$E/F$ a field extension, $[E:F]=4$ then $E=F(\alpha)$ for some $\alpha$

Question

$char(E)\not= 2\not=char(F)$.

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 $\alpha\in E$.

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1. Why does $\deg(Irr(\gamma, F; x))$ divide $[E:F]$?
2. Why is the condition $char(E)\not=2\not=char(F)$ enough to ensure that $f'\not=0$? ($f=Irr(\gamma, F; x)$)
3. 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.

Answer 1

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.

Answer 2

1. By the Tower Law, $[E:F]=[E:F(\gamma)][F(\gamma):F]$, so $[F(\gamma):F]$, which is the same thing as the degree of the irreducible polynomial for $\gamma$ over $F$, divides $[E:F]$.

2. $f(x)=x^a+$ terms of lower degree, where $a$ is $1$, $2$, or $4$, so the leading term of $f'(x)$ is $ax^{a-1}$, which is not zero if the characteristic isn't two, so $f'$ isn't identically zero.

Answer 3

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.