Extension of places to an algebraic closure Let $K$ be a number field and $v$ a place of $K$. One often extends $v$ to a place $\overline{v}$ of an algebraic closure $\overline{K}$ of $K$, which is equivalent to choosing an embedding $\overline{K} \hookrightarrow \overline{K_v}$, where $\overline{K_v}$ denotes an algebraic closure of the completion $K_v$. However, I haven't found an explanation of how this extension process is done concretely. Can you help me on this one? 
 A: The method suggested by Brandon Carter in his comments will surely work, but I think an explicit call to the Axiom of Choice will be necessary to make the process succeed.
You seem to recognize that there are two ways of attacking the question: by finding an extension of a $K$-place to $\overline K$, and embedding $\overline K$ into an algebraic closure of $K_v$, the completion of $K$ with respect to $v$. In my opinion, whichever approach you use, you’ll need AC.
I had rather use the second approach: from $K\hookrightarrow\overline{K_v}$ and $K\hookrightarrow\overline K$ you conclude that there is a map $\overline K\to\overline{K_v}$ making the triangle commutative. It’s a standard fact that a morphism $\phi$ from $K$ into an algebraically closed field $\Omega$ (not necessarily an algebraic closure of $K$) may be extended to any algebraic extension $K'$, that is, there is (in fact are many) $\phi':K'\to\Omega$.
You prove this standard fact by setting up the partially ordered set of all $(L,\psi)$ for which $K\subset L\subset K'$ and $\psi$ extends the given $\phi$. The ordering is the natural one of extension, and you just do a standard Zorn’s Lemma argument on it. You get a maximal member $(\Lambda,\Psi)$, and you then do the inductive step of arguing that $\Lambda$ must equal $K'$.
If this was too sketchy, don’t hesitate to e-mail me for details.
