Why ${\rm Gal}\left (\overline{\mathbb{Q}_p}/ \mathbb{Q}_p\right )$ action on $ \overline{\mathbb{Q}_p}$ extends to its action on $\mathbb{C}_p$

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Why ${\rm Gal}\left (\overline{\mathbb{Q}_p}/\mathbb{Q}_p\right )$ action on $ \overline{\mathbb{Q}_p}$ extends to ${\rm Gal}\left (\overline{\mathbb{Q}_p}/ \mathbb{Q}_p\right )$ action on $\mathbb{C}_p$?

$a\in \mathbb{C}_p$ can be regarded as a limit of a Cauchy sequence in $\overline{\mathbb{Q}_p}$. I think I just need to define $\sigma (a)=\lim \limits _{n\to \infty}\sigma (a_n)$, but my book writes we use$$|\sigma (a)|=|a|\text{ for all } a\in \overline{\mathbb{Q}_p}.\qquad ①$$I don't see how to use condition $①$ to prove $\sigma (a)=\lim \limits _{n\to \infty}\sigma (a_n)$ is in $\mathbb{C}_p$. Or I may have some misunderstanding? We can extend group action without using $①$, but we use $①$ to prove the action is continuous?
Thank you in advance.

Best Answer

Technically speaking, mere continuity doesn't necessarily imply the existence of extension. For example, the continuous function $x\mapsto \frac{1}{x}$ on $(0,\infty)$ cannot be extended to the completion $[0, \infty)$. Some kind of uniformality is needed.

But this is a much easier problem to handle with the given fact. Since $\mathbb C_p=\widehat{\overline{\mathbb Q_p}}$, we may define $\mathbb C_p$ as equivalence classes of Cauchy sequences. Given Cauchy sequence $\{a_n\}\in \overline{\mathbb Q_p}$, we have $\{\sigma(a_n)\}$ must also be Cauchy for $|\sigma(a_n)-\sigma(a_m)|=|\sigma(a_n-a_m)|=|a_n-a_m|$. It remains to show two equivalent Cauchy sequences are sent to equivalent ones, which follows basically in the same way.