Algebraic number fields in which all rational primes are inert

Question

Is there an algebraic number field $F\supsetneq\mathbb{Q}$ such that all rational primes are inert in $\mathcal{O}_F$?

Answer 1

A sledgehammer proof of the fact that there is no such number field uses Minkowski's Theorem to the effect that every number field $F\neq\mathbf Q$ is ramified over $\mathbf Q$. This means that there is at least one rational prime that ramifies in $F$. Such a prime is not inert.

Answer 2

There are many good reasons why such a number field cannot exist, but here it is an elementary one: if $F=\mathbb Q(\alpha)$, with $\alpha$ a root of $f\in \mathbb Z[x]$ monic irreducible of degree $>1$ and $p$ is an inert prime in $F$, then $f$ is irreducible modulo $p$. However, $f(\mathbb Z)$ must be strictly bigger than $\{0,\pm1\}$ since otherwise it would be constant. Thus, there exists $n\in\mathbb Z$ such that $f(n)\neq 0,\pm1$. But then by the fundamental theorem of algebra there exists a prime number $p$ such that $p\mid f(n)$. This shows that $f$ has a root modulo $p$, hence it is not irreducible!