Given a Galois extension $K \supseteq \mathbb{Q} $, prove that if there is only one unramified prime number $p$ over $K$ then there is only one prime ideal $\mathfrak{p} \subseteq O_K$ containing $p$ and the ramification index $e_p = [K:Q]$
I was given a hint which tells me to consider the field extension $K^{I_{\mathfrak{p}}}/\mathbb{Q}$, where $I_{\mathfrak{p}}$ is the inertia subgroup of $\mathfrak{p}$. Based on what I have done, I can see that once we have $K^{I_{\mathfrak{p}}}=\mathbb{Q}$ or $ I_{\mathfrak{p}} = \text{Gal}(K|\mathbb{Q})$, then the ramification index has to be $e_{p}$, but I can't see how I can use the condition of having a unique unramified prime to conclude, so can anyone tell me how to prove it?
Hint: suppose the inertial field is a nontrivial extension of $\Bbb Q$. We then know its discriminant is nonunital (by an argument involving juggling the terms in Minkowski's bound) and has prime divisors, so we know at least one prime ramifies in the inertial field. Is that a problem?