0
$\begingroup$

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

  • 0
    @MarianoSuárez-Alvarez, thanks for your reply, thats the answer I want.2012-10-18

1 Answers 1

3

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.

  • 0
    Thanks-I knew I meant that, but I guess that notation really means product, not sum. @user33263: that's right.2012-10-18