# Uniqueness of the decomposition of finitely generated modules over Dedekind domains

If $R$ is a Dedekind domain and $M$ a finitely generated $R$-module, then $M$ splits as a direct sum of a torsion and a projective $R$-module. Is such a splitting unique? And what if we ask about uniqueness up to isomorphism (of the submodules that occur in the splitting)?

I understand the second question might seem trivial at first since by the classification all localizations of the submodules will be isomorphic... but who guarantees that this the isomorphisms of the localized submodules all come from an isomorphism of the original submodules?

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Your initial statement is not correct. What if $M$ is torsion? –  Mariano Suárez-Alvarez Dec 1 '12 at 3:26
Sorry, I meant torsion not torsion-free (the torsion-free part is the projective part). I fixed it above. Thank you! –  Bernard Dec 1 '12 at 3:28
I suggest first thinking about your question in the simplest case where $R$ is the integers. Then you're asking if writing a finitely generated abelian group as $T \oplus F$, where $T$ is torsion and $F$ is finite free, is unique. There are easy counterexamples. Try $\pm 2^{\mathbf Z} = \pm(-2)^{\mathbf Z}$, so you can take $F = 2^{\mathbf Z}$ or $(-2)^{\mathbf Z}$. –  KCd Dec 1 '12 at 5:20

The decomposition of $M$ as a direct sum of a torsion module and a projective module is not unique. Given any such decomposition $M=T\oplus P$, and any $R$-module morphism $\varphi: P\to T$, we can form the "twisted" decomposition $M=T\oplus P'$, where $P'=\{(t,p)\in T\oplus P=M: t=\varphi(p)\}$.

These decompositions have $P\cong P'$.

Because projective modules over Dedekind domains are torsion-free, the torsion submodule is unique (it is the set of all torsion elements of $M$). The "projective complement" will always be abstractly isomorphic to the quotient of $M$ by the torsion submodule.

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Suppose $M=N\oplus P$ with $P$ and $N$ arbitrary. Pick any $R$-linear map $\phi:P\to N$ and consider the submodule $P'=\{p+\phi(p):p\in P\}\subseteq M$. Then you can easily check that $M=N\oplus P'$ and that $P\cong P'$.

If $P$ is projective, then there are non-zero maps $\phi:P\to N$ and for those choices then $P'\neq P$.

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