# $p$-roots of unity

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 !

• 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? Nov 12, 2019 at 20:22
• Yes sorry I've forgot complete in the hypothesis. Yes, I've tried some computations with the Hensel lemma but didn't get anything :/ Nov 13, 2019 at 8:22
• @TorstenSchoeneberg If you have any solution, it would really help me. Sorry I'm a newbie in this topic Nov 13, 2019 at 16:41
• 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? Nov 14, 2019 at 1:32
• Yes I has the same problem. It comes from an article of Yves André and Lucia Di Vizio "q-difference p-adic differential equations". Nov 14, 2019 at 8:52

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.

• Wo Thank you very much ! I didn't knew about this lemma. Nov 18, 2019 at 10:11
• Krasner’s Lemma is a powerful tool. You can certainly pull this directly out of Hensel. Jul 17, 2020 at 21:34
• @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] Jul 17, 2020 at 22:12
• 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. Jul 18, 2020 at 1:57