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Let $p$ a prime, and the p-adic norm $|x|_p = (\frac{1}{p})^{v_p(x)}$, with $v_p$ the p-adic valuation. Show that $$\{ |x|_p : x \in \mathbb{Q}_p \} = \{ p^k : k \in \mathbb{Z}\}$$ My question is how to show $\supseteq$? $\subseteq$ is easy!

Thanks!

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Hint:

$$\forall\;k\in\Bbb Z\;,\;\;p^k=\left|\;\frac{1}{p^k}\;\right|_p\ldots$$

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  • $\begingroup$ Thanks, this way is easier! $\endgroup$ – P. M. O. Mar 24 '13 at 21:32
  • $\begingroup$ Yup, I think so...it's just a matter of not getting confused with the fractions, positive/negative exponents and stuff. $\endgroup$ – DonAntonio Mar 24 '13 at 21:35
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Hint: think about powers of $p$! $\mathbb{Q}\subseteq\mathbb{Q}_p$, and you should know that $|\mathbb{Q}|_p$ is your desired set.

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You want to show that given $k$ there is a $x$ such that $\lvert x\rvert_p = p^k$. How about considering $x = p^{n}$. What would $n$ have to be?

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