Does the determinant give you the index over $\mathcal{O}_k$ as well as over $\mathbb{Z}$? It is a standard fact that if $M$ is a nonsingular $n\times n$ integer matrix, the index of the $\mathbb{Z}$-span of its columns as an abelian group in $\mathbb{Z}^n$ is $|\det M|$. What happens if we replace $\mathbb{Z}$ by the ring of integers in some number field? More precisely:

Let $K$ be a finite extension of $\mathbb{Q}$ with ring of integers $\mathcal{O}_K$. Let $M$ be an $n\times n$ nonsingular matrix with entries in $\mathcal{O}_K$, and let $\Lambda$ be the sub-$\mathcal{O}_K$-module of $\mathcal{O}_K^n$ spanned by the columns of $M$. What can we say about the index of $\Lambda$ as an abelian group inside $\mathcal{O}_K^n$? Is it $|\det M|$? If so, what's the proof? (And if not, what's really going on?)

EDIT: The question is naive in a way. A priori, $\det M$ is an element of $\mathcal{O}_K$, so $|\det M|$ isn't defined without specifying a particular archimedean place of $K$. If there are several, then $|\det M|$ will depend on an arbitrary choice whereas $[\mathcal{O}_K^n:\Lambda]$ will not. So this leads me to expect that in general, the answer is no. On the other hand, what if, say, $\det M \in \mathbb{Z}$ and/or $M$ is unitary? 
2nd EDIT: A propos of Mariano's comment, maybe the right answer is actually $|N_{K/\mathbb{Q}}(\det M)|$, at least for number fields of class number $1$ and maybe more generally?

I believe that Propositions 1.10 and 1.16 of this paper of mine give (a generalization of) what you are looking for.  The notation may take some getting used to.  Indeed all the key ingredients of the discussion above appear in this somewhat abstracted context...and they comprise the proof.  
