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).

Title	extension of Krull valuation
Canonical name	ExtensionOfKrullValuation
Date of creation	2013-03-22 14:55:57
Last modified on	2013-03-22 14:55:57
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	13
Author	pahio (2872)
Entry type	Theorem
Classification	msc 13F30
Classification	msc 13A18
Classification	msc 12J20
Classification	msc 11R99
Related topic	GelfandTornheimTheorem