Proving $ord_p(ζ_p-1)=1/(p-1)$

After proving this, I was able to deduce an even more general result that $$ord_p(ζ_p-1)=ord_p(ζ_p^2-1)=...=ord_p(ζ_p^{p-1}-1)$$. Now, according to Lubin, $$ord_p(ζ_p-1)$$ should be $$1/(p-1)$$, but this isn't obvious to me so I need to prove it.

My thought was to multiply out the product $$(ζ_p-1)(ζ_p^2-1)...(ζ_p^{p-1}-1)$$ and hope that I get something with p-order $$1$$ (since this would finish the proof), but I'm stuck.

I know that $$ord_p(ζ_p^k)=0$$ for any $$k \geq 0$$. So if we multiply out the product, we get a sum with all of its terms having p-order $$0$$. I see from this that $$ord_p[(ζ_p-1)(ζ_p^2-1)...(ζ_p^{p-1}-1)] \geq 0$$.

• You have the cyclotomic polynomial $$\Phi_p(x)=x^{p-1}+x^{p-2}+\cdots+x+1=\prod_{k=1}^{p-1}(x-\zeta_p^k).$$ What do you get when plug in $x=1$ into the above equation? – Jyrki Lahtonen Dec 7 '18 at 22:19
• And, I'm positive we have covered this already somewhere on the site. Too late an hour for me to spend time searching though. – Jyrki Lahtonen Dec 7 '18 at 22:20
• This is nothing but says $p$ is totally ramified at $(1-\eta_{p^r})$ with ramification index $\phi(p^r)$. And it should $p-1$ instead of $1/(p-1)$. – user45765 Dec 7 '18 at 22:32
• @JyrkiLahtonen Wow, that's almost too good to be true! I think that answers my question. – Pascal's Wager Dec 7 '18 at 22:52

$$|a|_p = p^{-v_p(a)}$$

Since $$\zeta_p^p = 1$$ then $$|\zeta_p|_p=1$$

$$\sum_{m=0}^{p-1} x^m =\frac{x^p-1}{x-1}=\prod_{k=1}^{p-1} (x-\zeta_p^k)$$

$$\prod_{k=1}^{p-1} |1-\zeta_p^k| = |\sum_{m=0}^{p-1} 1^m|_p = p^{-1}$$ so there is some $$k$$ such that $$|1-\zeta_p^k|_p <1$$.

$$1-\zeta_p^{kn} =1- (1+ (1-\zeta_p^k))^n =1-(1+n(1-\zeta_p^k)+O((1-\zeta_p^k)^2)$$

whence (for $$p \nmid n$$)

$$|1-\zeta_p^{kn}|_p = |n|_p |1-\zeta_p^{k}|_p=|1-\zeta_p^{k}|_p$$ and $$|1-\zeta_p^{k}|_p^{p-1} = p^{-1}$$

letting $$kn \equiv 1\bmod p$$ you get your result $$|1-\zeta_p|_p =p^{-1/(p-1)}$$

Note this is also a proof that $$\sum_{m=0}^{p-1} x^m$$ is irreducible since we need to multiply all the $$1-\zeta_p^k, k \in 1 \ldots p-1$$ to obtain something of integer valuation