Prove that the sequence satisfying $a^{p^n}\equiv a^{p^{n-1}}$mod $p^n$ for $n\geq 1$, converges to the Teichmuller Representative

Question

Prove that for any $a\in\mathbb{Z}_p$, the sequence satisfying $a^{p^n}\equiv a^{p^{n-1}}$mod $p^n$ for $n\geq 1$, converges to the Teichmüller Representative congruent to $a$ mod $p$.

So a p-adic integer has general form $a = a_0+a_1p+a_2p^2+...$ , for $a_i\in\{0,...,p-1\}$.

Now if $a^p\equiv a$ mod $p$, then $a^p$ and $a$ must have the same constant term. And if $a^{p^2}\equiv a^p$ mod $p$ then $a^{p^2}$ and $a$ must agree in their first two coefficients. Continuing on in this way, this sequence seems to be approaching $a$, which is what I started with, but this can't be right, what am I not understanding?

Edit: Ok I'm revising this, $a^{p^{n}}$ picks up its $p^{n-1}$ coefficient from $a^{p^{n-1}}$, could someone still give me some clarification on how one shows this becomes the Teichmuller Representative? I think I just need someone to tell me the 'right' way to think about it.

Answer 1

I answered this to a large extent earlier:

Suppose $a\in{\bf Z}_p$. Then using the binomial theorem, $$\left|\frac{p\cdot p\cdot p\cdots p}{1\cdot 2\cdot 3\cdots k}p^n\cdot(p^n-1)\cdots(p^n-(k-1))\right|_p\le\frac{1}{p^n}, $$ $$\left|(1+pa)^{p^n}-1\right|_p\le \max_{1\le k\le p^n}\left|{p^n\choose k}p^k\right|_p\le\frac{1}{p^n}\to0,$$ and hence $(1+pa)^{p^n}\to1$ (compare with $t^\epsilon\to1$ as $\epsilon\to0$ in $\bf R$ for $t$ in a nbhd of $1$). If $x\in{\bf Z}_p$, then $x^{p-1}\in1+p{\bf Z}_p$ (reduce mod $p$ and invoke Euler's theorem), hence $(x^{p^n})_{n=0}^\infty$ is Cauchy and has a limit in the $p$-adic integers, since $x^{p^{n+1}}-x^{p^n}=x^{p^n}\left((x^{p-1})^{p^n}-1\right)$. In an ultrametric space, the distance between successive terms tending to $0$ is sufficient to be Cauchy. Since multiplication and hence fixed powers are by definition continuous in a topological ring, $x\mapsto x^p$ is continuous, so $$\omega(x):=\lim\limits_{n\to\infty}x^{p^n}\implies \omega(x)^p=\left(\lim_{n\to\infty} x^{p^n}\right)^p=\lim_{n\to\infty}x^{p^{n+1}}=\omega(x),$$ in which case we have $\omega(x)^{p-1}=1$ when $\omega(x)$ is a unit. Since $x^p\equiv x\bmod p$, and ${\bf Z}_p\to{\bf Z}/p{\bf Z}$ is a continuous topological ring homomorphism, $\omega(x)\equiv x\bmod p$. This tells us the image of units under $\omega$ consists of the $(p-1)$th roots of unity.

It is not necessarily the case that $a$ and $a^{p^2}$ agree on the first two terms, or that $a^{p^n}$ shares more and more terms with $a$. Setting $p=3$, $a=2+1\cdot3$, we have $a^{3^2}=2+2\cdot3+\cdots\not\equiv 2+3\bmod 9$.

Answer 2

I'd separate this into two claims.

1. The limit of your sequence converges $p$-adically. This follows because it is Cauchy (and $\mathbb{Z}_p$ is complete). If $\ell$ denotes this limit, then $a \equiv \ell \mod p$. (This latter assertion is obvious.)

2. The limit $\ell \in \mathbb{Z}_p$ is a $(p-1)$-th root of unity (or $0$, I suppose). This is equivalent to the assertion that $\ell^p = \ell \in \mathbb{Z}_p$. This holds provided that $\ell \equiv \ell^p$ mod $p^k$ for all integers $k$, and this fact follows from the limit definition of $\ell$.