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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.

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Keith Conrad's blurb: www.math.uconn.edu/~kconrad/blurbs/gradnumthy/Qw2.pdf is right on the money here. –  Cam McLeman Jan 11 '12 at 19:43

2 Answers 2

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 $L$.

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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 $\mathbb{Z}_K$.

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Yes, I already showed that in my question. It's the general case that's hard. I mean if the ideal in the order can be embedded in an ideal in the ring of integers, then the same argument works. –  asdf Jan 11 '12 at 18:58
    
I believe we can extend the argument to the general case. Since $\mathbb{Z}_K$ is the unique (this being the important part) maximal order in $K$ we should be able to embed every ideal of an arbitrary order $\mathcal{O}$ into an ideal of $\mathbb{Z}_K$. –  Sebastian Schoennenbeck Jan 11 '12 at 22:42

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