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Homeproof of theorem on equivalent valuations

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# 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 $0\leq x^{c}<1$ if and only if $0<x\leq 1$.

Assume that the valuations $|\cdot|_{1}$ and $|\cdot|_{2}$ are equivalent. Let $b$ be an element of $K$ such that $0<|b|_{1}<1$. Because the valuations are assumed to be equivalent, it is also the case that $0<|b|_{2}<1$. Hence, there must exist positive constants $c_{1}$ and $c_{2}$ such that $|b|_{1}^{{c_{1}}}={1\over 2}$ and $|b|_{2}^{{c_{2}}}={1\over 2}$.

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 $0<|a|_{1}<1$. Assume that $|a|_{1}^{{c_{1}}}\neq|a|_{2}^{{c_{2}}}$. Then either $|a|_{1}^{{c_{1}}}<|a|_{2}^{{c_{2}}}$ or $|a|_{1}^{{c_{1}}}>|a|_{2}^{{c_{2}}}$. We may assume that $|a|_{1}^{{c_{1}}}<|a|_{2}^{{c_{2}}}$ 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 $2^{n}|a|_{2}^{{mc_{2}}}>1$. Then we have

$2^{n}|a|_{1}^{{mc_{1}}}<2^{{n-1}}|a|_{2}^{{mc_{2}}}<1<2^{n}|a|_{2}^{{mc_{2}}}.$ |

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

$\left|{a^{m}\over b^{n}}\right|_{1}^{{c_{1}}}<1<\left|{a^{m}\over b^{n}}\right% |_{2}^{{c_{2}}},$ |

but then

$\left|{a^{m}\over b^{n}}\right|_{1}<1$ |

and

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

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

Q.E.D

## Mathematics Subject Classification

13A18*no label found*

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