# proof of theorem on equivalent valuations

It is easy to see that $|\cdot |$ and $|\cdot {|}^{c}$ are equivalent valuations for any constant $c>0$ — it follows from the fact that $$ if and only if $$.

Assume that the valuations^{} $|\cdot {|}_{1}$ and $|\cdot {|}_{2}$ are equivalent^{}.
Let $b$ be an element of $K$ such that $$. Because the valuations are assumed to be equivalent, it is also the case that $$. Hence, there must exist positive constants ${c}_{1}$ and ${c}_{2}$ such that ${|b|}_{1}^{{c}_{1}}=\frac{1}{2}$ and ${|b|}_{2}^{{c}_{2}}=\frac{1}{2}$.

We will show that show that ${|x|}_{1}^{{c}_{1}}={|x|}_{2}^{{c}_{2}}$ for all $a\in K$ by contradiction^{}.

Let $a$ be any element of $k$ such that $$. Assume that ${|a|}_{1}^{{c}_{1}}\ne {|a|}_{2}^{{c}_{2}}$. Then either $$ or ${|a|}_{1}^{{c}_{1}}>{|a|}_{2}^{{c}_{2}}$. We may assume that $$ without loss of generality.

Since ${|a|}_{2}^{{c}_{2}}/{|a|}_{1}^{{c}_{1}}>1$, there exists an integer $m>0$ such that ${({|a|}_{2}^{{c}_{2}}/{|a|}_{1}^{{c}_{1}})}^{m}>2$. Let $n$ be the least integer such that ${2}^{n}{|a|}_{2}^{m{c}_{2}}>1$. Then we have

$$ |

Since $2={|{b}^{-1}|}_{1}^{{c}_{1}}={|{b}^{-1}|}_{2}^{{c}_{2}}$, this implies that

$$ |

but then

$$ |

and

$${\left|\frac{{a}^{m}}{{b}^{n}}\right|}_{2}>1,$$ |

which is impossible because the two valuations are assumed to be equivalent.

Q.E.D

Title | proof of theorem on equivalent valuations |
---|---|

Canonical name | ProofOfTheoremOnEquivalentValuations |

Date of creation | 2013-03-22 14:55:40 |

Last modified on | 2013-03-22 14:55:40 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 12 |

Author | rspuzio (6075) |

Entry type | Proof |

Classification | msc 13A18 |