TITLE: An automorphism of the field of $p$-adic numbers
QUESTION [24 upvotes]: Is an automorphism of the field $\mathbb{Q}_p$ of $p$-adic numbers the identity map?
If yes, how can we prove it?
Note:We don't assume an automorphism of $\mathbb{Q}_p$ is continuous.

REPLY [12 votes]: The following proof is basically the same as Mr. Dietrich Burde's answer.
Let $\mathbb{Q}_p$ be the field of $p$-adic numbers.
Let $\nu$ be the canonical additive valuation on $\mathbb{Q}_p$, i.e. $\nu(p) = 1$.
Let $\sigma$ be an automorphism of $\mathbb{Q}_p$.
Since $\mathbb{Q}$ is dense in $\mathbb{Q}_p$ and $\sigma$ induces the identity map on $\mathbb{Q}$, it suffices to prove that $\sigma$ is continuous.
Let $U$ be the set of $p$-adic units.
We first show that $\sigma(U) \subset U$.
Let $\alpha \neq 0$ be a $p$-adic number.
Suppose the set $S = \{n \in \mathbb{Z}, n \gt 0\mid x^n = \alpha$ has a solution in $\mathbb{Q}_p\}$ is infinite.
If $x^n = \alpha$, then $n\nu(x) = \nu(\alpha)$.
Hence if $\nu(\alpha) \neq 0$, then $\nu(\alpha)$ is divisible by infinite numbers of rational integers. This is absurd.
Therefore $\nu(\alpha) = 0$, which means $\alpha \in U$.
Now suppose $\epsilon$ is a $p$-adic unit.
The set $S = \{n \in \mathbb{Z}, n \gt 0\mid x^n = \epsilon$ has a solution in $\mathbb{Q}_p\}$is infinite as shown in here.
Hence the similar set for $\sigma(\epsilon)$ is infinite.
Therefore $\sigma(\epsilon) \in U$ by what we have shown above.
This means $\sigma(U) \subset U$.
Let $\alpha \neq 0$ be an element of $\mathbb{Z}_p$.
Let $n = \nu(\alpha)$.
Then $\alpha = p^n \epsilon$, where $\epsilon \in U$.
Since $\sigma(\alpha) = p^n\sigma(\epsilon)$, $\sigma(\alpha) \in \mathbb{Z}_p$.
Hence $\sigma(\mathbb{Z}_p) \subset \mathbb{Z}_p$.
Hence $\sigma(p^n\mathbb{Z}_p) \subset p^n\mathbb{Z}_p$ for every positive integer $n$.
This means $\sigma$ is continuous.