TITLE: $E/F$ a field extension, $[E:F]=4$ then $E=F(\alpha)$ for some $\alpha$
QUESTION [1 upvotes]: $char(E)\not= 2\not=char(F)$.
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.

REPLY [0 votes]: This solution is along the lines of separable closure of a field extension. First of all if $ch(F)=0$, then we are done by direct application of primitive element theorem. Now if we consider $ch(F)=p>0$, then by hypothesis we know $p\neq 2$. Consider $S:=\text{separable}\;\text{closure}\;\text{of}\;E/F$, so that $E/S$ becomes purely inseparable extension. Since $|E:S|<\infty$ it must be the case $|E:S|=p^n$ for some nonnegative integer $n$. But $p\neq 2$ and $|E:S|$ divides $4$, implies $|E:S|=1$ or $S=E$. Therefore $E/F$ is separable and again we are done by primitive element theorem.
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