Showing the lemma $\operatorname{ord}_p(1+ζ_p)=0$ if $p>2$

Question

Just to give some background regarding my motivation, I'm trying to prove a lemma to help me solve How do we prove p-order of $g_k$ is $\frac {k} {p-1}$?

Let $Z_p$ denote the p-adic integers, and let us adjoin a primitive p-th root of unity $ζ_p$. Assume $p>2\DeclareMathOperator{\ord}{ord}$.

I want to show that $\ord_p(1-ζ_p)=\ord_p(1-ζ_p^2)$, but I see that it is enough to prove $\ord_p( \frac {1-ζ_p^2} {1-ζ_p})=0$ or, equivalently, $\ord_p(1+ζ_p)=0$.

I have tried applying the properties from here. (Even though they are for rational numbers, I'm going to assume for now that they hold in $Z_p(ζ_p)$ also. If not, please correct me.)

I see that, if $\ord_p(ζ_p) \neq \ord_p(1)=0$, we will have $\ord_p(1+ζ_p)=inf \{\ord_p(1), \ord_p(ζ_p)\} \leq 0$ and we will get a contradiction if $<0$ .

But if instead $\ord_p(ζ_p)=0,$ then I don't know what to do. We will have $\ord_p(1+ζ_p) \geq \inf \{\ord_p(1), \ord_p(ζ_p)\}=0$, but I don't see how $>0$ would give us a contradiction.

Answer 1

Note that $\frac {ζ^2-1} {ζ-1}=1+ζ$ and $\frac {ζ-1} {ζ^2-1} =1+ζ^2+ζ^4+...+ζ^{p-1}$.

In the case that $ord_p(ζ)=0$, we will have $ord_p(ζ^2)=ord_p(ζ^3)=...=0$.

This would imply $ord_p(ζ^2-1)-οrd_p(ζ-1)=ord_p(1+ζ) \geq inf \{ord_p(1),ord_p(ζ)\}=0 $ and also $οrd_p(ζ-1)-ord_p(ζ^2-1)=ord_p(1+ζ^2+ζ^4+...+ζ^{p-1})\geq inf\{ord_p(1),...,ord_p(ζ^{p-1})\}=0$.

From these two inequalities, it should be easy to see $ord_p(ζ^2-1)=οrd_p(ζ-1)$ as needed.

Answer 2

Let $$(x-1)^p-1 = f(x)g(x) \in \mathbf{Z}_p[x]$$ where $f(x)$ is the minimal polynomial of $1+\zeta_p$ so $$f(x) = \prod_{\alpha \in Gal(\overline{\mathbf{Q}_p}/\mathbf{Q}_p)\cdot (1+\zeta_p)} (x-\alpha)$$

by definition $|.|_p$ is $Gal(\overline{\mathbf{Q}_p}/\mathbf{Q}_p)$ invariant and$$|1+\zeta_p|_p=|f(0)|_p^{1/\deg(f)}$$

and since $g(0) \in \mathbf{Z}_p$ $$1 \ge |f(0)|_p \ge |f(0)g(0)|_p = |(-1)^p-1|_p = 1$$

whence $|f(0)|_p= 1$ and $|1+\zeta_p|_p = 1$.

Answer 3

Here’s yet another argument:
Start with the minimal polynomial for $\zeta_p$ , \begin{align} \text{Irr}(\zeta_p,\Bbb Q_p)&=\frac{X^p-1}{X-1}\\ f=\text{Irr}(\zeta_p-1,\Bbb Q_p)&=\frac{(X+1)^p-1}X&\text{($p$-Eisenstein, root $\pi$)}\\ v_p(\pi)=v_p(\zeta_p-1)&=\frac1{p-1}\\ v_p(\pi+2)=v_p(\zeta_p+1)&=v_p(2)=0&\text{(’cause $p\ne2$)}\,. \end{align}