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 $|a-1-b|_p<p^{-1/(p-1)}$, 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?
 A: 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 $\frac{\zeta_p-1}\pi$.
A: 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}}. $
