Example of application of Krasner's Lemma

Question

I'm learning about valued fields at the moment, and I stumbled upon these notes

http://www-personal.umich.edu/~wuyifan/ExpositoryArticles/NumberTheory/LocalFields/Krasner%27s_Lemma.pdf

As proposition 2.1 it states that when $\eta$ is a primitive $p$-th ($p$ odd) root of unit then $\mathbb{Q}_p(\eta)$ contains all $(p-1)$-st roots of $-p$.

I have to compute $v(\sqrt[{p-1}]{-p}-\eta)$ as well as $v(\sqrt[p-1]{-p}-\zeta \cdot \sqrt[p-1]{-p})$ where $\zeta$ is a $(p-1)$-st root of unity.

However I am unable to see how to do that computation.

Edit: I am trying to learn how to use Krasner's Lemma, this result is just an example I found, of something I'm currently unable to do.

Answer 1

It is easier to check first the elementary proof.

Let $K = \Bbb{Q}_p(\zeta_p)$ and $O_K$ its ring of integers with residue field $O_K/(\pi)$.

• From $$\zeta_p^p - 1 \equiv (\zeta_p-1)^p \equiv 0\bmod \pi \implies v_p(\zeta_p-1)> 0$$ $$\implies v_p(\sum_{l=0}^{a-1}\zeta_p^l) = v_p(a)=0\implies v_p(\zeta_p-1)=v_p(\zeta_p^a-1)$$ and $$\prod_{a=1}^{p-1} (1-\zeta_p^a)= \sum_{k=0}^{p-1} 1^k = p$$ we know that $$v_p(\zeta_p-1)= \frac1{p-1} \implies O_K=\Bbb{Z}_p[\zeta_p-1], \pi = \zeta_p-1$$ and $K/\Bbb{Q}_p$ is totally ramified of degree $p-1$.

Its uniformizer $\pi=\zeta_p-1$ satisfies $$v(\pi^{p-1})=1\implies \pi^{p-1} = u p, u\in \Bbb{Z}_p[\zeta_p]^\times \implies u= \zeta_{p-1}^b (1+r),v(r)> 0$$

Letting $U=(1+r)^{-1/(p-1)} =\sum_{n\ge 0} {-1/(p-1)\choose n} r^n\in \Bbb{Z}_p[\zeta_p]^\times$ we have $$(\pi U)^{p-1} = p\zeta_{p-1}^b \implies \Bbb{Z}_p[\zeta_p] = \Bbb{Z}_p[ (p\zeta_{p-1}^b)^{1/(p-1)}]$$

• To find $b$ use that $\zeta_p^a-1 = (\zeta_p-1) \sum_{l=0}^{a-1}\zeta_p^l=(\zeta_p-1)a (1+R_a),v(R_a)> 0$ thus $$(\zeta_p-1)^{p-1}=\prod_{a=1}^{p-1}\frac{\zeta_p^a-1}{a (1+R_a)} = \frac{p}{- (1+s)}, v(s) > 0 \implies \zeta_{p-1}^b = -1$$