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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? – martini Oct 18 '12 at 7:17
How about $R$ as an $R$-module (the so called left-regular $R$-module)? It is free, which means it is also projective. – Oliver Braun Oct 18 '12 at 7:17
@Oliver Thanks, but how to produce another projective module other than $R$ – Ram Oct 18 '12 at 7:36
@martini edited question, thanks – Ram Oct 18 '12 at 7:37
In general, you cannot produce projectives which are not free, because there exists rings such that all projectives are free. – Mariano Suárez-Alvarez Oct 18 '12 at 7:57
up vote 3 down vote accepted

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 $R^k$ – Ram Oct 18 '12 at 8:00
$R^\kappa$ for an infinite cardinal $\kappa$ is not free in general. You mean $R^{\oplus \kappa}$. – Zhen Lin Oct 18 '12 at 8:23
Thanks-I knew I meant that, but I guess that notation really means product, not sum. @user33263: that's right. – Kevin Carlson Oct 18 '12 at 8:31

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