If $K$ is a field of characteristic $0$ in which every polynomial of degree $\geq3$ is reducible, then $\bar{K}/K$ has degree $1$ or $2$

Question

Is it true that if $K$ is a field of characteristic $0$ in which every polynomial of degree 3 is reducible, then $\bar{K}/K$ has degree 1 or 2? I have thought of taking $\alpha$ algebraic over $K$. Then $K\subset K(\alpha)\subset\bar{K}$. From the condition on the degree of irreducible polynomials $[K(\alpha):K]=1,2$ and $[\bar{K}:K]=[\bar{K}:K(\alpha)][K(\alpha):K]$. Can I guess something about $[\bar{K}:K(\alpha)]$? The whole situation looks just like $\mathbb{R}$ or $\mathbb{C}$...

Answer 1

The Primitive Element Theorem says that every finite separable extension is simple; that is, if $L/K$ is a field extension that is finite, and $L$ is separable over $K$, then there exists $\alpha\in L$ such that $L=K(\alpha)$. Extensions of fields of characteristic $0$ are necessarily separable, so finite extensions of fields of characteristic zero are always simple.

If $\alpha\in \overline{K}$, then $\alpha$ is algebraic over $K$, so its characteristic polynomial has degree either $1$ or $2$. If you always get degree $1$, then $\alpha\in K$ for all $\alpha\in\overline{K}$, so you have $\overline{K}=K$.

If there is at least one $\alpha$ with $[K(\alpha):K]$ of degree $2$, let $\beta\in \overline{K}$. Then we have a tower $K\subseteq K(\alpha)\subseteq K(\alpha,\beta)$. Since $K(\alpha,\beta)=K(\gamma)$ for some $\gamma$ by the Primitive Element Theorem, we have $[K(\beta,\alpha):K] = [K(\gamma):K]\leq 2$. Therefore, $$2\geq [K(\gamma):K] = [K(\gamma):K(\alpha)][K(\alpha):K] =2[K(\gamma):K(\alpha)]\geq 2$$ so $[K(\gamma):K(\alpha)]= 1$. This proves that $\gamma\in K(\alpha)$. Thus, $\overline{K}\subseteq K(\alpha)$, hence they are equal. Thus $[\overline{K}:K] = [K(\alpha):K] = 2$.

In summary, if $K$ has characteristic $0$ and irreducible polynomials over $K$ always have degree $1$ or $2$, then either $[\overline{K}:K]=1$ or $[\overline{K}:K]=2$.