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If $p\in\mathbb{Z}$ is prime, prove that:

$1)$ If $a\in\mathbb{Q}_p^{\times}$, then $a\in\mathbb{Z}_p^{\times}\iff \exists\, b_n\in\mathbb{Q}_p^{\times}$ such that $a=b_n^n$ for all $n>0$ with $\gcd(n, p(p-1))=1$

$2)$ $\text{Aut}(\mathbb{Q}_p)=\{id_{\mathbb{Q}_p}\}$

Here's what I've done:

$1)$ $(\Leftarrow)$ suppose that $a=b_n^n$ for all such $n$. Letting $|a|_p=p^k$ and $|b_n|_p=p^{e_n}$ with $k, e_n\in\mathbb{Z}$, we have that $k=ne_n$, so $e_n\to 0$ when $n\to\infty$, which means $e_{n_0}=0$ for some $n_0$. Therefore $k=n_0e_{n_0}=n_0(0)=0$, so $|a|_p=1$, which means $a\in\mathbb{Z}_p^{\times}$.

for $(\Rightarrow)$, I still have no clue.

Now $2)$: assume $1)$ is true. Taking $a\in\mathbb{Z}_p^{\times}$ and $\phi\in\text{Aut}(\mathbb{Q}_p)$, we prove that $\phi(a)\in\mathbb{Z}_p^{\times}$. If $a\tilde{a}=1$, then $\tilde{a}\in\mathbb{Z}_p^{\times}$ so we have that $\exists\,\tilde{b}_n$ with $\tilde{a}=\tilde{b}_n^n$ for every $n \geq 0$. That way, $a\tilde{a}=1\Rightarrow\phi(a)\phi(\tilde{a})=1$ $\Rightarrow \phi(a)^{-1}=\phi(\tilde{b}_n^n)=\phi(\tilde{b}_n)^n$ for all $n\geq 0$, which means $\phi(a)^{-1}\in\mathbb{Z}_p^{\times}$, thus $\phi(a)\in\mathbb{Z}_p^{\times}$. So, basically, $\phi(\mathbb{Z}_p^{\times})=\mathbb{Z}_p^{\times}$, which looks like I'm getting somewhere. But I don't know how to conclude $\phi=id_{\mathbb{Q}_p}$.

Any tips? Thanks!

1 Answers 1

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For (1), hint: Hensel's lemma.

For the claim (2), it suffices to show that any automorphism $ \phi : \mathbf Q_p \to \mathbf Q_p $ must be continuous. This follows; because the fact that $ \phi $ preserves $ p $-adic units implies that it preserves $ p $-adic valuations (and by linearity, it suffices to check continuity at $ 0 $). Now, $ \phi $ fixes the dense subset $ \mathbf Q $, so it must be the identity.

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    Thanks! The only thing I'm still can't figure out is why does $\phi(\mathbb{Z}_p^{\times})=\mathbb{Z}_p^{\times}$ imply that $|\phi(a)|_p=|a|_p$ for all $a\in\mathbb{Q}_p$ – 2017-02-20
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    Any $ p $-adic number can be written as $ p^k u $, where $ k $ is an integer and $ u $ is a $ p $-adic unit. $ \phi $ maps this to $ p^k \phi(u) $, and since $ \phi(u) $ is a $ p $-adic unit, this has $ p $-adic valuation $ k $; equal to the one of $ p^k u $. – 2017-02-20
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    but how do you know that $\phi(p)=p$? I guess that should be my question – 2017-02-20
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    @AguirreK Because $ \phi(1) = 1 $, and $ p $ is a sum $ 1 + 1 + 1 + \ldots + 1 $. If $ \phi(1) = 1 $ is not explicitly required in your definition of "automorphism", note that an automorphism preserves roots of $ X^2 - X $, thus either $ \phi(1) = 1 $ or $ \phi(1) = 0 $; and $ \phi $ is trivial in the latter case, i.e not an automorphism. – 2017-02-20
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    Oh, of course! The problem feels much more natural now, thanks! – 2017-02-20
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    What’s the argument that $\varphi$ must preserve the $p$-adic units? Even when $\varphi$ isn’t assumed to be continuous? – 2017-02-28
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    @Lubin If you pick a number $ n $ coprime to $ p(p-1) $, as outlined in the question, then $ p $-adic units, and only $ p $-adic units are $ n $th powers for arbitrarily large $ n $; and $ \phi $ preserves powers, of course. (The condition on $ n $ guarantees that the unit group $ \mathbf Z_p^{\times} \cong C_{p-1} \times \mathbf Z_p $ is $ n $-divisible.) – 2017-02-28
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    On units: right, that’s just the argument I’d cooked up for myself. I was just wondering whether you had a different one. – 2017-03-01