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Let $K$ be a field of characteristic zero complete with respect to a non archimedian absolute value with a residue field of characteristic $p>0$. I would like to show that if $K$ contains the $p$-roots of unity then it also contain $p-1$ distinct non zero roots of the equation $X^p=-pX$. Thanks for the help !

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  • $\begingroup$ What have you tried? Is it already clear to you that you're actually looking at $X^{p-1}-p$, and that you just have to show that $K$ contains one root $\alpha$ of that (because the others will be multiples of $\alpha$ with $p-1$-th roots of unity which exist in $\mathbb Q_p \subset K$)? Then for the existence of one such root, have you tried using the general version of Hensel's lemma and calculations like in the answers to math.stackexchange.com/q/2977896/96384? $\endgroup$ Nov 12, 2019 at 20:22
  • $\begingroup$ Yes sorry I've forgot complete in the hypothesis. Yes, I've tried some computations with the Hensel lemma but didn't get anything :/ $\endgroup$
    – Pierre21
    Nov 13, 2019 at 8:22
  • $\begingroup$ @TorstenSchoeneberg If you have any solution, it would really help me. Sorry I'm a newbie in this topic $\endgroup$
    – Pierre21
    Nov 13, 2019 at 16:41
  • $\begingroup$ I've tried to write up an answer but got stuck. I thought one can use $\zeta_p-1$ as an approximate root of the polynomial, which via Hensel would lift to an actual root, but I get the condition for Hensel working only for $p \le 3$ (where $p=2$ is trivial anyway, and for $p=3$ one already has $\sqrt{-3} \in \mathbb Q(\zeta_3)$ by basic algebra). I almost have doubts whether it's true for $p \ge 5$. Where does the question come from? $\endgroup$ Nov 14, 2019 at 1:32
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    $\begingroup$ Yes I has the same problem. It comes from an article of Yves André and Lucia Di Vizio "q-difference p-adic differential equations". $\endgroup$
    – Pierre21
    Nov 14, 2019 at 8:52

1 Answer 1

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This follows from Krasner's lemma.

The 'right-hand (larger) side' of Krasner's lemma:

If $\alpha$ of $\beta$ are distinct roots of $X^{p-1}=-p$ then $\beta = \zeta_{p-1}\alpha$, where $\zeta_{p-1}$ is a $(p-1)$th root of unity (which, of course, belongs to ${\mathbb Q}_p$), and not equal to one. Therefore $$|\alpha-\beta| = |\alpha|\cdot|1-\zeta_{p-1}|.$$ Now, the second norm on the right is equal to one [$ x^{p-1} -1 $ has distinct non-zero roots $\pmod p$], therefore $$|\alpha-\beta| = |p|^{1/(p-1)}.$$

The 'left-hand (smaller) side' of Krasner's lemma:

Now, set $\pi = \zeta_p -1$, with $\zeta_p$ a primitive $p$th root unity, and take the (shifted cyclotomic) polynomial $$f(x) = {(x+1)^p-1 \over x}. $$ Then $f(x) \equiv x^{p-1} \pmod p$, $f(\pi)=0$, and $f(0 )=p$.

Therefore $-\pi^{p-1} = p\pi (\cdots) + p.$ Hence $$\alpha^{p-1} - \pi^{p-1} = -p + p + p\pi (\cdots),\tag{*}$$ and $$|\alpha^{p-1} - \pi^{p-1}|\le |p\pi|.$$ Now the left of $({}^*)$ can be completely factored, with factors of the form $\alpha -\zeta_{p-1}\pi$. At least one of the factors has norm (strictly) less than $ |p|^{1\over (p-1)}$.

Hence Krasner's lemma applies.

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  • $\begingroup$ Wo Thank you very much ! I didn't knew about this lemma. $\endgroup$
    – Pierre21
    Nov 18, 2019 at 10:11
  • $\begingroup$ Krasner’s Lemma is a powerful tool. You can certainly pull this directly out of Hensel. $\endgroup$
    – Lubin
    Jul 17, 2020 at 21:34
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    $\begingroup$ @Lubin Thanks! Your answer (math.stackexchange.com/questions/3760336/… ) is certainly more direct. And I had been so pleased to have thought of and used K's lemma 'in real life'... But, to paraphrase your answer, I must conclude that no pile-driver brain have I, alas; rather, one of the simpler, peanut variety - not that I needed a reminder... Sniff, sob - and +1 for your answer. Ah well - dem da breaks... [This comment is only for your entertainment] $\endgroup$
    – peter a g
    Jul 17, 2020 at 22:12
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    $\begingroup$ Please don’t feel bad. I’ve been at this racket for a hell of a long time, and I have a crazy-strong prejudice for Hensel. $\endgroup$
    – Lubin
    Jul 18, 2020 at 1:57

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