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

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.

• That $|\zeta_p|_p = 1$ is obvious from $\zeta_p^p = 1$ – reuns Dec 7 '18 at 20:15

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 $$|1+\zeta_p|_p = 1$$.

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.

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}