Completion of global field = extension of completed sub-field?

Question

Given a global field $K = F(\alpha)$ as a finite seperable extension of $F$, where $F = \mathbb{Q} $ or $\mathbb{F}_q(t)$, suppose that $u$ is a fixed place of $F$, and that $v$ is an extension of $u$ over $K$. Let $p$ be the minimal polynomial of $\alpha$ over $F$.

I am trying to show that the completion of $K$ under $v$, $K_v = F_u(\beta)$ for some root $\beta$ of $p(x)$.

(i) Now, unless I'm wrong (please correct me if I am), $F_u(\beta) \cong F_u(\alpha) \hookrightarrow K_v$, proving one side of the inclusion.

(ii) For the other side, I want to say that $F_u(\beta)$ is already $"v"$-complete, so that (i) it contains $F$, (ii) it contains $\beta = im(\alpha)$, (iii) it is complete with respect to $v \implies$ it exactly contains $(F(\alpha))_v = K_v$.

Now, in doing this, I am able to note that (i) $F_u$ is $u$-complete, and since $u = v|_F$, $F_u$ is $v$-complete. Now, I'm not sure how this shows that $F_u(\beta)$ is $u$-complete. Moreover, if we can show this, does this yield $v$-completeness? Or can we stop bothering with $u$-completeness because we have shown that $F_u$ is $v-$complete already?

The textbook I'm following mentions "$F_u(\beta)$ is finite dimensional over $F_u$ and hence is locally compact, so it is a complete field" - but I am not sure why (i) it is locally compact, and (ii) why it is complete.

Answer 1

To show $F_u(\beta)$ is $v$-complete, it suffices to show that a Cauchy sequence $\{x_i\}_{i\ge0}$ of elements of $F_u(\beta)$ has a limit. If $F_u(\beta)/F_u$ has degree $d$, then each $x_i$ can be uniquely written as $x_i=\sum_{j=0}^{d-1}a_{ij}\beta^j$ for some $a_{ij}\in F_u$. Now, you can check (this is an exercise):

1. For each $0\le j\le d$ the sequence $\{a_{ij}\}_{i\ge0}$ is a Cauchy sequence.
2. From 1 (and the completeness of $F_u$), $\{a_{ij}\}_{i\ge0}$ has some limit $\alpha_j$. Then $\sum_{j=0}^{d-1}\alpha _j\beta^j$ is the limit of $\{x_i\}$ (in the $v$-adic topology).