Let $K$ over $\mathbf{Q}_p$ be an algebraic extension, not complete for $|\cdot|_K$ (the unique extension of $p$-adic norm $|\cdot|_p$ to $K$).
(For example, $K=\cup_n\mathbf{Q}_p(\zeta_{p^n})$ the cyclotomic extension of $\mathbf{Q}_p$.)
Denote $G_K$ the absolute Galois group of $K$. Denote $\widehat{K}$ the completion of $K$ under the norm $|\cdot|_K$, and $G_{\widehat{K}}$ its absolute Galois group.
The question is :
Question (a): Is it true that $G_K\simeq G_{\widehat{K}}?$
This statement itself is very strange for me as I never heard it before (I should have if it is true…). But I wrote a proof, not difficultly, as follows.
So please give me some remarks or let me know whether I made some mistakes in the following proof of Lemma (b) and hence proof of (a).
My proof of (a):
Let $L$ over $K$ be a Galois extension with Galois group $G$ and denote similarly $|\cdot|_L$ the unique extension of $p$-adic norm to L. Then I claim it suffices to prove
Lemma (b): We have $|\cdot|_L\circ g=|\cdot|_L$, for any $g\in G$, i.e. $G$ acts isometrically over $L$.
Let me first explain why Lemma (b) is enough: (the proof is classical) By Lemma (b) we will have $G_K$ acts isometrically over $\overline{K}$, hence any $g \in G_K$ maps Cauchy sequences in $\overline{K}$ again to Cauchy sequences. Hence it extends to an automorphism of $\widehat{\overline{K}}$ (the completion of $\overline{K}$ with respect to the unique extension of $|\cdot|_p$). This gives an injective map
$G_K \to Aut_{K}(\widehat{\overline{K}})$. Moreover this is surjective (and hence an isomorphism), since an automorphism must maps algebraic elements over $K$ again to algebraic elements. Finally by continuity one notices
$Aut_{K}(\widehat{\overline{K}})\simeq Aut_{\widehat{K}}(\widehat{\overline{K}})\simeq Aut_{\widehat{K}}(\widehat{\overline{\widehat{K}}})\simeq G_{\widehat{K}}$.
(If we denote typically $\widehat{\overline{K}}$ by $C$, then summarize:
$G_K \simeq G_{\widehat{K}}\simeq Aut_{\widehat{K}}(C)$.)
Now we prove Lemma (b):
Proof of (b): Observe that $|\cdot|_L\circ g$ and $|\cdot|_L$ give two norms over $K$ that both extends $|\cdot|_K$ over $K$, and hence they both in particular extends $|\cdot|_p$ over $\mathbf{Q}_p$. Hence they must be the same by the unique extension of norm (not for $|\cdot|_K$ but for $|\cdot|_p$.)
Remark: Intuitively this Lemma (b) is wrong, since usually to prove this kind of arguments one always uses the completeness of the base field $K$ and then argue that the two norms (comes from the two sides of the equality in Lemma (b)) must be the same by the unique extension of norms. While here exactly we assume the base field $K$ to be not complete.
Edit: From the excellent comment of Torsten Schoeneberg, this "phenomenon" could be explained naturally by the basic fact "algebraic extensions of Henselian fields are still henselian".
Best Answer
Note that you can rewrite your proof of b to showing that for every algebraic extension of a local field, the valuation still extends uniquely to ("further") algebraic extensions. The concept that will clarify this is "Henselian". In particular, local fields are Henselian, and algebraic extensions of Henselian fields are Henselian.