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 $ord_p[(ζ_p-1)(ζ_p^2-1)...(ζ_p^{p-1}-1)] \geq 0$.

  • 3
    $\begingroup$ 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? $\endgroup$ – Jyrki Lahtonen Dec 7 '18 at 22:19
  • 1
    $\begingroup$ And, I'm positive we have covered this already somewhere on the site. Too late an hour for me to spend time searching though. $\endgroup$ – Jyrki Lahtonen Dec 7 '18 at 22:20
  • $\begingroup$ 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 $1/(p-1)$. $\endgroup$ – user45765 Dec 7 '18 at 22:32
  • $\begingroup$ @JyrkiLahtonen Wow, that's almost too good to be true! I think that answers my question. $\endgroup$ – 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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.