extension of Krull valuation

The Krull valuation  $|\cdot {|}_K$  of the field $K$ is called the 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 $k$.

Theorem 1.

The trivial valuation is the only of the trivial valuation of $k$ to an algebraic extension field $K$ of $k$.

Proof.  Let’s denote by  $|\cdot |$  the trivial valuation of $k$ and also its arbitrary Krull to $K$.  Suppose that there is an element $\alpha$ of $K$ such that  $|\alpha |>1$.  This element satisfies an algebraic equation

$${\alpha }^n+a_1{\alpha }^{n-1}+\mathrm{\dots }+a_n=0,$$

where  $a_1,\mathrm{\dots },a_n\in k$.  Since  $|a_j|\leqq 1$  for all $j$’s, we get the impossibility

$$0=|{\alpha }^n+a_1{\alpha }^{n-1}+\mathrm{\dots }+a_n|=\max \{{|\alpha |}^n,|a_1|\cdot {|\alpha |}^{n-1},\mathrm{\dots },|a_n|\}={|\alpha |}^n>1$$

(cf. the sharpening of the ultrametric triangle inequality).  Therefore we must have  $|\xi |\leqq 1$  for all  $\xi \in K$,  and because the condition    would imply that  $|{\xi }^{-1}|>1$,  we see that

$$|\xi |=1\mathit{\hspace{1em}}\forall \xi \in K\setminus \{0\},$$

which 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):

Theorem 2.

Every Krull valuation of a field $k$ can be extended to a Krull valuation of any field of $k$.

References

[1] Emil Artin: .   Lecture notes.  Mathematisches Institut, Göttingen (1959).