How many primes stay inert in a finite (non-cyclic) extension of number fields? In the following suppose L/K is a finite Galois extension of number fields, (maybe it works for other cases also, I don't know) By the Chebotorev density theorem when Gal(L/K) is cyclic,  there are infinitely many primes in K that stay inert during this extension (cf Janus p136, Algerbaic Number Fields.) When L/K is non cyclic, an exercise from Neukirch (somewhere in Chap I) says there are at most finitely many primes that stay inert. I want to say that there are none. The reason is by a cycle description from Janus, p101, Prop 2.8, 
In short, that proposition says when $\delta:=Frob(\frac{L/K}{\beta})$, $\beta|p$ is a prime in L, consider $\delta$ act on the cosets of H in G, H=Gal(L/E), $K\subset E\subset L$, then every cycle of length i corresponds to a prime factor in E with residue degree i. In particular, for inert guys we want there is only one cycle in the action. When we take H to be trivial, E=L is Galois over K, and the cosets are just the elements of G themselves. 
So we want that there exist an element (the Frobenius element above p) act transitively on G, thus G is cyclic. 
I wonder if this is true, then more people should have been aware of it. If it is not, is there a counter example?   
 A: Does "inert" mean that there is only one prime over $p$, or does it mean that there is only one prime over $p$ and that prime is unramified? As the other answers have explained, unramified primes can not remain inert. However, it is possible that there is only prime over $p$. For example, let $p$ be an odd prime, take $K = \mathbb{Q}$ and $L= \mathbb{Q}(\sqrt{p}, \sqrt{a})$ where $a$ is not a quadratic residue modulo $p$.
A: It's worth noting that if $L/K$ is abelian, then you can easily see, using class field theory, that a non-cyclic extension can have no inert primes: the inertia degree of a prime is the order of it's Frob in the class group.  
A: If $L/K$ is a finite, Galois extension of number fields such that $\text{Gal}(L/K)$ is not cyclic, then no prime of K remains inert L. Indeed, one always has an isomorphism $D_p/I_p\cong \text{Gal}(L_p/K_p)$ of the Decomposition group modulo the Inertia group with the Galois group of the corresponding residue field extension. The latter group is the Galois group of a finite extension of finite fields, hence is cyclic. If the prime p were to remain inert in L, then by definition the Inertia group would be trivial and the Decomposition group would be all of $\text{Gal}(L/K)$. But this would imply that $L/K$ was a cyclic extension - a contradiction.
[Edit] I can't help but mention a cute application of this. Let $n$ be any positive integer for which $(\mathbb{Z}/n\mathbb{Z})^*$ is not cyclic. Then the cycotomic polynomial $\Phi_n(x)$ is reducible modulo $p$ for every rational prime $p$. Indeed, suppose that $p$ is a rational prime for which $\Phi_n(x)$ is irreducible modulo $p$. Then by the Dedekind-Kummer theorem, $p$ is inert in the cyclotomic field $\mathbb{Q}(\zeta_n)$. Then the Galois group of the residue class field extension, which is cyclic, is isomorphic to the Decomposition group, which in this case is $\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})\cong(\mathbb{Z}/n\mathbb{Z})^*$. But the latter group is not cyclic - contradiction. Thus $\Phi_n(x)$ is reducible modulo $p$ for all rational primes $p$.
