Let $F$ be a field, $f(X)\in F[X]$ with $\deg f=2$ and $E$ is its splitting field.
We want to show that there are no proper intermediate fields in the extension $E/F$.
My thought. We know that $f$ has at most $2$ roots in its splitting field. Let $\alpha,\beta$ be its roots. Then, is is $E=F(\alpha,\beta)$. So, we have the Tower of Fields $F \leq F(\alpha)\leq F(\alpha,\beta)=E$. So, from the Tower Law, $$[F(\alpha,\beta):F]=[F(\alpha)(\beta):F(\alpha)][F(\alpha):F].$$ Since $f(\alpha)=0_F$, we have $$m_{(\alpha,F)}(X)|f(X) \implies \deg m_{(\alpha,F)} \leq \deg f \iff [F(\alpha):F]\leq 2.$$ Now, consider the extension $F \leq F(\alpha)$. Then, we have $m_{(\beta,F(\alpha))}|m_{(\beta,F)}$.
But like before, since $f(\beta)=0_F$, we have $m_{(\beta,F)}(X)|f(X) $. So, $$\deg m_{(\beta,F(\alpha))} \leq \deg m_{(\beta,F)} \leq \deg f \implies [F(\alpha,\beta):F(\alpha)]\leq 2.$$ Therefore, $[E:F]\in \{1,2,3,4\}$. If it is $1$, trivially our claim holds and if $[E:F]\in \{2,3\}$, since $2,3$ are prime, there are no proper intermediate fields in $E/F$. So, we take the case that $[E:F]=4$. By above the only possible case is to consider $[F(\alpha)(\beta):F(\alpha)]=[F(\alpha):F]=2$
But how can we continue from that? Is this proof right/in the correct direction?
Thank you.
It seems like the easier approach is to note that if $x^2-ax+b=(x - \alpha)(x - \beta)$, then $a=\alpha+\beta$ (and $b=\alpha \beta$) so $a \in F \subseteq F[\alpha]$ and $\alpha \in F[\alpha] \Rightarrow \beta = a - \alpha \in F[\alpha]$ and $E=F[\alpha]$.