TITLE: Can all polynomials of a given degree be reducible?
QUESTION [11 upvotes]: Let $n > 1$ be a fixed integer. Does there exist a field $F$ with the following properties?

$F$ is not algebraically closed.
Every polynomial $f(x) \in F[X]$ of degree $n$ is reducible.

I cannot think of any such field, even for $n=2$. All I can prove is that $F$ must be infinite.

REPLY [2 votes]: Over a finite field $\mathbb F_p$, fix an algebraic closure $\bar{\mathbb F}_p$ and consider the (increasing) union $F$ of the subextensions ${\mathbb F}_{p^{n^d}}$, $d\ge 0$. Let $f(x)\in F[x]$ of degree $n$. Then for some $d\ge 0$, 
$$f(x)\in F_d:=\mathbb F_{p^{n^d}}[x].$$ 
If $f(X)$ is irreducible over $F$, then it is irreducible over $F_d$. The extension of $F_d$ generated by a root $\alpha$ of $f(x)$ in $\bar{\mathbb F}_p$ has degree $n$ over $F_d$, so $\alpha\in F_{d+1}\subseteq F$. Thus $f(x)$ is reducible in $F[x]$. Finally, $F\ne \bar{\mathbb F}_p$: take a number $\ell$ prime to $n$, then $\mathbb F_{p^\ell}\cap F=\mathbb F_p$. In particular, $\mathbb F_{p^\ell}\not\subseteq F$.