From Robert Ash (Basic Abstract Algebra)
Suppose that $E=F(\alpha)$ is a finite Galois extension of $F$, where $\alpha$ is a root of the irreducible polynomial $f \in F[x]$. Assume that the roots of $f$ are $\alpha_1 = \alpha, \alpha_2, ... , \alpha_n$. Describe, as best you can from the given information, the Galois group of $E/F$.
Answer: $G = \{\sigma_1, ... ,\sigma_n\}$ where $\sigma_i$ is the unique $F$-automorphism of $E$ that takes $\alpha$ to $\alpha_i$.
There are two things that confuse me about this answer:
1) What if some of the $\alpha_i$'s are in F? Then how would we have an $F$-automorphism that takes $\alpha$ to $\alpha_i$? Because we know that any $F$-automorhpism must fix all elements of $F$, right?
2) Don't we need $E$ to be a normal extension? Because how else would we know that the other roots of $f$ are also contained in $E$?
Thank you in advance