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

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

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

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