4 There are quite few-equivalent-definitions of projective module. I would like to think about projective modules in terms of the Hom Hom functor and short exact sequences as both structures are …

In my opinion, the very basic projective non-free modules are (fractional) ideals in number rings (and Dedekind domains); this particular case of Pic is very intriguing for those who are interested in …

Being projective is indeed a local property for the Zariski topology. In fact, it is even local for the fpqc topology --- this is a famous theorem of Raynaud and Gruson (see MR0308104).

Dec 2, 2021 · The two ways I know of come down to the fact that projective modules, being summands of free modules which are flat, must also be flat; or, to use derived functors using projective …

Oct 25, 2018 · A free module is rather intuitive: there exists a subset satisfying suitable properties which may act as a basis for elements. Taking as example an $\\mathbb R$-module, or a vector space, we …

9 I'm able to prove it for finitely generated modules, by appealing to the characterization of a projective module as a summand of a free module, and the fact that finite-rank free modules are isomorphic to …

Mar 2, 2013 · Is every projective $\mathbf {Z} [x]$-module free? Ask Question Asked 12 years, 8 months ago Modified 12 years, 8 months ago

Projective modules are direct summand of free modules. In particular they are submodules of a free module and hence they are free because the ring is a PID (this last theorem is usually proved in the …

The wikipedia article on projective modules contains some nice information, in particular sketching an example of such a module over a Boolean ring. For a Boolean ring though the localization at every …

Nonfree projective module over a regular UFD? Ask Question Asked 14 years, 10 months ago Modified 2 years, 11 months ago