What is the $p$-adic valuation of $\zeta_p-1$?

Question

Let $\zeta_p$ be the $p^{th}$ root of unity in $p$-adic field.

I know that $v(\zeta_p-1)=\frac{1}{p-1}$ but I couldn't prove it.

I have tried in the following way:

$f(x)=(x-\zeta_p)(x-\zeta_p^2) \cdots (x-\zeta_p^{p-1})=\sum_{j=0}^{p-1}x^j=1+x+\cdots+x^{p-2}+x^{p-1}=\prod_{i=1}^{p-1}(x-\zeta_p^{i}).$

Now putting $x=1$, we get $$f(1)=(1-\zeta_p)(1-\zeta_p^2) \cdots (1-\zeta_p^{p-1}),$$ and also $f(1)=p$. Thus $$1-\zeta_p=\frac{f(1)}{(1-\zeta_p^2) \cdots (1-\zeta_p^{p-1})}=\frac{p}{(1-\zeta_p^2) \cdots (1-\zeta_p^{p-1})}.$$ So $v(1-\zeta_p)=v(p)-v[(1-\zeta_p^2) \cdots (1-\zeta_p^{p-1})]=1-v(1-\zeta_p^2)- \cdots-v(1-\zeta_p^{p-1})$

How to finish the proof ?

Help me in the above proof ?

Answer 1

For $p\nmid n$, $v(\zeta_p-1)=v(\zeta_p^n-1)$ because $$\frac{\zeta_p^n-1}{\zeta_p-1}= \sum_{m=0}^{n-1} \zeta_p^m \equiv \sum_{m=0}^{n-1} 1 \bmod\mathfrak{m}$$ Next $$\Phi_p(x)=\prod_{n=1}^{p-1}(x-\zeta_p^n)=\sum_{m=0}^{p-1} x^m, \qquad \Phi_p(x+1) = \prod_{n=1}^{p-1}(x-(\zeta_p^n-1))$$ $$v(p) = v(\Phi_p(1)) = \sum_{n=1}^{p-1} v(\zeta_p^n-1)=(p-1) v(\zeta_p-1)$$

This implies the irreducibility of $\Phi_p(x)\in \Bbb{Q}_p[x]$ and that $\Bbb{Q}_p(\zeta_p)/\Bbb{Q}_p$ is totally ramified.

A similar idea applies to any Eisenstein polynomial, an element and its $\Bbb{Q}_p$-conjugates have the same valuation.