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

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

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 $$\alpha\in E$$.

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.

• There's no such thing as a polynomial with infinite degree. There is such a thing as an infinite degree separable extension, even an infinite degree algebraic separable extension, but it won't be a simple extension, it won't be $F(\alpha)$ for some $\alpha$. – Gerry Myerson Jan 10 '20 at 2:16

For question 1. If $$\gamma \in E$$ then $$F(\gamma)$$ is a sub-field of $$E$$. So multiplicative law says $$[E:F(\gamma)]=[E:F(\gamma)][F(\gamma):F]$$.
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.
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.