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Let $p$ be a prime, $a$ a primitive $p$-th root of unity in $\overline{\mathbb{Q}_p}$ and $b$ a root of $X^{p-1}+p$ in $\overline{\mathbb{Q}_p}$. How can I show that $\mathbb{Q}_p(a)=\mathbb{Q}_p(b)$?

I have a feeling that Krasners Lemma might be helpful, because the distance of $a$ to any of its conjugates is $p^{-1/(p-1)}$ and the same holds also for $b$ (and also for $a-1$). Hence if one could show that $|a-1-b|_p<p^{-1/(p-1)}$, then Krasners Lemma would imply $\mathbb{Q}_p(a)=\mathbb{Q}_p(b)$. However, I have no idea how to tackle the computation of $|a-1-b|_p$. Is this the right path? If yes, how can one compute $|a-1-b|_p$? If not, how to tackle the problem?

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  • $\begingroup$ Since $b$ satisfies $X^{p-1}+p=0$, we get $b=(-p)^{\frac{1}{p-1}}$, which is a uniformizer, Dwork's favorite choice. $\endgroup$
    – MAS
    Jul 17, 2020 at 16:28
  • $\begingroup$ Strongly related: math.stackexchange.com/q/3432658/96384 $\endgroup$ Jul 17, 2020 at 17:45
  • $\begingroup$ @RedundantAunt - I suppose you looked at 'my' K lemma argument on the page to which Torsten linked above? Of course, Lubin's K-free here answer is nice - I never would have thought of it. $\endgroup$
    – peter a g
    Jul 18, 2020 at 15:27
  • $\begingroup$ @TorstenSchoeneberg Thanks for sharing the very helpful link! $\endgroup$ Jul 19, 2020 at 10:52
  • $\begingroup$ @peterag Yes I saw your solution, and I have to say I really like it! Of course Lubin's answer might be considered more elegant as it uses less heavy machineary, but I really liked your trick that one of the factor has to have a norm strictly less than $p^{-1/(p-1)}$; really ingenious as well! $\endgroup$ Jul 19, 2020 at 10:54

2 Answers 2

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A peanut as simple as this should not require a pile-driver like Krasner to crack it open. Hensel should be plenty strong enough.

I’ll show that a primitive $p$-th root of unity $\zeta_p$ can be found in $\Bbb Q_p(\pi)$, where $\pi=\sqrt[p-1]{-p}$. Since this field has the same degree over $\Bbb Q_p$ as $\Bbb Q_p(\zeta_p)$, that will suffice.

As you know, or can calculate, the minimal $\Bbb Q_p$-polynomial for $\zeta_p-1$ is $G(X)=X^{p-1}+pX^{p-2}+\frac{p(p-1)}2X^{p-3}+\cdots\frac{p(p-1)}2X+p$. Thus a polynomial with $\frac{\zeta_p-1}\pi$ for a root is $$ \frac{G(\pi X)}{\pi^{p-1}}=X^{p-1}+\frac p\pi X^{p-2}+\cdots\frac{p(p-1)}{2\pi^{p-2}}X-1\equiv X^{p-1}-1\pmod \pi\,. $$ Since $X^{p-1}-1$ factors into linears over $\Bbb Z/(p)$, Hensel says that $G(\pi X)/\pi^{p-1}$ factors into linears over $\Bbb Z_p[\pi]$, and this ring therefore contains $\frac{\zeta_p-1}\pi$.

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  • $\begingroup$ Thanks for the very insightful response! What confused me about this problem is that at first I couldn't really see an algebraic connection between a $p$-th root of unity and a $p-1$-th root of $-p$, but your clever use of Hensel shows the connection. Nicely done! $\endgroup$ Jul 19, 2020 at 10:56
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    $\begingroup$ Something about this beautiful answer made me realise though what you're proving here is also that the "first layer" of the Lubin-Tate extension of $\mathbb Q_p$ w.r.t. the formal group law corresponding to $(1+X)^p-1$ is the same as the one coming from $pX+X^p$, which of course follows from the standard uniqueness/independence results in Lubin-Tate theory. Now if Krasner is a "pile-driver" for this "peanut", how would you call such an invocation of Lubin-Tate, Professor Lubin? $\endgroup$ Jul 20, 2020 at 3:50
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    $\begingroup$ Indeed, @TorstenSchoeneberg , that is how I always understood the equality pf $\Bbb Q_p(\zeta_p)$ and $\Bbb Q_p((-p)^{1/(p-1)})$, but that would not have been appropriate to OP’s level of achievement at this point, a far too advanced argument. Whether that has the same power as Krasner, I don’t know, especially since I’ve never fully absorbed Krasner. $\endgroup$
    – Lubin
    Jul 20, 2020 at 4:58
  • $\begingroup$ @Lubin, Can please check the 2nd line from the bottom? You said $ \text{$X^{p-1}-1$ factors into linears over $\mathbb{Z}/(p)$}$. Isn't it $\mathbb{Z}_p/(p)$ or $\mathbb{Z}_p/(\pi)$ instead of $\mathbb{Z}/(p)$ ? $\endgroup$
    – MAS
    Dec 9, 2020 at 14:58
  • $\begingroup$ Well, @Why , aren’t the three all the same? $\endgroup$
    – Lubin
    Dec 9, 2020 at 20:18
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Since $a$ is primitive $p^{th}$ root of unity, the $p$-adic valuation of $(a-1)$ is $1/(p-1)$ i.e., $|a-1|_p=p^{-1/(p-1)}.$

Also, as $b$ satisfies the equation $x^{p-1}+p=0$, we have $$b=(-p)^{1/(p-1)}.$$ So that we have $|b|_p=p^{-\frac{1}{p-1}}.$

Thus, from your equality $|a-1-b|_p \leq \max \{|a-1|_p, \ |b|_p \} =p^{-\frac{1}{p-1}}. $

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  • $\begingroup$ Isn’t $|a-1|_p=p^{-1/(p-1)}$? $\endgroup$ Jul 17, 2020 at 17:19
  • $\begingroup$ @RedundantAunt, yes my mistake $\endgroup$
    – MAS
    Jul 17, 2020 at 17:25
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    $\begingroup$ Ok but I don’t see how this helps; can you elaborate? For Krasners Lemma one would need a strict inequality. $\endgroup$ Jul 17, 2020 at 18:15

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