Absolute Galois group of $\mathbb{Q}_p$

Question

Let $\sigma$ be an element in the absolute Galois group of the $p$-adic numbers $\mathbb{Q}_p$. We denote the algebraic closure by $\mathbb{A}_p$. If we have for any algebraic integers $x\in\mathcal{O}_{\mathbb{A}_p},\sigma(x)-x\in(p)\subseteq\mathcal{O}_{\mathbb{A}_p}$. Then to show is $\sigma$ is the identity.

I want to prove $\sigma(x)=x$ for any algebraic integers. It can be reduced to show that Galois conjugates can not be congruent mod $(p)$.

Answer 1

Given $\sigma\in Gal(\overline{\Bbb{Q}}_p/\Bbb{Q}_p)$ such that $\forall x\in O_{\overline{\Bbb{Q}}_p}^\times, v(x-\sigma(x)) \ge v(p)=1$. $\tag{1}$

Let $a\in p\, O_{\overline{\Bbb{Q}}_p}$, for every $k$ $$(1+a)^{p^k}=\sum_{m=0}^{p^k}{p^k\choose m} a^m\equiv 1\bmod p^k \tag{2}$$

Assume $v(x)=0,x\ne \sigma(x)$ then $v((\frac{x}{\sigma(x)})^{1/p^k}-1)=v(x^{1/p^k}-\sigma(x^{1/p^k}))> 0$, let $$r = \inf_k v( x^{1/p^k}-\sigma(x^{1/p^k}))$$

• If $r<1$ then we have a contradiction with $(1)$.

• Thus $r \ge 1$ and for all $k$ we have $v( (\frac{x}{\sigma(x)})^{1/p^k}-1)\ge 1$,

  $(2)$ implies that for all $k$, $v( \frac{x}{\sigma(x)}-1)\ge k$, letting $k\to \infty$ we get $v( \frac{x}{\sigma(x)}-1)=\infty$ ie. $$x=\sigma(x)$$

And hence $\sigma$ is the identity.