# JS Milne Algebraic Number Theory 8.6 : normalized absolute values for local fields

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 $$K$$.

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 $$|N_{L/K}a|_K=q^{-v_K(N_{L/K}a)}=q^{-fv_L(a)}=(q^f)^{-v_L(a)}=|a|_L$$.