Why do we fix an extension of valuation of $K$ to $\overline{K}$?

Question

Let $K$ be a valuation field. Let $v$ be a valuation of $K$.　Let $K_v$ be completion of $K$ at $v$. I often encounter an expression.

Fix an extension of $v$ to $\overline{K}$, which serves to fixes an embedding $\overline{K}\subset \overline{K_v}$.

but cannot understand meaning of this by two reasons.

$1$. Extension of $v$ to any algebraic extension is unique. Why do we have to 'fix' it ?

$2$. The map $\overline{K}\subset \overline{K_v}$ is given by $a \to (a,a,a,・・・)$ and this has nothing to do with $v$. Why 'fixing extension of $v$' serves to define this map ?

Thank you in advance.

Answer 1

1. The extension is not unique unless $K$ is complete with respec to $v$. Take a global field $K$, with $v$ being the valuation for a maximal ideal $\mathfrak m$, then for any algebraic extension $K\subset K'$, the extension of $v$ to $K'$ is not necessarily unique, as the ideal $\mathfrak m$ may split in $\mathcal O_{K'}$.
2. Since $K_v$ is complete, for any finite extension $K_v\subset K'$, there is indeed a unique extension of $v$ to $K'$, hence there is a unique extension of $v$ to $\overline{K_v}$. An embedding $\overline{K}\hookrightarrow\overline{K_v}$ obviously induces a valuation on $\overline{K}$. On the other hand, fix a valuation on $\overline{K}$. For any algebraic extension $K\subset K(\alpha)$, we have $K_v\subset K(\alpha)_v$, and since $\alpha$ is algebraic over $K$, it's also algebraic over $K_{\alpha}$, therefore $K_v(\alpha)$ is finite dimensional over $K_v$, there is a unique extension of $v$ to $K_v(\alpha)$ that must match the one on $K(\alpha)_v$. That is, $K(\alpha)$ can be isometrically embeded into $\overline{K_v}$. By Zorn's lemma or whatever, $\overline{K}$ can be isometrically embeded into $\overline{K_v}$.

More details can be found in the 8th chapter of Milne's notes, or Theorem 8.1 (Chapter II, Section 8) of Neukirch's book.