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Process of completion of $\mathbb Q$ using the absolute value $|x|$ does not touch to the non-real complex numbers which are added to $\mathbb Q$ via extensions fields. However completion of $\mathbb Q$ using another kind of absolute value can aggregate complex numbers to $\mathbb Q$. (The process is formally the same: the ring $R$ of Cauchy sequences of $\mathbb Q$, in which addition and multiplication are defined termwise as usually, has the maximal ideal $M=\{(r_n)_{n\in\mathbb N}\in R: r_n \to 0\}$ so $R/M$ is a field, which gives a definition of $\mathbb R$ when$|x|=\max\{x,-x\}$, the Euclidean absolute value).

According to the sometimes called Ostrowski’s Theorem, all the (not trivial) absolute values on $\mathbb Q$ (except equivalent ones) are only the p-adic absolute values (if the non-zero rational $x$ is written as $x= p^n \frac{a}{b}$,with $a,b,p$ pairwise coprime and $n$ rational integer, its definition is $|x|_p = p^{-n} $ and $|0|_p=0$).

The question is about the following known result: “All the $(p-1)$-th roots of $1$ are elements of $\mathbb Q_p$” (it can be proved using residue field of $\mathbb Q_p$ as isomorphic to $\mathbb F_p$ and applying Hensel’s lemma to the polynomial $x^{p-1}-1$ of $\mathbb F_p[x]$)

As much as I try I fail to give me an entirely satisfactory understanding about why the completions of $\mathbb Q$ via $p$-adic absolute values contain a lot of non-real complex and the ordinary completion of $\mathbb Q$ does not contain any of them. I know there are quite "pathology" in $\mathbb Q_p$ but topological and metric, not of “inclusion of elements”. I do not even know by now how to write the $p$-adic expansion of $e^\frac {2\pi i k}{n}$ for adequate $k$ and $n$. Some help?

Piquito
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    It's not "pathology" in $\mathbb Q_p$, but simply "different from $\mathbb R$-ness". As in the answer(s), the problem is "judging" p-adic completions in terms of real and complex numbers... while, in reality, there is no genuine comparison. – paul garrett Jun 26 '15 at 22:59
  • @paul garret: I think it is usual in mathematics the word "pathology" for properties bizarre enough; this is not pejorative (all point of an open ball is its center, is a p-adic example of "pathology"....) – Piquito Jun 27 '15 at 00:36
  • Undoubtedly you know this, but adding this just to make sure (reiterating an old comment by Lubin). The roots of unity of order a factor of $p-1$ are gotten as follows. Let $a\in{1,2,\ldots,p-1}$. Then consider the limit $$\xi_a=\lim_{n\to\infty}a^{p^n}.$$ It is an easy exercise to show that this sequence is Cauchy, and hence converges to a limit $\xi_a\equiv a\pmod p$. After that we trivially have $\xi_a^p=\xi_a$, so... – Jyrki Lahtonen Jun 27 '15 at 08:29
  • @JyrkiLahtonen: Thank you very much.There are in fact, as you say, more complex than the (p-1)-th roots but I gave only these ones . I wanted to give only the imaginary unit but was unsure of their inclusion (for example $i$ is not in $\mathbb Q_3$ where $\sqrt-3$ is an element. – Piquito Jun 27 '15 at 12:26
  • This is an old post and I hope OP has become more careful than they were back then in this question and comments, but for the record, I argue against the conflation of $p$-adic with real or complex numbers in https://math.stackexchange.com/a/4007515/96384. Basically, Lubin says the same in his answer here, and so did Harry Altman in https://math.stackexchange.com/a/174827/96384 9 years ago. But it cannot be said often enough. – Torsten Schoeneberg Aug 12 '21 at 16:42
  • @Torsten Schoeneberg: Thanks a lot. It is common in mathematics of a certain level that the sources do not say enough about what they are about, that they are not very explicit. This happens a lot, for example, in treatises on elliptic curves and algebraic geometry. – Piquito Aug 13 '21 at 17:19

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When you’re speaking of the $p$-adic domain, the concepts of “real” and “complex” don’t make sense. Those concepts relate only to the archimedean metric on $\Bbb Q$.

You’re correctly observing that every root of the $n$-th cyclotomic polynomial is complex, but what about $X^6-5$? Irreducible, all the roots are equivalent in a Galois-theoretic viewpoint, and if you look at the abstract field $\Bbb Q(\xi)/(\xi^6-5)$, is $\xi$ real or complex? The right way to look at this is that the abstract field has two inequivalent real absolute values, and four complex, though the latter fall into equivalent pairs. If you know algebraic number theory, $r_1=r_2=2$ here.

As to your question about $e^{i\pi/n}$, this is doubly meaningless in the $p$-adic domain. First, there’s no $\pi$ there; and second, although there is an exponential function defined on a relatively small open neighborhood of $0$, there’s no reason to think that your “$i\pi/n$” would lie in it.

I’m sorry for this diatribe. Lang told us, when I was in college, that the $p$-adic domains were just as good as the real and complex domains. Here’s something that is a wonderful tool $p$-adically, but has no equivalent in the complex domain. It’s the $p$-adic logarithm, which is defined for all $\xi$ with $|\xi-1|<1$. This set, a pretty large neighborhood of the multiplicative identity $1$, is already a group (!!) in the $p$-adic domain, even when you make algebraic extensions to $\Bbb Q_p$. And the logarithm, defined by the very same power series that you learned in Calculus, is a homomorphism (!!) defined on that group. This is something that’s not true in the complex domain: the logarithm, in whichever way you define it on an open subset of $\Bbb C$, is not a homomorphism, because it can’t be defined on an open multiplicative subgroup of $\Bbb C$.

Lubin
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  • Not entirely correct. If you closed $\Bbb Q_p$, you get a field isomorphic to $\Bbb C$; if you also take the metric completion, you get a topological field homeomorphic to $\Bbb C$. So the complex numbers do lurk in the background. – Asaf Karagila Jun 26 '15 at 22:57
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    @AsafKaragila, it may mislead the questioner to say "isomorphic to $\mathbb C$"... and to say "metric completion" ... which requires similar understanding of "what metric?". Although non-canonical (algebraic) isomorphisms among these things have been used in important work (Grothendieck-Deligne-etal), it is sort-of "over-clocking" the ideas, I think. – paul garrett Jun 26 '15 at 23:03
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    Dear @Asaf, To what metric completion are you referring? The absolute value on $\mathbf{Q}_p$ extends to an algebraic closure uniquely, and you can complete the algebraic closure for this absolute value, but the resulting topological field is totally disconnected. – Keenan Kidwell Jun 26 '15 at 23:29
  • @Lubin: Thanks for your answer. $e^\frac {2\pi ik}{n}$ is for adequate values of $k$ and $n$ a $(p-1)$nt root of $1$ so is an element of $\mathbb Q_p$ as well as it is so for rational. Where is the "doubly meaningless"? (on the other hand, we are obviously speaking in language of isomorphisms: who can manipulated $\sqrt 2$ as an equivalence class of Cauchy sequences?) – Piquito Jun 27 '15 at 00:28
  • @paulgarret: "Grothendieck-Deligne", perhaps you knew that the first hated his disciple Deligne? – Piquito Jun 27 '15 at 03:23
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    @LuisGomezSanchez, I don't think there is good evidence for any such thing. Sure, "it makes a better story" that way, but by that standard arguably almost all academic mathematicians "hate" all others, because they would have like to prove all the theorems themselves, etc. – paul garrett Jun 27 '15 at 14:34
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    @LuisGomezSanchez, sorry for the delay, but I was away from wifi. If $\Phi_n$ is the $n$-th cyclotomic polynomial, whose roots are the primitive $p$-th roots of unity, and if $\xi$ is a root of $\Phi_n$, then $\xi$ may be sent into $\Bbb C$ to an $\exp(k\pi i/n)$, but which $k$? Similarly, $\xi$ may be sent into $\Bbb Q_p$ if $n|(p-1)$, but in which of the possible $\phi(n)$ ways? There is simply no way to identify which element of a $p$-adic field $\exp(i\pi/n)$ corresponds to. – Lubin Jun 29 '15 at 09:34
  • @Lubin: It may be that I did not express well because of my deficient English. But I think it should be understood that by saying $ p $ -adic expansion, $ p $ was fixed and then "adequate" $ k $ and $ n $ have a good sense. Greetings. – Piquito Jun 30 '15 at 15:19