$char(E)\not= 2\not=char(F)$.
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.
Best Answer
For question 1. If $\gamma \in E$ then $F(\gamma)$ is a sub-field of $E$. So multiplicative law says $[E:F]=[E:F(\gamma)][F(\gamma):F]$.
For question 2. Question 1 implies our minimal polynomial has degree $1,2$ or $4$ (Check Dummit & foote sec 13.1 theorem 14). $f'=0$ only if all coefficient are $0$, if $char(F)\neq 2$ and $f$ has degree 2, that's impossible because leading coefficient of $f'$ will be $1$ (f is a monic polynomial). When $f$ has degree 1, and 4 it's the same reasoning.
For question 3. There is no something like an infinite minimal polynomial for field extensions of infinite degree.