# Norm of $\zeta_n^k-1$ in $\Bbb Z[\zeta_n]$

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?

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

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.

• Steps are not good. Commented Aug 11 at 17:47
• 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.
– KCd
Commented Aug 11 at 21:22
• Thanks for your answer. I wanted to have an elementary solution which uses only the definition of norm. Commented Aug 13 at 9:26
• 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.
– KCd
Commented Aug 13 at 13:45