For a field extension $E/F$, the degree $[E:F]=4$ is given.
Then, show $E=F(\alpha)$ for some $\alpha$.
The answer that is given to me proves first that $E/F$ is separable, finishes in the Primitive Element Theorem.
More specific, for $\gamma\in E$, let $Irr(\gamma, F; x)$ be the minimal irreducible polynomial of $\gamma$ over $F$.
Then, since $\deg(Irr(\gamma, F; x)) | [E:F]$, $\deg(Irr(\gamma, F; x))=1,2,$ or $4$.
And the characteristics of the fields are not $2$, so the derivative of $Irr(\gamma, F; x)$ is not $0$, hence $Irr(\gamma, F; x)$ is separable.
Thus $E/F$ is separable and $E=F(\alpha)$ for some $\alpha\in E$.
- Why does $\deg(Irr(\gamma, F; x))$ divide $[E:F]$?
- Why is the condition $char(E)\not=2\not=char(F)$ enough to ensure that $f'\not=0$? ($f=Irr(\gamma, F; x)$)
- This is a little off topic, but can a polynomial such that has infinite degree, but square-free be called "separable"? So basically can infinite degree separable extension be possible?
I apologize for asking too many questions. Any hint would be greatly appreciated.