$Norm_{(\mathbb Q(\zeta_p)|\mathbb Q)}(1-\zeta_p)=p$

Question

I want to show that $N=Norm_{(\mathbb Q(\zeta_p)|\mathbb Q)}(1-\zeta_p)=p$. I know that this is an elementary fact but I need a final suggestion:

We have the identity $1 + X + \cdots + X^{p-1} = (X - \zeta)(X - \zeta^2)\cdots (X - \zeta^{p-1})$. Replacing $X$ by $1$ we get that $p=(1-\zeta_p)...(1-\zeta_p^{p-1})$. Now how can I conclude? We have that $N(p)=p^{p-1}=N(1-\zeta_p)...N(1-\zeta_p^{p-1})$. If we show that $1-\zeta_p$ cannot be a unit we are done for example.

Answer 1

By (one) definition, $N(\alpha) = \prod_{\sigma \in G} \sigma(\alpha)$ where $G$ is the Galois group. So $$N(1 - \zeta_p) = (1 - \zeta_p)(1 - \zeta_p^2)\cdots(1 - \zeta_p^{p-1}) = p.$$

Answer 2

$$N(1-\zeta_p)=\lim_{x\to 1}\prod_{k=1}^{p-1}(x-\zeta_p^k)=\lim_{x\to 1}\Phi_p(x)=\lim_{x\to 1}\frac{x^p-1}{x-1}\stackrel{d.H.}{=}\lim_{x\to 1}px^{p-1}=p.$$

Answer 3

A primitive $p$-th root $\zeta$ of $1$, $p$ odd, is a root of the irreducible polynomial $f(X)= \frac {X^p -1}{X-1}$, so $\zeta - 1$ is a root of $g(X)=f(X+1)=X^{p-1} +...+p$, and $N(\zeta - 1) = p$.