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Suppose $a$ is an integer.

If $a$ is divisible by $p$ then clearly the sequence converges to $0$ since the $p$-adic norm of $a^{p^n}$ tends to $0$.

If $(a, p) = 1$ then we can still show the sequence is Cauchy by $$\left|a^{p^n}-a^{p^m}\right|_p = \left|a^{p^m}\right|_p\left|a^{p^n - p^m}-1\right|_p = \left|a^{p^m(p-1)\frac{p^{n-m}-1}{p-1}}-1\right|_p \leq \frac{1}{p^{m+1}}$$ since $a^{p^m(p-1)} = a^{\phi\left(p^{m+1}\right)} = 1 \pmod{p^{m+1}}$. And $m \leq n$.

Since $\mathbb{Q}_p$ is a complete field, this shows the sequence $\{a^{p^n}\}$ converges to some $\alpha \in \mathbb{Q}_p$.

My question is, is there a more explicit description of this element $\alpha$, and does it satisfy some additional interesting properties?

So far I can see that the differences $a^{p^n} - a^{p^{n-1}}$ are divisible by $p^n$ and so we can form the series $$\alpha = a+ \sum\limits_{n=1}^{\infty}\left(\frac{a^{p^n} - a^{p^{n-1}}}{p^n}\right)p^n$$ which gives a $p$-adic expansion series representation of $\alpha$ [not actually a $p$-adic expansion because the terms might not be between $0$ and $p-1$].

Based on this, I would expect also $\alpha \in \mathbb{Z}_p^*$. Is this correct?

Tob Ernack
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3 Answers3

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Letting $f(\alpha)=\alpha^p$ then you have a recursive sequence, $a,f(a),f(f(a)),\dots$, so any limit must be a fixed point of $f$. So $\alpha=\alpha^p$, with $\alpha\equiv a\pmod{p}$.

There is only one such root for each $a\pmod{p}$.

If $a$ is not divisible by $p$ then neither is $\alpha$ and $\alpha$ is a $p-1$th root of unity in $\mathbb Q_p$, again with $\alpha\equiv a\pmod{p}$.

Hensel's lemma can be used to solve $x^{p-1}-1=0$ with $x\equiv a\pmod{p}$. It won't give much different results.

You'll have $a$ has multiplicative order $k$ modulo $p$, then $\alpha$ will have multiplicative order $k$ in $\mathbb Q_p$.

It converges starting with any $a.$ This is because if $a\equiv b\pmod{p^k}$ for $k>1,$ thjen $$a^p-b^p=(a-b)\left(a^{p-1}+a^{p-2}b+\cdots + b^{p-1}\right)$$

Now, $p^k\mid a-b$ and:

$$\begin{align}a^{p-1}+a^{p-2}b+\cdots+ b^{p-1}&\equiv a^{p-1}+a^{p-1}+\cdots+a^{p-1}\pmod{p^k}\\ &=pa^{p-1}\end{align}$$

So we get that $p^{k+1}\mid a^p-b^p.$

This means, in particular, since $a^{p}\equiv a\pmod{p}$ you get, by induction:

$$a^{p^n}\equiv a^{p^{n-1}}\pmod {p^n}$$

And hence the sequence is Cauchy in $\mathbb Q_p.$

Thomas Andrews
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This operation is called the Teichmüller character, and the elements of its image are called the Teichmüller representatives.

Qiaochu Yuan
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You can show it by induction in a simple way that's essentially just Hensel lifting.

Start with the inductive step, for some $\omega^{p-1}=1$ and $x_i \in \mathbb{Z}_p$ by expanding the binomial,

$$(\omega+p^nx_n)^p = \omega^p + p \omega^{p-1} p^nx_n + p^{n+1} * \text{extra terms}$$

Since $\omega^p=\omega$ and the term with the binomial coefficient $\binom{p}{p-1}=p$ (the only term that doesn't naturally get smaller by a power of $p^n$ from the binomial expansion) gets one anyways, we can safely factor out $p^{n+1}$ leaving a p-adic integer behind,

$$(\omega+p^nx_n)^p = \omega + p^{n+1}x_{n+1}$$

This completes the inductive step, since it means every power of p will always "clean off" one extra digit of the expansion of $\omega$. All that's left is to confirm the base case, which means deciding on your starting digit.