# 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

 $\alpha^{n}+a_{1}\alpha^{n-1}+...+a_{n}=0,$

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 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 ExtensionOfKrullValuation 2013-03-22 14:55:57 2013-03-22 14:55:57 pahio (2872) pahio (2872) 13 pahio (2872) Theorem msc 13F30 msc 13A18 msc 12J20 msc 11R99 GelfandTornheimTheorem