Extensions of a complete valued field

Question

Let $K$ be a field which is complete with respect to an absolute value $| \cdot |_K : K \to \mathbb{R}$, and let $E$ be a finite extension of $K$. I know that there is at most one absolute value $|\cdot|_E$ on $E$ which extends the given one on $K$.

On the other hand, there is such an absolute value on $E$ provided that $E/K$ is separable, $K$ is locally compact and $|\cdot|_K$ non-archimedean, or if $|\cdot|_K$ is discrete.

Are all these assumptions really necessary, i.e. are there examples for $K$ and $E$, where $|\cdot|_K$ cannot be extended? If yes: Is there a more general extension theorem which includes the above cases?

Answer 1

Here is a sketch of the proof that $|a|_E := \sqrt[n]{|N_{E/K}(a)|_K}$ is an absolute value on $E$ for any extension field $E$ of degree $n$:

It clearly satisfies the first two axioms of an absolute value, so it remains to show the (non-archimedean) triangle inequality. Since the first two axioms hold, it suffices to show that $|a|_E \leq 1$ implies $|a+1|_E \leq 1$ for all $a \in E$. Obviously, $|a|_E \leq 1$ is equivalent to $|N_{E/K}(a)|_K \leq 1$. Now we can apply the following result which is a corollary of the theory of Newton polygons, and which also follows from a specific version of Hensel's Lemma:

For a polynomial $f = \sum_{i=0}^m b_i X^i \in K[X]$ set $|f|_K = \max_{0 \leq i \leq m} |b_i|_K$. If $f$ is a power of an irreducible polynomial then $|f|_K = \max \{ |b_0|_K, |b_m|_K \}$.

By this result $|N_{E/K}(a)|_K \leq 1$ implies $|f|_K = 1$ for $f = \mathrm{charpol}(a)$. But then $|N_{E/K}(a+1)|_K = |f(-1)|_K \leq |f|_K = 1$ and hence $|a+1|_E \leq 1$.