This is from JS Milne's notes on Algebraic Number Theory, lemma 8.6
Let $K$ be a local field with normalized absolute value $|.|_K$. Let $L$ be a finite separable extension of degree $n$, and $|.|_L$ be the unique extension of $|.|_K$ to $L$.
Let $||.||$ be the normalized absolute value that is equivalent to $|.|_L$. The claim is that for $a \in L$, $||a|| = |a|_L^n$.
First question: How is the normlized absolute value is defined? After theorem 7.14, it's defined only for number fields (but $K$ is not a number field) as $\frac{1}{\mathbb{N}p}$ if $|.|_K$ is the $p$-adic absolute value for some prime ideal $p$ in $O_K = \{x \in K : |x| \leq 1 \}$, where $\mathbb{N}p$ is the numerical norm of $p$. But again, this is only defined for number fields. So How is the normalized absolute value defined for a local field $K$?
Now, the proof: For Archimedean $K$, this is obvious. For non-Archimedean $K$, since $||.||$ and $|.|_L$ are equivalent, we have $||.|| = |.|^c$ for some constant $c$. Let $\pi$ be a prime element of $K$ and $\Pi$ be a prime element of $L$. $(\pi)$ factors in $O_L$ as $ u \Pi^e$ where $u$ is a unit in $O_L$.
Therefore, $||\pi|| = ||\Pi^e|| = \frac{1}{\mathbb{N} \Pi}^e = \frac{1}{\mathbb{N}{\pi}}^{ef} = |\pi|^n$ so the constant is $n$.
Second question: Why does $ \frac{1}{\mathbb{N} \Pi} = \frac{1}{\mathbb{N}{\pi}}^{f} $ ? I think $f$ is the inertia degree of the factorization. I think it has to do with how $\mathbb{N}\Pi$ is defined. But again , I don't understand how it's defined for the non-number field $K$.
Best Answer
The normalized absolute value can be defined on a local field $K$ in the same way as for number fields.
Indeed, let $k$ be the residue field of $K$, and $q$ its cardinality. Let $\varpi$ be a uniformizer of $K$. The normalized absolute value $|\cdot|_K$ on $K$ is the one such that $|\varpi|_{K}=q^{-1}$.
Now, let $L/K$ be finite separable with degree $n$, ramification degree $e$ and residual degree $f$, and $|\cdot|_L,|\cdot|_K$ are the respective normalized absolute values and $\varpi_L,\varpi_K$ are the uniformizers.
If $a \in L^{\times}$, then $v_K(N_{L/K}a)=e^{-1}v_L(N_{L/K}a)=e^{-1}nv_L(a)=fv_L(a)$. So $|N_{L/K}a|_K=q^{-v_K(N_{L/K}a)}=q^{-fv_L(a)}=(q^f)^{-v_L(a)}=|a|_L$.