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Let $K$ be a number field. I want to show that

Every Archimedean absolute value of $K$ is equivalent to the absolute value $|x|:=|\sigma(x)|_\infty$ for $x\in K$ where $\sigma$ is an embedding of $K$ into $\mathbb R$ or $\mathbb C$

I'm trying to show this using Ostrowski's theorem:

Given an Archimedean absolute value $|\cdot |$ on $K$ with $\hat{K}$ the completion of $K$ wrt $|\cdot |$ then there is an isomorphism $\sigma$ from $\hat{K}$ onto $\mathbb R$ or $\mathbb C$ satisfying $|x|=(|\sigma(x)|_\infty)^s$ for all $x\in \hat{K}$ for some fixed $0<s\leq 1$, where $|\cdot |_\infty$ is the usual absolute value on $\mathbb C$.

First of all I want to assume that $s=1$, but this is not easy to see.

What is the line of reasoning I need to take?

Thank you for your help!

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    $\begingroup$ It is good practice to include a link to your related previous question. $\endgroup$ – A.P. May 31 '15 at 22:26
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    $\begingroup$ You don't need to assume that $s = 1$, because two absolute values $|\cdot|_1$ and $|\cdot|_2$ on $K$ are equivalent if and only if there is an $s > 0$ such that $|x|_1 = |x|_2^s$ for every $x \in K$. Further, the isomorphism in Ostrowski's theorem isn't from $K$, but from the completion $\hat{K}$. $\endgroup$ – A.P. May 31 '15 at 22:35
  • $\begingroup$ Thanks I've corrected the error. $\endgroup$ – eddie May 31 '15 at 23:13
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You're almost done. There is a natural embedding $i \colon K \hookrightarrow \hat{K}$ such that $|x| = |i(x)|$ for every $x \in K$. As you say, by Ostrowski's theorem you have an isomorphism of fields $\sigma$ from $\hat{K}$ to $\Bbb{R}$ or $\Bbb{C}$ such that $$ |x| = |\sigma(x)|_{\infty}^s $$ for some fixed $0 < s \leq 1$. Furthermore, $\tau := \sigma \circ i$ is an embedding of $K$ into $\Bbb{R}$ or $\Bbb{C}$, thus for every $x \in K$ we have $$ |x| = |i(x)| = |\sigma(i(x))|_{\infty}^s = |\tau(x)|_{\infty}^s $$ This is enough to conclude the proof, because we know that two absolute values $|\cdot|_1$ and $|\cdot|_2$ on $K$ are equivalent if and only if there is a constant $s > 0$ such that $|x|_1 = |x|_2^s$ for every $x \in K$ (for example, see proposition II.3.3 of Neukirch's book).

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