# Need for projective modules

I wanted to ask why we require projective modules. After studying all the essential ingredients my guess is -

Firstly, we worked with vector spaces (say modules over field $F$) (which are free modules) and in that we could extend any basis of a submodule to get a basis for whole $V$ and thus property P

P- "any submodule of $V$ is a direct summand of $V$" is satisfied.

But in general for $R$-modules we could not express any submodule to be a direct summand, and thus we coined semisimple rings whose definition is those rings $R$ for which every $R$ module $M$ satisfies P

But what inspired people to go for projective modules, what do they generalize?

Two equivalent definitions of a projective module P are-

• P is isomorphic to a direct summand of a free $R$ module.

• every exact sequence of the form $$0 \to M'\to M\to P\to 0$$ splits.

I was looking for the inspiration that led to the study of projective modules and how do they help in simplifying studies of modules?

• Interesting question. I guess one unsatisfying answer (that you've essentially already said) is that they generalize free modules in the sense that projective modules are summands of free modules. And furthermore, a projective module is free iff $R$ is a PID. – Derek Allums Jun 11 '15 at 17:16
• One thing you can do with finitely generated projective modules is taking traces of endomorphisms (and, if I am not mistaken, also determinants and characteristic polynomials). Also, the Hom and tensor functors are exact (not just left-exact, resp. right-exact). I think a good, if imprecise, conception of projective modules is "the kind of modules for which almost all basis-free linear algebra works". And by "basis-free", I mean "stateable without bases", not "provable without bases". – darij grinberg Jun 11 '15 at 17:21
• Taking Hom is a natural way to get a new $R$-module from two other $R$-modules (when $R$ is commutative). And fixing the first argument results in a functor. But this functor is not exact in general. Projective modules are those for which it is exact, and further they can be used to calculate what is "missing" in order to get an exact sequence from applying Hom (if I get time tomorrow, I may expand this into a proper answer). – Tobias Kildetoft Jun 11 '15 at 18:29
• @darijgrinberg Can you expand your answer? I want to know about properties of finitely generated projective modules as they are also used in defining grothendieck group, which gave me the motivation to ask this question. Also expand some on basis free linear algebra works. It ll be really helpful. Thanks! – Bhaskar Vashishth Jun 12 '15 at 8:03
• @TobiasKildetoft Thanks but I am aware of them making Hom functor exact – Bhaskar Vashishth Jun 12 '15 at 8:08

What the theorem says is that, for any compact Hausdorff space $X$, there is a one-to-one correspondence between vector bundles over $X$ and projective modules over the ring $C(X)$ of real-valued continuous functions on $X$. Specifically, the set of all continuous sections of a vector bundle is a projective $C(X)$-module, and every projective $C(X)$-module has this form.
The same statement holds on a smooth manifold if you replace $C(X)$ by $C^\infty(X)$ (the ring of smooth real-valued functions on $X$) and “vector bundle” by “smooth vector bundle”. There is also an algebraic geometry version over affine varieties involving the structure sheaf.