Can all polynomials of a given degree be reducible? 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.
 A: 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, $\mathbb F_{p^\ell}\not\subseteq F$.
A: Here's an example of historical importance for $n=2$:
Let $F$ be the subfield of $\mathbb C$ consisting of all numbers constructible with ruler and compass if $0$, $1$ and $i$ are given. ($F$ is a field because adding, negating, multiplying, taking reciprocals can be done with ruler and compass).
Since it is possible to take square roots and solve quadratics with ruler and compass, we see that all quadratic polynomials are reducible.
On the other hand, the old classic problem of doubling the cube (i.e. finding a root of $x^3-2=0$) is not solvable with ruler and compass, hence $F\ne \overline F$.
