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Are there some examples of projective modules which are not free modules? Thank you very much.

Pete L. Clark
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LJR
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    Take a non-principal ideal of a Dedekind domain. It is projective over the domain but is not free. More generally, projective modules feel like vector bundle, which is certainly not trivial in most cases. –  Mar 20 '13 at 06:42
  • For an explicit example of the general result described by Sanchez see my answer here. – Jyrki Lahtonen Mar 20 '13 at 08:18

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Let $k$ be your favorite field and consider the ring $R = k \times k$. Give $k$ the structure of an $R$-module via $$(a, b)\cdot c = ac.$$ Then $k$ is a projective $R$-module because it is a direct summand of the free module $R$ (it's the left factor: $k \times 0$) but it is not isomorphic to a direct sum of copies of $R$ so it is not free.

Jim
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    +1: In my opinion this is indeed the simplest example and should be given first. I also like to talk about Swan's Theorem and vector bundles in this context, and that perspective helps here: the geometric explanation is that a disconnected space yields very cheap examples of nontrivial vector bundles, namely those with different fiber dimensions over the connected components. – Pete L. Clark Mar 20 '13 at 07:36
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There is a discussion of projective versus free modules in $\S 3.5.4$ of my commutative algebra notes. See also $\S 6$ on Swan's Theorem, especially $\S 6.4$ for a discussion and examples of modules which are stably free but not free.

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Pete L. Clark
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