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Let $K$ be an infinite algebraic extension of $\mathbf{Q}_p$ (not the algebraic closure). Let $L$ be the algebraic closure of the completion $\widehat{K}$.

Let $\mathbf{C}_p$ be the completion of the algebraic closure $\overline{\mathbf{Q}}_p$.

We have a map $\mathbf{C}_p \longrightarrow \widehat{L}$. Is this map always an isomorphism?

Bun
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  • Maybe remind us where that map comes from. Either way, I would think Kürschak's Theorem (https://math.stackexchange.com/a/123962/96384) is relevant here. – Torsten Schoeneberg Oct 23 '22 at 02:58
  • If I'm not mistaken, $\mathbf{Q}_p \to K$, therefore $\overline{\mathbf{Q}}_p \to \overline{K} \to L$, therefore $\widehat{\overline{\mathbf{Q}}}_p = \mathbf{C}_p \to \widehat{L}$. – Bun Oct 23 '22 at 05:43
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    I misunderstood that you said $\Bbb{C}_p\to \widehat{L}$ not $\Bbb{C}_p\to L$. Your question is a bit trivial once you know that $\Bbb{C}_p$ is algebraically closed, where are you stuck at proving that $\widehat{L}=\Bbb{C}_p$ ? – reuns Oct 23 '22 at 06:24
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    I think I may have just been confused by not drawing out the diagram carefully. Now it seems that $K \to \mathbf{C}_p$, therefore $\widehat{K} \to \mathbf{C}_p$, therefore $L \to \mathbf{C}_p$ (using that $\mathbf{C}_p$ is algebraically closed), therefore $\widehat{L} \to \mathbf{C}_p$. I guess I haven't checked that the composition of $\widehat{L} \to \mathbf{C}_p$ and $\mathbf{C}_p \to \widehat{L}$ gives automorphisms on both sides. – Bun Oct 23 '22 at 07:30

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