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Let $w$ be an n-th root of unity, I have two questions

1) What are the conditions on the prime $p$ such that $w\in \mathbb{Z}_p$, and if it is the case what is the p-adic expansion of an n-th root of unity in that case (do we have a closed formula of this expansion)

2) What about the other cases i.e when $w$ does not belong to $\mathbb{Z}_p$ and belongs to a finite extension of $\mathbb{Q}_p$, do we have an expression in terms of generators of this extension.

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    $\begingroup$ I think this question is more suited to math.stackexchange.com. But briefly, you can find what you need in most books about $p$-adic numbers. For example, have a look at Gouvea's "p-adic Numbers: An introduction" book. Specifically, Proposition 3.4.2. $\endgroup$
    – Leray Jenkins
    Commented Aug 14, 2019 at 9:55
  • $\begingroup$ Sure, what about the p-adic expansion ? $\endgroup$
    – wkm
    Commented Aug 14, 2019 at 10:00
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    $\begingroup$ No, what I'm saying is that you can find the coefficients, in a explicit way, by solving the equation modulo higher powers of $p$. Have a think of how to do this (or look at the book). So, if you know how to solve equations modulo some power of $p$ then you know how to find your coefficients. $\endgroup$
    – Leray Jenkins
    Commented Aug 14, 2019 at 10:43
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    $\begingroup$ Another approach, 143983, is to post to mathstackexchange, as has been suggested to you. But be sure to include everything you already know about the question, or else it will get closed, and be sure to include your objections to Hensel, or else you'll have to go through this all over again. $\endgroup$
    – Gerry Myerson
    Commented Aug 14, 2019 at 13:17
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    $\begingroup$ I've had interest in algebraic $p$-adic expansions using $p^\mathbb{Q}$ e.g. $\sqrt{-1}=1+2^{1/2}+2^{3/4}+2^{7/8}+ ... + \zeta_{3}2 + ...$ $\endgroup$
    – David Lampert
    Commented Aug 14, 2019 at 16:00

1 Answer 1

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For $p$ odd

  • $\Bbb{Z}_p^\times = \langle\zeta_{p-1}\rangle \times (1+p)^{\Bbb{Z}_p}$ where $\zeta_{p-1} = \lim_{n \to \infty} g^{p^n}$ for $g \in \Bbb{Z}$ of order $p-1$ in $\Bbb{Z}/p\Bbb{Z}$.

  • If $K/\Bbb{Q}_p$ is a finite extension whose residue field is $O_K/(\pi) \cong \Bbb{F}_{p^f}$ then take $g\in O_K$ of order $p^f-1 $ modulo $(\pi)$ you'll have $\zeta_{p^f-1} = \lim_{n \to \infty} g^{p^{fn}}$.

  • For $p \nmid m$ let $f$ be the order of $p \bmod m$, $O_{\Bbb{Q}_p(\zeta_{p^f-1})}=O_{\Bbb{Q}_p(\zeta_m)}= \sum_{l=0}^{f-1} \zeta_m^l\Bbb{Z}_p$, it is a complete DVR with uniformizer $p$ of valuation $1$ and residue field $\Bbb{F}_{p^f}$. For the Galois actions you might prefer a normal basis for $\Bbb{F}_{p^f}/\Bbb{F}_p$.

  • $O_{\Bbb{Q}_p(\zeta_m,\zeta_{p^r})} =\sum_{m=0}^{(p-1)p^{r-1}-1}(\zeta_{p^r}-1)^m O_{\Bbb{Q}_p(\zeta_m)}$ with uniformizer $\zeta_{p^r}-1$ of valuation $(p-1)p^{r-1}$ and residue field $\Bbb{F}_{p^f}$. In particular $\Bbb{Q}_p(\zeta_{p^r})/\Bbb{Q}_p$ is totally ramified of degree $(p-1)p^{r-1}$.

Knowing a finite extension of $\Bbb{Q}_p$ means knowing its uniformizer, residue field and how the Galois group acts on both.

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  • $\begingroup$ I'm looking for the general expression of coefficients! $\endgroup$
    – wkm
    Commented Aug 14, 2019 at 15:55
  • $\begingroup$ That's the general expression. The "coefficients" of what ? ${}{}$ $\endgroup$
    – reuns
    Commented Aug 14, 2019 at 16:00
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    $\begingroup$ I gave it : it is $\zeta = g^{p^n} \bmod p^n$. Do you understand that $p$-adic series and $p$-adic limit is the same ? $\endgroup$
    – reuns
    Commented Aug 14, 2019 at 16:08
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    $\begingroup$ What is not explicit to you ? $\endgroup$
    – reuns
    Commented Aug 14, 2019 at 16:11
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    $\begingroup$ There isn’t a formula for the digits of a root of unity, any more than there’s a formula for the real digits of $\sqrt2$. But you can start with any integer $n$ prime to $p$, and go $z_0=n\mapsto n^p=z_1$ and then $z_n\mapsto z_n^p=z_{n+1}$. You get more digit with each iteration. The limit is the root of unity $\equiv n\pmod p$. $\endgroup$
    – Lubin
    Commented Aug 14, 2019 at 16:58

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