A unit of seventh cyclotomic field

Question

I have troubles with the following problem about units.

Show that $1+\zeta $, $1+\zeta+\zeta^2$ are units in the field $\mathbb{Q[\zeta]}$, where $\zeta$ is a seventh primitive root of unit ($\zeta^7=1$).

Is it possible to prove this by straightforward calculation with norms?

Answer 1

The minimal polynomial of a primitive seventh root of unity is: $$\Phi_7(x) = 1+x+x^2+x^3+x^4+x^5+x^6 $$ hence: $$ -1 = \color{red}{(1+\zeta)}(\zeta+\zeta^3+\zeta^5) $$ as well as: $$ -1 = \color{red}{(1+\zeta+\zeta^2)}(\zeta+\zeta^4).$$

Answer 2

Another way to approach this is to remember that an algebraic integer $z\in{\cal O}_K$ ($K$ a number field) is a unit if and only if $N_{K/\Bbb Q}(z)=\pm1$.

Consider $z=1+\zeta$ as in the question (now $K=\Bbb Q(\zeta)$ and ${\cal O}_K=\Bbb Z(\zeta)$). If $\Phi(X)$ is the cyclotomic polynomial (as in Jack's answer) we have $$ N_{K/\Bbb Q}(1+\zeta)=\prod_{j=1}^6(1+\zeta^j)=\prod_{j=1}^6(-1-\zeta^j) =\Phi(-1)=1. $$ A similar computation holds for $1+\zeta+\zeta^2$.