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I posted the question on mathexchange as well, but realized that my chances would be higher posting here; In the paper "Almost diagonal matrices over Dedekind Domains" by L. Levy a specific decomposition of modules over Dedekind rings $D$ is used:

Let $M$ be submodule of $D^n$, then there exists a simultaneous decomposition

$D^n \cong D y_1 \oplus \dots \oplus D y_{r-1} \oplus H^{-1} \oplus H \oplus D^{n-r-1}$

$M \cong L_1 y_1 \oplus \dots \oplus L_{r-1} y_{r-1} \oplus L_r H^{-1}$

where $L_i$are integral ideals and $H$ a fractional ideal, $H^{-1}$ being its inverse.

This decomposition is not proved, Levy just mentions that it is scattered through the works of Krull. I want to ask, if someone knows a reliable (preferably modern) reference for this decomposition.

Furthermore, is it possible to derive it from the "ordinary" decomposition of f.g. modules $N$ over Dedekind domains (which can for instance be found in Narkiewicz book)?: $N \cong \bigoplus_{k=1}^s R/a_i \oplus H \oplus R^k$ where the $a_i$ and $H$ are ideals of $D$.

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Link to the other question:… – András Bátkai Feb 6 '14 at 7:46
It is always good to wait a few days before cross-posting. – András Bátkai Feb 6 '14 at 7:46
you are right, but I waited some weeks on a similar question at mathexchange. – user75148 Feb 6 '14 at 10:24

The basic idea is simple, and carrying it out requires just a bit of bookkeeping. We will first pass to the case of an inclusion of torsion-free $D$-modules of equal rank $r > 0$ (possibly losing freeness of the larger module). There will then be nothing to do if $r=1$, and if $r > 1$ we use rank-induction to describe these modules compatibly as direct sums of invertible modules in such a way that for the ambient $D$-module all but one of the invertible direct summands is free.

Now for the actual argument. Let $M' \subset D^n$ be the saturation of $M$ (i.e., $D^n \cap (K \otimes_D M)$ inside $K \otimes_D D^n$, where $K = {\rm{Frac}}(R)$), so $D^n/M'$ is torsion-free and hence projective. Thus, $D^n \twoheadrightarrow D^n/M'$ splits, which is to say $D^n = M' \oplus N$ for some $D$-submodule $N$ of $D^n$. Passing to top exterior powers, the invertible $D$-modules $\det M'$ and $\det N$ are inverse to each other. Define $H = \det N$ as a $D$-module; this can be identified with a fractional ideal by choosing a $K$-basis of $K \otimes_D \det N$.

Letting $r = \dim_K (K \otimes_D M)$, so $n-r = \dim_K (K \otimes_D N)$, if $n-r > 0$ then we have $N \simeq D^{n-r-1} \oplus H$ and $M' \simeq D^{r-1} \oplus H^{-1}$. Thus, we can replace $D^n$ with $M'$ to reduce to showing that if $M_1 \subset M_2$ is an inclusion of finitely generated projective $D$-modules with common generic rank $r > 0$ then an isomorphism $M_2 \simeq D^{r-1} \oplus L$ with invertible $L$ (which always exists, necessarily with $L \simeq \det M_2$ as $D$-modules) can be chosen so that the direct sum decomposition is compatible with $M_1$ in the sense that $M_1$ is the direct sum of its intersections with each of those rank-1 direct summands of $M_2$; i.e., it identifies $M_1$ with $J_1 \oplus \dots \oplus J_{r-1} \oplus J_r L$ for nonzero ideals $J_1, \dots, J_r \subset D$.

The case $r=1$ is obvious, so we may assume $r > 1$. Since $D^{r-1}$ occurs as a direct summand of $M_2$, we can find a quotient map $q_2:M_2 \twoheadrightarrow D$. Let $N_2 = \ker q_2$, and $N_1 = M_1 \cap N_2 = \ker q_1$ where $q_1 = q_2|_{M_1}$. Note that $q_1(M_1) \subset D$ is a nonzero integral ideal $I$.

Pick any $m_2 \in M_2$ so that $q_2(m_2) = 1 \in D$, so $I m_2 \subset M_1$. This latter $D$-submodule provides a compatible splitting of $q_1:M_1 \twoheadrightarrow I$. Continuing via induction on $r > 1$, we get compatible decompositions of $M_2$ and $M_1$ as direct sums of invertible $D$-modules in such a way that all but one of the invertible $D$-modules using for $M_2$ is actually free.


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well I appreciate your answer, but I can't really work it out. I'm not even sure that it proves, what we set out to prove - i'd indeed be thankful for a reference though! – user75148 Feb 10 '14 at 15:39
@user75148: It does prove what you asked. I'm not aware of any specific reference with this result. If you have a friend or colleague who knows a bit more commutative algebra, perhaps if they look at this answer then they can try to explain it to you in terms that are more accessible (I don't know your background). As a student, I found that many puzzling facts about Dedekind domains became far more transparent once I learned some more commutative algebra, to enable me to attack questions about ideals in the more robust framework of abstract projective and invertible modules. – user76758 Feb 10 '14 at 16:02

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