Recall from the last post that if is a commutative ring, we define to be the Grothendieck group of the isomorphism classes of finitely generated projective -modules. It is natural to ask what happens if we replace finitely generated projective modules with countably generated projective modules. Let us write for this group. It turns out that understanding is extremely easy.
Theorem: For any commutative ring , .
Proof: We have to show that if , , , and are countably generated projective -modules, there is some countably generated projective -module so that . Define . Hence .
A similar construction shows up in the theory of group rings. Here’s an exercise from T.Y. Lam’s Exercises in Classical Ring Theory:
Exercise 8.16: Let and be any two groups. Show that there is some ring so that . (Here is the ring of finite -linear combinations of elements of , and multiplication is defined by the group multiplication of .)
Solution: Let , and set . Then .
Lam makes the comment that, although consideration of the group rings and are very useful for determining properties of (for instance, the modules over these rings are the objects of study in group cohomology and representation theory, respectively), the group ring for an arbitrary ring might not give us much information about .
There’s an interesting article I found on more general Eilenberg swindles, but the authors don’t define progenerators, so I’ll include that here.
Let be a ring and a right -module. Define and . Then and are and bimodules, respectively. Furthermore, we can define multiplications by and by . We say that is a progenerator if and .