## Projective Modules (Pt. II)

**Point of Post: **This is a continuation of this post.

Now that we know projective modules (over any ring) always exist, the next obvious question is: is everything projective? Indeed, it is entirely plausible at first sight that all modules over a given ring are projective. To see why this is not true suppose that is a projective -module and suppose that we an epimorphism . Then, letting , in gives us a map such that . In other words, the map has a backmap. Thus, by the splitting lemma given any short exact sequence we can conclude this sequence splits so that, among other things, . From this we see that any time a ring has the property that there exists non-splitting short exact sequences in then there exists non-projective -modules. For example, since the usual short exact sequence is non-splitting in we have that is not a projective abelian group. Of course, we have already alluded to the fact that all short exact sequences ending in a given module splitting is actually equivalent to projectivity. Let’s see if we can figure out why this is true. We are now stuck with a diagram as in , and we want some kind of surjection onto . Of course, nothing in our diagram is going to give us what we want, and so we must introduce some outside source to surject onto . So, take to be some free module such that there exists an epimorphism . Now, since is free we know there exists some map such that . But, now we have something to work with! Namely, we know that there exists a map such that and so define . Then, is an -map and . Ta-da! Thus, we can conclude that

**Theorem: ***Let be a given ring and a left -module. Then, is projective if and only if every short exact sequence ending in splits in .*

Cool, so that’s one equivalence down. But, it leads us very naturally into the next equivalence. I mean, suppose that is projective, then any time we have epimorphisms onto we have valuable information. But, what is the richest source of epimorphisms onto ? Free modules of course! Namely, we know from first principles that we can find an epimorphism where is some free module, and if is the kernel of that map then inclusion into gives us the beautiful short exact sequence . But! We know that this splits, so that . Thus, every projective module is a direct summand of a free module. Conversely, suppose that is a module such that we can find a module with free. Suppose then that we have an epimorphism and a map . Note then that if denotes the usual projection then pre-composing with gives us a map . Since is projective we know there exists such that . Now, if is the canonical map consider . then, is a map and and so everything works out great! In particular, we have the following:

**Theorem: ***Let be a left -module. Then, is projective if and only if there exists a module such that is free. Moreover, if is finitely generated we can take to be finitely generated.*

This is, by and far, the most “functional” of the equivalent definitions of projectivity–it allows us to prove a lot of theorems concerning projective modules. For example:

**Theorem: ***Let and be projective left -modules. Then, any direct summand of is projective, as is .*

It also completely solves for us the case of projective modules over PIDs. Indeed, we can recall that submodules of free modules, and so considering our most recent characterization of projective modules we have the following:

**Theorem: ***Let be a PID, then a left -module is projective if and only if is free.*

So, this clinches it for abelian groups, for example. Interestingly, the ‘converse’ of this theorem is not true. Namely, there are rings which are not PIDs but which have the property that every projective module is free. It is a celebrated theorey of Kaplansky that local rings are amongst such rings. Moreover, this enables us to prove that, in general, the product of projective modules is not projective. Indeed, for abelian groups this would be equivalent to the statement that the product of free modules is free, which we have shown before this is not true in general.

Using our most recent characterization of projective modules (once again!), we can show that, in general, there are projective modules which aren’t free. For example, note that, as -modules, one has that so that both and are projective, but evidently not free.

We are almost completely done with proving the equivalence formulations for projectiveness in modules. The one conspicuous one we have yet to mention is the fact that projective modules are precisely those for which the covariant Hom functor is fully exact. Let’s see why this makes sense. Well, since we know the Hom functor is left exact the meat of the problem is that if we fix the first entry of the Hom functor to be projective, then epis should be taken to epis. Why is this true? Well, let’s see exactly what this should mean. Suppose for a second that we had some epi and some module . Then, we see that we have the induced map given by . We want this to be surjective, right? Well, this means that given some map we should be able to find a map such that . Looking familiar, isn’t it? Indeed, we see that a module makes exact if and only if whenever we have an epi and a map we can lift to find a map making our favorite diagram commute! Thus, we see, just by merely unraveling the definitions, that:

**Theorem: ***Let be a ring and a left -module. Then, is projective if and only if the covariant Hom functor is exact*

**References:**

[1] Dummit, David Steven., and Richard M. Foote. *Abstract Algebra*. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. *Advanced Modern Algebra*. Providence, RI: American Mathematical Society, 2010. Print.

[3] Blyth, T. S. *Module Theory.* Clarendon, 1990. Print.

[4] Lang, Serge. *Algebra*. Reading, MA: Addison-Wesley Pub., 1965. Print.

[5] Grillet, Pierre A. *Abstract Algebra*. New York: Springer, 2007. Print.

[…] are going to discuss now the “dual” notion to projective modules which, as one would expect, are just the modules one gets by dualizing the lifting axioms for […]

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