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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?

Ben Blum-Smith
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    The proof of that fact about the index is done using the Smith Normal Form of matrices over the integers, which in turn depends on the fact that the integers is a principal ideal domain. The usual proof will therefore break for most number fields. – Mariano Suárez-Álvarez Feb 02 '16 at 22:33
  • @MarianoSuárez-Alvarez - but it will work over fields of class number 1? – Ben Blum-Smith Feb 03 '16 at 02:38
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    For $1$-by-$1$ matrices $[\alpha]$, the answer is $\mathbf N(\alpha)$. This area is far from my meat, but the problem looks local in nature to me. Can’t you do it at a prime $\mathfrak P|p$ of $K$ and combine all the nonunit results? – Lubin Feb 03 '16 at 02:51
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    @Ben, I think so, replacing the absolute value by the field norm. Over such a field there is a SNF, so we can assume that the matrix is diagonal, and then the result follows from multiplicativity of the cardinal in direct sums and Lubin's observation for 1-by-1 matrices. – Mariano Suárez-Álvarez Feb 03 '16 at 02:58
  • @Lubin - that sounds like the right idea to me - I don't believe the result should depend on class number issues. I'm missing a couple steps though. I guess the idea is that one captures the $p$-valuation of $[R^n:MR^n]$ (letting $R=\mathcal{O}K$ and observing that $\Lambda = MR^n$) by taking a $\mathfrak{P}\mid p$ and then taking the $p$-valuation of $[R\mathfrak{P}^n:MR_\mathfrak{P}^n]$, and then combining over all $p$ for which the valuation is positive. (Since $R_\mathfrak{P}$ is a PID, Mariano's SNF argument applies for each $\mathfrak{P}$.) I have 2 questions to finish this: – Ben Blum-Smith Feb 03 '16 at 18:16
  • (1) Is it true that for $R_\mathfrak{P}$ the $p$-valuation of $[R_\mathfrak{P}:\alpha R_\mathfrak{P}]$ is given by that of $\mathbf{N}(\alpha)$? (Certainly there is no hope that these numbers will match beyond their $p$-valuations.) What's the argument? and (2) ... hmmm I thought I had another question but I guess (1) was really the only uncertainty left. – Ben Blum-Smith Feb 03 '16 at 18:20
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    Are you aware of the theorem in algebraic number theory that the index of $Ra$ in $R$ is $|\mathbf N(a)|,$? And then, locally, one ordinarily defines $v_{\mathfrak P}(a)$ exactly to be $v_p(\mathbf N(a))$. In the latter case, I’m thinking of the complete situation, where one then shows that this $v_{\mathfrak P}$ is a valuation by a Hensel’s-Lemma argument. – Lubin Feb 03 '16 at 19:59
  • As per Lubin's last comment it may well be that the norm of the determinant always gives the index (because it does so for 1x1 matrices). I have a vague recollection of having done this exercise for imaginary quadratic fields, but it's getting late here, so recalling it more precisely will have to wait. Anyway, the business with class numbers is IMVHO more about the fact that if $h>1$ you won't get all the $\mathcal{O}_K$ submodules of $\mathcal{O}_K^n$ as spans of columns of some square matrix. Again, the case of 1x1 matrices shows this already. – Jyrki Lahtonen Feb 03 '16 at 22:29
  • (cont'd) Not even those submodules that contain $n$ linearly independent vectors. – Jyrki Lahtonen Feb 03 '16 at 22:30
  • @Lubin - Thank you, yes I know that theorem. What I was unsure about is the relationship of $N(a)$ to $[R_\mathfrak{P}:a R_\mathfrak{P}]$. How does one prove that the $p$-valuation of $[R_\mathfrak{P}:aR_\mathfrak{P}]$ is related to the norm of $a$ in this way? Is your point that the $p$-valuation of $[R_\mathfrak{P}:aR_\mathfrak{P}]$ is the same as the $\mathfrak{P}$-valuation of $a$? (Since $aR_\mathfrak{P} = \pi^{v_\mathfrak{P}}R_\mathfrak{P}$ for $\pi$ a uniformizer for the DVR $R_\mathfrak{P}$?) – Ben Blum-Smith Feb 04 '16 at 06:18
  • Yes, I guess that was my point. – Lubin Feb 04 '16 at 06:31
  • Related (about your standard fact) : https://math.stackexchange.com/questions/317065/question-on-determinants-of-matrices-changing-between-integer-matrices. It works directly if $O_K$ is a PID. – Watson Nov 19 '18 at 11:30

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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.

Pete L. Clark
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