Why does the automorphism mapping $\omega$ to 1, not an element of Galois group

Question

Let $L/\mathbb{Q}$ be a field extension, where $L=\mathbb{Q}(\omega)$ and $\omega=e^{\frac{2\pi i}{7}}$.

In my textbook it states that $Aut(L/\mathbb{Q})=\{\sigma_i|\sigma_i(\omega)=\omega^i, 1\leq i\leq 6\}$. But, what about the map $\sigma(\omega)=\omega^7=1$. Do the automorphisms have to be injective, hence we cannot include this as an automorphism since we already have $\sigma(1)=1$.

Answer 1

Automorphisms are always bijective, for any structure (the general definition being: a homomorphism for which an invrese homomorphism exists). Moreover we deal with fields here, that is rings with only two ideals: the trivial ideal and the full field. Thus even if we loosen the definition of automorphim, the only non-injective homomorphism is the trivial homomorphism that sends everything to $0$.

Also, $1$ is not a root of $X^6+X^5+X^4+X^3+X^2+1$.

Answer 2

If what you mean is: "Why $\sigma$ is not an element of $Aut(L/\mathbb{Q})$?", then the answer is that $\sigma(\omega)$ must be an element of order $7$ (as $\omega$). $1$ has not that order.

(More generally, $\omega$ must be mapped to another root of its minimal polinomial).