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# extension of Krull valuation

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 $k$.

###### Theorem 1.

The trivial valuation is the only extension 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 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):

###### Theorem 2.

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

# References

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

## Mathematics Subject Classification

13F30*no label found*13A18

*no label found*12J20

*no label found*11R99

*no label found*

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