Prove that this quadratically closed extension of $\mathbb{F}_2$ remains quadratically closed after considering cubic extensions In Conway's book On Numbers and Games, Conway builds the algebraic closure of $\mathbb{F}_2$ (with ordinals, but that doesn't matter here) by first considering the smallest algebraic extension that is quadratically closed, and then considering cubic extensions of it until he get's a cubically (is this the right word?) closed.
He argues that as the Galois group of every finite field is abelian, the quadratically closed field remains quadratically closed after considering cubic extensions, and so on for every prime. I don't know how to prove it in general.
My try (to go from $2$ to $3$) was this:
suppose $G$ is the smallest quadratic closure of $\mathbb{F}_2$ and take $\alpha$ such that $[G(\alpha):G]$. Conway says that it suffices to consider $\alpha$ such that $\alpha^3\in G$ and I believe him (for now).
Now suppose there exists $\beta \in \bar{\mathbb{F}_2}$ such that $[G(\alpha)(\beta):G(\alpha)] = 2$. Moreover, we can assume that $\beta^2 \in G(\alpha)$. Now, $G(\alpha \cdot \beta)$ is a subfield, so the degree of the extension must divide $6$. Maybe I can prove that $[G(\alpha^3\cdot\beta^3):G] =2$ so $\alpha^3\beta^3 \in G$ as it's quadratically closed. But as $\alpha^3\in G, \beta$ is also there. Abs!
This is not a formal proof and I don't understand yet how to go from that $n$-th prime to the $n+1$-th prime.
Any help or hints? They only way of using the abelianity of the Galois group was by saying that all sufields are normal, by the way.
For references, it's page 60 but it starts a bit earlier.
 A: Here's a general theorem along these lines.  Let $p$ be a prime and suppose $K$ is a field such that no polynomial of degree $p$ is irreducible and every algebraic extension of $K$ is abelian.  Let $L$ be an algebraic extension of $K$.  Then no polynomial of degree $p$ over $L$ is irreducible.
To prove this, first note that we may assume $L$ is a finite extension of $K$, since every polynomial over $L$ has coefficients contained in some subfield which is a finite extension of $K$.  Now suppose there is some irreducible polynomial of degree $p$ over $L$ and let $E$ be the extension of $L$ obtained by adjoining a root of it.  Then $[E:L]=p$, so $[E:K]$ is divisible by $p$.  Now $E$ is abelian over $K$, and any finite abelian group of order divisible by $p$ has an index $p$ subgroup (this follows from the classification of finite abelian groups, for instance).  So let $H\subset Gal(E/K)$ be an index $p$ subgroup.  Then the fixed field $E^H$ is an degree $p$ extension of $K$.  But this is impossible, since there are no irreducible polynomials of degree $p$ over $K$.

So, for instance, taking $K$ to be the quadratic closure of $\mathbb{F}_2$, this means every algebraic extension of $K$ is still quadratically closed, and similarly for all other primes.
