If $R$ is domain, as a projective module always exist over R. But how to produce such a module over $R$.
Projective module over a ring
-
0Is 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
-
3How 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
-
0@Oliver Thanks, but how to produce another projective module other than $R$ – 2012-10-18
-
0@martini edited question, thanks – 2012-10-18
-
4In general, you *cannot* produce projectives which are not free, because there exists rings such that all projectives are free. – 2012-10-18
-
0@MarianoSuárez-Alvarez, thanks for your reply, thats the answer I want. – 2012-10-18
1 Answers
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.
-
0thanks. so it is in general not possible to produce projective modules over $R$ except $R^k$ – 2012-10-18
-
0$R^\kappa$ for an infinite cardinal $\kappa$ is _not_ free in general. You mean $R^{\oplus \kappa}$. – 2012-10-18
-
0Thanks-I knew I meant that, but I guess that notation really means product, not sum. @user33263: that's right. – 2012-10-18