# How to show that $\mathbb{Q}_p(a)=\mathbb{Q}_p(b)$ where $a^p=1$ and $b^{p-1}=-p$?

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 $$|a-1-b|_p, then Krasners Lemma would imply $$\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?

• Since $b$ satisfies $X^{p-1}+p=0$, we get $b=(-p)^{\frac{1}{p-1}}$, which is a uniformizer, Dwork's favorite choice.
– MAS
Jul 17, 2020 at 16:28
• Strongly related: math.stackexchange.com/q/3432658/96384 Jul 17, 2020 at 17:45
• @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. Jul 18, 2020 at 15:27
• @TorstenSchoeneberg Thanks for sharing the very helpful link! Jul 19, 2020 at 10:52
• @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! Jul 19, 2020 at 10:54

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 $$\frac{\zeta_p-1}\pi$$.

• 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! Jul 19, 2020 at 10:56
• 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? Jul 20, 2020 at 3:50
• 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. Jul 20, 2020 at 4:58
• @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 $\mathbb{Z}/(p)$ ?
– MAS
Dec 9, 2020 at 14:58
• Well, @Why , aren’t the three all the same? Dec 9, 2020 at 20:18

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

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