Some questions about ramifications of primes 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.
 A: I will slightly modify your problem as follows.
Let $L/K$ be an extension of algebraic number fields with the rings of integers $\mathcal{O}_L\supset \mathcal{O}_K$.
Let $\mathcal{O}$ be a subring of $\mathcal{O}_L$ such that $\mathcal{O}_K \subset \mathcal{O} \subset \mathcal{O}_L$.
Let $\mathfrak{p}$ be a prime ideal of $\mathcal{O}_K$.
Since $\mathcal{O}$ is integral over $\mathcal{O}_K$, by the lying-over theorem, there exists a prime ideal $\mathfrak{P}$ of $\mathcal{O}$ lying over $\mathfrak{p}$.
Suppose $\mathfrak{p}\mathcal{O}\subset \mathfrak{P}^e$ for some $e>1$.
Since $\mathcal{O}_L$ is integral over $\mathcal{O}$, by the lying-over theorem, there exists a prime ideal $\mathfrak{Q}$ of $\mathcal{O}_L$ lying over $\mathfrak{P}$.
Since $\mathfrak{P} \subset \mathfrak{Q}$, $\mathfrak{P}^e \subset \mathfrak{Q}^e$.
Hence $\mathfrak{p} \subset \mathfrak{Q}^e$.
Hence $\mathfrak{p}$ ramifies in $L$.
A: In the special case of your question I would say that the answer is "yes, $(2)$ is (completely) ramified in your extension field. I would reason that $\sqrt[3]{2}$ is definitely an element of the ring of integers in $K$ and since $\sqrt[3]{2}^3=2$ we have $(\sqrt[3]{2})^3=(2)$ in $\mathbb{Z}_K$. On the other hand we have $[K:\mathbb{Q}]=3$. 
So we found an ideal $p$ with the property $p^{[K:\mathbb{Q}]}=(2)$ and $(2)$ is there for completely ramified in $K$ regardless of the complete structure of $\mathbb{Z}_K$.
