Exercise 4.3.2: Let $K$ be a field and $f(x) \in K[x]$ a separable polynomial. Show that the Galois group of $f(x)$ acts transitively on the roots of $f(x)$.
You're looking for solutions to Chapter 4 of "Abstract Algebra" by David S. Dummit and Richard M. Foote! abstract algebra dummit and foote solutions chapter 4
($\Leftarrow$) Suppose every root of $f(x)$ is in $K$. Let $\alpha_1, \ldots, \alpha_n$ be the roots of $f(x)$. Then $f(x) = (x - \alpha_1) \cdots (x - \alpha_n)$, showing that $f(x)$ splits in $K$. Exercise 4
Solution: Let $a \in K$. If $a = 0$, then $\sigma(a) = 0$. If $a \neq 0$, then $a \in K^{\times}$, and $\sigma(a)$ is determined by its values on $K^{\times}$. Dummit and Richard M
Exercise 4.2.1: Let $K$ be a field and $f(x) \in K[x]$. Show that $f(x)$ splits in $K$ if and only if every root of $f(x)$ is in $K$.
Exercise 4.3.1: Show that $\mathbb{Q}(\zeta_5)/\mathbb{Q}$ is a Galois extension, where $\zeta_5$ is a primitive $5$th root of unity.