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If $R$ is domain, as a projective module always exist over R. But how to produce such a module over $R$.

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    Is there a typo somewhere? You write (markup added) "If $R$ is any **domain** ..." and "... arbitrary **Domain**." Sure, you meant domain in both places?2012-10-18
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    How about $R$ as an $R$-module (the so called left-regular $R$-module)? It is free, which means it is also projective.2012-10-18
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    @Oliver Thanks, but how to produce another projective module other than $R$2012-10-18
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    @martini edited question, thanks2012-10-18
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    In general, you *cannot* produce projectives which are not free, because there exists rings such that all projectives are free.2012-10-18
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    @MarianoSuárez-Alvarez, thanks for your reply, thats the answer I want.2012-10-18

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Every free module over $R$, that is to say $R,R^n,$ or $R^{\oplus\kappa}$ for any cardinal $\kappa$, is projective.

We can't give any other examples in general. Since projective modules are submodules of free modules, in a principal ideal domain every projective is free, since submodules of free modules are direct sums of ideals, and principal ideals in an integral domain are isomorphic to $R$ as modules. The same equivalence of projective and free holds in local rings.

There are lots of specific examples of projective-but-not-free modules over at Wikipedia. I think the most interesting ones are $R^n$ as an $M_n(R)$ module under left multiplication and every (direct sum of) non-principal ideal(s) in a Dedekind domain such as a ring of algebraic integers.

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    thanks. so it is in general not possible to produce projective modules over $R$ except $R^k$2012-10-18
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    $R^\kappa$ for an infinite cardinal $\kappa$ is _not_ free in general. You mean $R^{\oplus \kappa}$.2012-10-18
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    Thanks-I knew I meant that, but I guess that notation really means product, not sum. @user33263: that's right.2012-10-18