If $R$ is domain, as a projective module always exist over R. But how to produce such a module over $R$.
2026-03-30 12:28:39.1774873719
Projective module over a ring
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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.