$A \otimes_{\mathbb{Z}_p} \mathbb{Q}_p$$ has nilpotent elements?

Can anyone help me?

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• Where exactly are you stuck? Have you tried the basic case of $A= \mathbb Z_p$? Jun 23 at 16:42
• The problem is how to use the fact that $A \otimes_{\mathbb{Z}_p} \mathbb{Q}_p$ has no nilpotent element. I have tried to use $A \cong \mathbb{Z}_p ^n$ as module but it did not bring results. Jun 24 at 8:52
• Use that for what? To show that it's a direct product of finite extensions of $\mathbb Q_p$? Jun 24 at 15:06