Why does an element of the Galois group of $K(\mu_n)/K$ maps an $n$-th root of unity to another $n$-th root of unity?

Question

I am working through Keith Conrad's article on Cyclotomic extensions and have a question regarding the proof in Lemma 2.1.

Let $K$ be any field and $\mu_n \subseteq K^\times$ be the multiplicative subgroup containing all $n$-th roots of unity. We assume $T^n-1$ to be separable over $K$, so $K(\mu_n)/K$ is Galois.

Let $\sigma$ be in the Galois group of $K(\mu_n)/K$ and $\zeta_n$ be a generator of $\mu_n$ (i.e. $\zeta_n$ is a primitive $n$-th root of unity). Then $\zeta_n^n = 1$ and $\zeta_n^j\neq 1 $ for all $j$ with $1\leq j <n$.

Question: Next, it is stated that $\sigma(\zeta_n)^n = 1$ and $\sigma(\zeta_n^j) \neq 1$ for all $j$ with $1\leq j <n$. Why is this true?

My attempt for a proof

• It is $\sigma(\zeta_n)^n = \sigma(\zeta_n^n) = \sigma(1) = 1$,
• Similarly, we have $\sigma(\zeta_n)^j = \sigma(\zeta_n^j)$. I know that $\zeta_n^j \neq 1$ but how can I exclude the case $\sigma(\zeta_n^j) = 1$?
  

It would be cool if you could help me explaining this step.

Answer 1

Every field homomorphism is injective, so if $\sigma(\zeta_n^j) = 1 = \sigma(1)$, we would have $\zeta_n^j = 1$.