The probrem I have now is
Let $F$ be a Field, and $f(X) = X^3+aX+b \in F[X]$ be a irreducible polynomial. Also, $\alpha, \beta, \gamma \in E$ ($E$ is the smallest splitting field of $f$) are roots of $f$, and $d$ is discriminant of $f$. Then, the followings are equivalent.
- (1). $F(\alpha)/F$ is normal extension.
- (2). A square root of $d$ is in $F(\alpha)$.
- (3). A square root of $d$ is in $F$.
My approch are followings.
- $(1) \Rightarrow (2)$ follows from $d = (\alpha-\beta)^2(\beta-\gamma)^2(\gamma-\alpha)^2$ and $\alpha, \beta, \gamma \in K(\alpha)$.
- $(3) \Rightarrow (2)$ follows from $\exists r \in F \subset F(\alpha)$ s.t. $d = r^2$.
- $(2) \Rightarrow (1)$ Let $g(X) \in F(\alpha)[X]$ be a polynomial with $\beta$ and $\gamma$ as roots. Then, by (2), $$ (\alpha-\beta)(\beta-\gamma)(\gamma-\alpha) = g(\alpha)(\gamma-\beta) \in F(\alpha) $$ and $g(\alpha) \in F(\alpha)$. So, $\beta-\gamma \in F(\alpha)$. Also, $\beta+\gamma = -\alpha \in F(\alpha)$ by Vieta's formula. Therefore, $(2) \Rightarrow (1)$.
I want to prove $(1) \Rightarrow (3)$ or $(2) \Rightarrow (3)$ and $X^2-d \in F[X]$ is key to solve this in my thought, but I have come up with nothing.
Please some hints or solution. Thankyou for your time!