The Krull valuation $|\cdot|_K$ of the field $K$ is called the extension of the Krull valuation $|\cdot|_k$ of the field $k$ , if $k$ is a subfield of $K$ and $|\cdot|_k$ is the restriction of $|\cdot|_K$ on $k$ .

Proof. Let's denote by $|\cdot|$ the trivial valuation of $k$ and also its arbitrary Krull extension valuation to $K$ . Suppose that there is an element $\alpha$ of $K$ such that $|\alpha| > 1$ . This element satisfies an algebraic equation $$\alpha^n+a_1\alpha^{n-1}+...+a_n = 0,$$ where $a_1,\, ..., \,a_n\, \in k$ . Since $|a_j| \leqq 1$ for all $j$ 's, we get the impossibility $$0 = |\alpha^n+a_1\alpha^{n-1}+...+a_n| = \max\{|\alpha|^n,\,|a_1|\!\cdot|\!\alpha|^{n-1},\, ...,\, |a_n|\} = |\alpha|^n > 1$$ (cf. the sharpening of the ultrametric triangle inequality). Therefore we must have $|\xi| \leqq 1$ for all $\xi\in K$ , and because the condition $0 < |\xi| < 1$ would imply that $|\xi^{-1}| > 1$ , we see that $$|\xi| = 1 \quad \forall\xi\in K\!\setminus\!\{0\},$$ which means that the valuation is trivial.

The proof (in [1]) of the next ``extension theorem'' is much longer (one must utilize the extension theorem concerning the place of field):

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