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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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  • $\begingroup$ Where exactly are you stuck? Have you tried the basic case of $A= \mathbb Z_p$? $\endgroup$ Jun 23 at 16:42
  • $\begingroup$ 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. $\endgroup$
    – MathMario
    Jun 24 at 8:52
  • $\begingroup$ Use that for what? To show that it's a direct product of finite extensions of $\mathbb Q_p$? $\endgroup$ Jun 24 at 15:06

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