Field Completions

Question

Let $(K,v)$ be a number field with an absolute value. Denote $K_v$ to be a completion of $K$ and $\overline{K_v}$ to be an algebraic closure of $K_v$. Let $E$ be a finite extension of $K$. Every embedding (over K) $\sigma: E\to \overline{K_v}$ gives rise to an absolute value on $E$ which extends $v$. Two embeddings gives rise to the same absolute value on $E$ if and only if they are conjugate over $K_v$. Why is it that every absolute value of $E$, extending $v$, must be of this type?

Answer 1

If you have an absolute value $w$ on $E$ extending $v$, extend it by continuity to a map $w':E\otimes_K K_v\to\mathbb{R}_+$. This map $w'$ will still be multiplicative and satisfy the triangle inequality. We have $E\otimes_K K_v\cong L_1\oplus\dots\oplus L_k$ for some finite extensions $L_i\supset K_v$. From the triangle inequality you can see that $w'$ is not identically zero on at least one $L_i$, and in fact on exactly one $L_i$, as multiplicativity shows., So $w'$ is a valuation on $L_i$. What we have now: $K_v\subset L_i$, $w'$ on $L_i$ extending $v$ on $K$ - it's the unique extension (as $K_v$ is complete); and $E\subset L_i$, $E$ dense in $L_i$, $w'$ extending (completing) $w$. Hence $w$ is obtained by $E\subset L_i\subset\overline{K_v}$.