I was trying to show that the ring of integers of $K=\mathbb{Q}(\sqrt[3]{2})$ is $\mathbb{Z}[\sqrt[3]{2}]$ and came up with the following question. Computing the discriminant of $\mathbb{Z}[\sqrt[3]{2}]$, we see that the possible discriminants are
$-3,-12,-27,-27\times 4$
and the first two would cause the Minkowski bound to be smaller than $1$, so we know the discriminant is one of $-27,-27\times 4$ and showing it's $-27\times 4$ would imply what we want. Since $\sqrt[3]{2}$ is not a unit in the ring of integers of $K$ (compute the norm), we know that some prime $\mathfrak{p}$ in the ring of integers of $K$ contains $\sqrt[3]{2}$. Since $(2)=(\sqrt[3]{2})^3\subset \mathfrak{p}^3$, it follows that $\mathfrak{p}^3\mid 2$. Therefore $2$ ramifies and must divide the discriminant.
My question can be rephrased in a general form as follows: Let $L/K$ be an extension of (number/global) fields with rings of integers $\mathcal{O}_L\supset \mathcal{O}_K$. Assume that $\mathcal{O}$ is an order of $L$ and $\mathfrak{P}$ a prime of $\mathcal{O}$ lying above some prime $\mathfrak{p}$ of $\mathcal{O}_K$. If $\mathfrak{p}\mathcal{O}\subset \mathfrak{P}^e$ for some $e>1$ can we deduce that $\mathfrak{p}$ ramifies in $L$? If $\mathcal{O}=\mathcal{O}_L$, then the claim is of course precisely that $\mathfrak{P}^e$ divides $\mathfrak{p}\mathcal{O}_L$.
I seem to be getting confused here, since any order $\mathcal{O}\neq \mathcal{O}_L$ of $L$ is not Dedekind, so ideals don't necessarily satisfy nice properties with respect to multiplication. In addition, since $\mathcal{O}_L$ is a larger ring than $\mathcal{O}$, there might not be a prime of $\mathcal{O}_L$ lying above $\mathfrak{P}$, since an element of the ideal might become invertible in the larger ring.
Thanks for any help.