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I would like to have a classification of absolute values on the field $\Bbb Q_p$ of $p$-adic numbers (for any prime $p$).

I guess that such an absolute value is either equivalent to the trivial one, or equivalent to the usual $p$-adic absolute value. (Recall that two absolute values $|\cdot|_1$ and $|\cdot|_2$ on a field $F$ are equivalent if and only if there is a real number $s > 0$ such that $|x|_1 = |x|_2^s$ for every $x \in F$. See Neukirch's Algebraic Number Theory, proposition 3.3, chapter II.)

I know that if $| \cdot |$ is an absolute value on $\Bbb Q_p$, it gives an absolute value on $\Bbb Q$, hence is equivalent to the trivial one or to $| \cdot |_p$ for some $p \leq \infty$ by Ostrowski's theorem. But why doesn't $| \cdot |_q$ extend to $\Bbb Q_p$ when $q \neq p$? I see that I cannot extend it directly, because of a continuity issue (here I wanted to use density of $\Bbb Q$ is $\Bbb Q_p$), but after all, an absolute value is not required to be continuous in any sense.

Alphonse
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    $|\cdot|_q$ does extend to $\Bbb Q_p$ for $q\ne p$. – Angina Seng May 27 '18 at 11:39
  • @LordSharktheUnknown : even if $q = \infty$? And is one (or the?) extension of $|.|_q$ equivalent to $| . |_p$? – Alphonse May 27 '18 at 11:42
  • Yes, no.${}{}{}$ – Angina Seng May 27 '18 at 12:22
  • @LordSharktheUnknown : thank you for your comment. Is it true that any two extensions of $|.|_q$ are equivalent absolute values of $\Bbb Q_p$? – Alphonse May 27 '18 at 12:26
  • I have seen that given a valuation on a field (or any absolute value (not necessarily non-archimedian?), one can extend it to any field extension (not necessarily algebraic!) ; 2) If $|.|_q$ is equivalent to $|.|_p$ on $\Bbb Q_p$ then $\Bbb Z_p$ has to be bounded for $|.|_q = |.|_p^s$, one can probably conclude that $q=p$ from this.
    • – Alphonse May 27 '18 at 12:29
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    Although one can extend $q$-adic valuations to $\Bbb Q_p$, they would be discontinuous there, and since one needs Zorn-type arguments to create them, there is little one can say about them. – Angina Seng May 27 '18 at 12:31
  • @LordSharktheUnknown : one more question, if you don't mind : on the other hand, is any absolute value $|.| : \Bbb Q_p \to \Bbb R$, which is continuous with respect to the $p$-adic topology, equivalent to $|.|_p$ ? – Alphonse May 27 '18 at 14:19
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    Answer to last question: yes. – Lubin May 27 '18 at 23:06
  • @Lubin : thank you very much! I see that $p^n \to 0$ implies (by continuity of $|.|$) $|p| < 1$, so WLOG $|p| = 1/p$. Then I only need to show that any unit $u$ of $\Bbb Z_p$ has absolute value $1$. Since $\Bbb Z_p$ is compact and $|.|$ is continuous, it has bounded image. But why does it follow that $|u| = 1$ ? – Alphonse May 28 '18 at 16:44
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    Well, take a unit with $|u|\ne1$, we may assume $>1$. Its powers form an unbounded set, contradicting the boundedness of the group of units. – Lubin May 28 '18 at 16:54
  • Related: https://math.stackexchange.com/q/338148/96384 – Torsten Schoeneberg May 31 '18 at 06:30