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Let $n\geq2$ be an integer, $\zeta_n=\exp(\tfrac{2\pi i}n)$ and $N(x)$ be the norm of $x\in\Bbb Z[\zeta_n]$.

Let $1\leq k\leq n-1$. Is it true that $$N(\zeta_n^k-1)=\left(\Phi_{\tfrac n{\gcd(n,k)}}(1)\right)^{\frac{\phi(n)}{\phi(\gcd(n,k))}}$$ where $\Phi_m(x)$ is the cyclotomic polynomial of $\zeta_m$ and $\phi(m)$ is the Euler's totient function. If not, what is the correct formula?

Thanks in advance.

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    $\begingroup$ I think the formula fails for $(n,k)=(16,4)$, and that you should replace $\phi(gcd(n,k))$ with $\phi(n)/\phi(n/gcd(n,k))$. $\endgroup$
    – Aphelli
    Commented Aug 11 at 12:24
  • $\begingroup$ Thank you very much. I edited. @Aphelli Can you write the proof? I am confused. $\endgroup$
    – Bob Dobbs
    Commented Aug 11 at 13:18

1 Answer 1

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Step 1: show the number $z := \zeta_n^k$ is a root of unity with order $m := n/(k,n)$.

Step 2: use transitivity of the norm mapping on field extensions to show $$ {\rm N}_{\mathbf Q(\zeta_n)/\mathbf Q}(z-1) = {\rm N}_{\mathbf Q(z)/\mathbf Q}(z-1)^{\varphi(n)/\varphi(m)}. $$ You wrote ${\rm N}_{\mathbf Q(\zeta_n)/\mathbf Q}(x)$ as $N(x)$, but that leaves out the field extension, which makes it harder to express transitivity of the norm mapping.

Step 3: show the minimal polynomial of $z-1$ over $\mathbf Q$ is $\Phi_m(T+1)$, which has constant term $\Phi_m(1)$.

Step 4: Put everything together.

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  • $\begingroup$ Steps are not good. $\endgroup$
    – Bob Dobbs
    Commented Aug 11 at 17:47
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    $\begingroup$ I outlined a way to solve the problem. If there is a step you can't work out, say what you tried and where you got stuck. $\endgroup$
    – KCd
    Commented Aug 11 at 21:22
  • $\begingroup$ Thanks for your answer. I wanted to have an elementary solution which uses only the definition of norm. $\endgroup$
    – Bob Dobbs
    Commented Aug 13 at 9:26
  • $\begingroup$ I suggest first getting some solution so you know what the answer is and only then worrying about getting a more elementary solution. By the way, there are multiple ways to define the norm, so without saying how you define the norm it will be unclear what counts as a solution only using the definition. $\endgroup$
    – KCd
    Commented Aug 13 at 13:45

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