Aut$(K/F)$ permutes roots of polynomial.

Question

Let Aut$(K/F)$ is the set of all automorphism from $F$ to $K$, where $K$ is a galois extension of $F$. Let $f(x) \in F[x]$ and $\alpha$ be a root of the polynomial $f(x)$. I am able to prove that for any $\sigma \in$ Aut$(K/F)$ then $\sigma(a)$ is a root of $f(x)$ but I am unable say that $\sigma$ is a one-one function on the set of all root (i.e. it permutes the root).

I think I am missing a very obvious step here!

Answer 1

Any polynomial $f \in F[x]$ with roots $a_1, a_2,...,a_n$ can be written as follows:

$$f(x) = (x-a_1)(x-a_2)\cdots(x-a_n)$$

Expanding this, we see its constant term is $\displaystyle \prod_{j=1}^n a_j$, an element of $F$. As such, any $\sigma \in \operatorname{Aut}(K/F)$ must fix this element. Knowing that any $\sigma \in \operatorname{Aut}(K/F)$ is sending roots to roots, if we suppose that this mapping is not injective, then we'd have $\displaystyle \sigma \left( \prod_{j=1}^n a_j \right) \neq \prod_{j=1}^n a_j$. Contradiction.

Answer 2

A (unital) ring homomorphism of a field is always injective. Since it maps roots of $f$ to roots of $f$, it induces an injection on the finite set of roots of $f$, thus a bijection.