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The definition of $p$-adic norm in most textbooks and here is not easy for me to understand and especially to implement in practice, but here it is the way I reworded it:

The norm of a $p$-adic number equals $\frac{1}{p^{n-1}}$ with $n$ being the index (the position) of the rightmost non-zero digit we can find, if the number is represented in base $p$

Is this definition correct?

Michael Hardy
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Yuriy S
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  • Are you talking about $\Bbb Z$ or $\Bbb Q$? – Gregory Grant Jul 26 '16 at 19:21
  • @GregoryGrant, I am talking of $\mathbb{Q}$ and the extensions as well. Why, is it different? – Yuriy S Jul 26 '16 at 19:23
  • If you're talking about extensions of $\Bbb Q$ then how do you write that in base $p$? I think it's easier to think about factoring a number $n$ into its prime factorization and if $n$ is the power of $p$ then $|n|=1/p^n$. Not sure how you extend that to extensions of $\Bbb Q$, you probably need to factor $p$ and base the norm on a prime factor of $p$ and not $p$ itself. – Gregory Grant Jul 26 '16 at 19:25
  • Here's a post that might help: http://mathoverflow.net/questions/30581/extension-of-valuation – Gregory Grant Jul 26 '16 at 19:27
  • @GregoryGrant, $p$ is always prime, what do you mean? Thanks for the link though – Yuriy S Jul 26 '16 at 19:29
  • $p$ is not always prime, in an extension of $\Bbb Q$ it might split. For example $2=(1+i)(1-i)$ in $\Bbb Q[i]$. – Gregory Grant Jul 26 '16 at 19:34
  • You said you are talking about $\Bbb Q$ "and the extensions as well". Well, $\Bbb Q[i]$ is an extension of $\Bbb Q$. – Gregory Grant Jul 26 '16 at 19:39
  • Furthermore if you just meant extensions of $\Bbb Q$ that are contained in $\Bbb R$, you still have the same problem, $p$ is not going to remain prime in all extensions even if they are real. – Gregory Grant Jul 26 '16 at 19:40
  • @GregoryGrant, more correctly, I mean the completion of $\mathbb{Q}$ with respect to $p-$adic norm, in the words used in mathworld. I.e. any possible $p-$adic numbers. I apologise for any confusion – Yuriy S Jul 26 '16 at 19:45
  • Yes I'm familiar with $p$-adics, very familiar as I did my PhD in number theory. I think you're confused about what an extension of $\Bbb Q$ is. An extension is obtained by adding roots of polynomials, for example $\sqrt{2}$. You can then extend the $p$-adic norm but you need to factor $p$ in the extension. – Gregory Grant Jul 26 '16 at 19:46
  • @GregoryGrant, I apologise again, I'm very sorry if I disrespected you in any way. In fact, I mean simply all possible $p-$adics, which can be explicitly written in base $p$ (or approximately, in the case of $p-$adic irrationals). I still don't understand why is $p-$adic norm not supposed to work like I described. – Yuriy S Jul 26 '16 at 19:48
  • Ah you mean the $p$-adic completion. How do you know for sure that $\Bbb Q[i]$ is not in the $p$-adic completion of $\Bbb Q$? Just because it's not in the usual completion doesn't imply it's not in the $p$-adic completion. – Gregory Grant Jul 26 '16 at 19:49
  • I'll bow out as I don't have an explicit answer to your question regarding extensions. But as far as $\Bbb Q$ goes I think your definition is equivalent. – Gregory Grant Jul 26 '16 at 19:51
  • @GregoryGrant, as far as I know, $p-$adics are completely different mathematical objects than the reals (or complex numbers). So even if we have $\sqrt{-1}$ for some $p$ (and we do), it's still not $i$. – Yuriy S Jul 26 '16 at 19:51
  • The $p$-adics are completely different topologically, but they're not different algebraically. In fact the $p$-adic completion of $\Bbb Q$ can be embedded (as a field) in $\Bbb C$ and always if there's an element whose square is $-1$ it maps under this embedding to $i$ (or $-i$). So $i$ is universal, just as is $\sqrt{2}$. The algebraic objects are identical. It's just a different topology that's all. – Gregory Grant Jul 26 '16 at 19:55
  • You might want to read thrrough this page: http://math.stackexchange.com/questions/338148/is-there-an-explicit-embedding-from-the-various-fields-of-p-adic-numbers-mathb – Gregory Grant Jul 26 '16 at 19:58

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Let me try to cut through the fog of the comments to your question.

Your conjecture is true, but only for natural integers (elements of $\Bbb Z$) if you’re thinking of the standard $p$-ary expansion of real numbers. Let’s take $5$, as a not-atypical prime, for our $p$.

Look at the $5$-ary representation of $20$ and $5/4$ here. Twenty is $40_5$, and your method works. But for the rational number $5/4$, the standard $5$-ary expansion is $1.11111\cdots_5$, and you can check this with the formula for geometric series, since you’ve written $1+\frac15+\frac1{25}+\cdots$, $a=1$, $r=\frac15$, so $a/(1-r)=1/(1-\frac15)=5/4$. For $5/4$, then, your method doesn’t work.

On the other hand, if you write $5$-adic numbers as $5$-ary expansions extending (potentially) infinitely to the left, then $\frac54$ has the expnsion $\cdots333340_5$, and your method works fine.

(You check that my $5$-adic expansion is correct by noticing that it represents $20+3(25+125+625+\cdots)$, and the infinite part evaluates as $\frac{75}{1-5}=-18\frac34$, just right.)

A couple of remarks more on your question: in the expansion of twenty, I would have called the rightmost digit the zero-th position, and the $4$ to appear in the first position. Then your formula would involve $n$ instead of $n-1$. And for algebraic extensions of $\Bbb Q_p$, like $\Bbb Q_5(\sqrt2\,)$, you will always need as many $p$-adic coordinates as the degree of your field over the base, just as we need two real numbers to describe a complex number.

Lubin
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  • Lubin, thank you, but of course I meant the correct p-adic expansion of a number, not it's p-ary form. I apologise for any confusion. And thank you for clarification about extensions, now I feel really bad for arguing with @GregoryGrant. But I meant the completion, not extensions – Yuriy S Jul 27 '16 at 13:28
  • The point is: I know how to write any rational number in its $p-$adic form, and I know how to write a "random" p-adic number (i.e. $2-$adic $\dots 11100100010100110110$), but I wouldn't know how to use the usual definition of the norm. My definition on the other hand, should work for any number if I know the first non-zero digit of its $p-$adic expansion from the right. – Yuriy S Jul 27 '16 at 13:32
  • And you're perfectly correct in your conjecture. The point is that factoring out the maximum power of $p$ will lead you to a $p$-adic unit, whose absolute value is $1$. Since $\mid p \mid=1/p$, you're good to go. – Lubin Jul 27 '16 at 23:30