Is there always a prime having only one prime above it? Given a number field $K / \Bbb Q$, can we find a prime $p \in \Bbb Z$ which has only one prime of $K$ above it (e.g. $p$ is inert or $p$ is totally ramified)? 
For instance, if $K$ is Galois over $\Bbb Q$ with non-cyclic Galois group, then no prime of $\Bbb Q$ is inert in $K$. But maybe we could have totally ramified primes, or a factorization like $p O_K = P^s$ with $1 \leq s \leq [K:\Bbb Q]$. (What happens if we assume $K/\Bbb Q$ to be Galois ?)
Thank you!
 A: Consider the cyclotomic field $K=\Bbb Q(\zeta_8)$ $\mathcal O_K = \Bbb Z[\zeta_8]$, this has a non-cyclic Galois group, so no prime is inert in $K$. The only prime that is ramified in $K$ is $2$, but $2$ is totally ramified, so this is not a counterexample yet.
Now consider $L=\Bbb Q(\zeta_7)$, where $7$ is the only ramified prime and there are two different primes above $(2)$ and conversely, there are two distinct primes above $(7)$ in $K$. (This is becaue the $8$th cyclotomic polynomial $x^4+1$ splits as $x^4+1=(x^2+3x+1)(x^2+4x+1) \pmod{7}$ and the $7$th cyclotomic polynomial $x^6+x^5+x^4+x^3+x^3+x^2+x+1$ splits as $x^6+x^5+x^4+x^3+x^2+x+1=(x^3+x+1)(x^3+x^2+1) \pmod{2}$)
So if we take the composite field $K \cdot L = \Bbb Q(\zeta_{56})$, we get an extension such that there are at least two primes above every prime in $\Bbb Z$. $K \cdot L$ is also Galois.
A: If there is only one prime of $K$ above $p$, then $\mathrm{Gal}(K/\mathbb Q)$ is isomorphic to the decomposition group of $p$. But the decomposition group is solvable (the higher ramification groups are a composition series).
In particular, any extension with non-solvable Galois group will work.
