Let $K$ be a number field, $\mathcal{O}_{K}$ it's ring of integers. Suppose $p$ is a prime in $\mathbb{Z}$ such that d = disc($\mathcal{O}_{K}$) is exactly divisible by p$^{m}$ with $m$ odd. Prove that $p$ is ramified in $K$ by considering the fact that the normal closure of K contains $\mathbb{Q}[\sqrt{d}]$. Show that disc($\mathcal{O}_{K}$) can be replaced by disc($\alpha_{1},...\alpha_{n}$) where those alphas form a basis for $K$ over $\mathbb{Q}$.
I found that $p$ is ramified in $\mathbb{Q}[\sqrt{d}]$ because $m$ is odd so $\sqrt{d} = \sqrt{p^{m}n}$ = $p^{\frac{(m-1)}{2}}\sqrt{pn}$, where we can assume $n$ is square free without loss of generality. So $\mathbb{Q}[\sqrt{d}]=\mathbb{Q}[\sqrt{pn}]$ thus $p = (p, \sqrt{pn})^{2}$ in that field (reasoning by using a previous Thm in my textbook). My question is with the relationship between $\mathbb{Q}[\sqrt{d}]$ and $K$. Is $K$ an extension of $\mathbb{Q}[\sqrt{d}]$, and so $p$ is ramified in $K$ as well? I'm just not sure where to go from here.
Also, for the second part concerning the basis, is it not true that disc($\alpha_{1},...\alpha_{n}$) = disc($\mathcal{O}_{K}$)?
Any hints are more than welcome! Thanks!