Normalised absolute values on $p$-adic extensions I have the following problem: show that if $L/K$ are finite extensions of the p-adic numbers $\mathbb{Q}_p$ with normalised absolute values $|\cdot|_K$ and $|\cdot|_L$, with $n=[L:K]$, then $|x|_L=|x|_K^n$ $\forall$ $x \in K$. 
The intention is to deduce from this that $|N_{L/K}(x)|_K = |x|_L$, but it's the first part I'm having trouble with. I think the issue is I don't know what 'normalised' means in this context: I know a normalised valuation has image $\mathbb{Z}$, but what does it mean in this absolute value situation? Something to do with $|p|_K$ and $|p|_L$ perhaps? I expect the problem is probably quite easy after that: something to do with just defining a basis of size $n$ in K for L and then playing around with the absolute value definition. Thank you for the help!
 A: There are various ways to normalize the absolute value on a local field.  As Justin Cambell mentions, one way is to choose the absolute value so that it agrees with a given absolute value on some natural choice of subfield (e.g. with the usual absolute value on $\mathbb R$, in the case of $\mathbb C$ over $\mathbb R$,
or with the $p$-adic absolute value on $\mathbb Q_p$, in the case of an extension
of $\mathbb Q_p$).
However, there is a more intrinsic way to normalize the absolute value, which does
not make any reference to a given choice on a subfield.  Namely, if $x$ is any
non-zero element of the local field $K$, then multiplication by $x$ scales additive Haar measure on $K$ by some amount; we declare this scaling factor to be
$|x|$.  This is the normalized absolute value on $K$ referred to in your question.
E.g. if $K = \mathbb R$, this will give the usual absolute value on $\mathbb R$.
If $K = \mathbb C$, it will give the square of the usual absolute value on $\mathbb C$.
If $K$ is an extension of $\mathbb Q_p$ with residue field of order $q$ and $\pi$ is a uniformizer, then (by comparing the
measure of $\mathcal O_K$ and $\pi \mathcal O_K$) one finds that $|\pi| = q$.
Now with this normalization, your problem is fairly easily solved.  
One approach
is to use the relation $[L:K] = e f$, as Justin Cambell suggests. 
An alternative approach is to use directly the fact that $L$ is a vector space over $K$ of dimension $[L:K]$, and think about how Haar measure
on $L$ is thus related to Haar measure on $K$, and how it will scale under
multiplication by an element of $K$.  (The analogy with the case of $\mathbb C$ over $\mathbb K$ can be helpful here.)
Note that using the second approach, and then considering the first approach in reverse, one can derive the formula $[L:K] = ef$.  This is a typical application of normalized absolute values.  (I.e. one begins with a more analytic viewpoint, and then uses it to derive arithmetic facts.)  Another is
to derive the product formula for a global field (that the product over all absolute values of an element of the global field equals $1$); and note that in this formula, it is crucial that normalized absolute values be used.
