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Show that if $R$ is a Dedekind domain, then every projective $R$-module (not necessarily finitely generated) is a direct sum of ideals of $R$.

I have spent a while on this problem and I wonder if it is true that every nonzero (integral) ideal of $R$ is projective as an $R$ module. If so, can we use this fact to prove the result above?

  • Some related discussion. The existence of a class group (of classes of fractional ideals) is the key there. It sounds like you are working in a more general setting though. (+1+1 :-) – Jyrki Lahtonen Mar 30 '15 at 19:20
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    @user_m You may want to look up "hereditary rings" which are characterized by having projective ideals. Hereditary domains are exactly the Dedekind domains. – rschwieb Mar 30 '15 at 19:21
  • @rschwieb Thanks! I will just avoid using the terms hereditary rings or domains just to avoid any confusion! –  Mar 30 '15 at 19:37

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I figured it out: First, one can show that R is Dedekind domain if and only if every (integral) ideal is invertible if and only if every (integral) ideal is projective when seen as a $R$-module. Following the proof outlined in Hilton and Stammbach's A Course in Homological Algebra for the fact that every submodule of a free $R$-module $M$ where $R$ is a PID is free, one can show every submodule of a $R$-module $M$ for $R$ a domain in which every ideal is projective when seen as an $R$-module, is isomorphic to a direct sum of the ideals of $R$, which then implies the result about projective modules over Dedekind domains we hope to prove.