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Hello I am stuck showing the following:

If $M$ is a finitely generated projective module over a Dedekind ring $R$, then $M\cong\bigoplus_{i=1}^k\mathfrak{a}_i$ for some ideals $\mathfrak{a}_i\subseteq R$. I have been told to try and localize and induct, however, I am unsure how to proceed on this.

I know that localization will give me that $M_\mathfrak{p}$ is projective, which will imply that $M_\mathfrak{p}$ is free over $R_\mathfrak{p}$ (as $R_\mathfrak{p}$ is a PID). Perhaps I could use this to get an ideal decomposition over the localization and use this to provide one over the original ring, but I am not sure how to go between them.

Thank you for any help.

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First prove that the localizations of $M$ are free of the same rank, say $k$. Then use induction on $k$. –  user26857 Mar 18 '13 at 22:02
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1 Answer 1

One can prove it directly without localizations.

Assume $M \neq 0$. Since $M$ is projective, there is a nontrivial homomorphism $M \to R$, whose image is an ideal $\mathfrak{a} \neq 0$. Since $R$ is Dedekind, $\mathfrak{a}$ is projective. Hence, there is an isomorphism $M \cong \mathfrak{a} \oplus M/\mathfrak{a}$. Now continue with $M/\mathfrak{a}$ (which is again f.g. proj. as a direct summand of $M$). This has to end since $M$ is noetherian.

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Thank you, but why is it that there is a homomorphism from $M$ to $R$? –  user67361 Mar 19 '13 at 2:37
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$M$ is a direct summand of some $R^n$, so you get $n$ projection homomorphisms $M \to R$. At least one of them is nontrivial because $M \neq 0$. –  Martin Brandenburg Mar 19 '13 at 12:36
    
Alternatively, this follows by Dual Basis Theorem. –  messi Jun 25 '13 at 9:05
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