Recall from the last post that if R is a commutative ring, we define K_0(R) to be the Grothendieck group of the isomorphism classes of finitely generated projective R-modules. It is natural to ask what happens if we replace finitely generated projective modules with countably generated projective modules. Let us write \mathfrak{K}_0(R) for this group. It turns out that understanding \mathfrak{K}_0(R) is extremely easy.

Theorem: For any commutative ring R, \mathfrak{K}_0(R)=0.

Proof: We have to show that if A, B, C, and D are countably generated projective R-modules, there is some countably generated projective R-module E so that A\oplus D\oplus E\cong B\oplus C\oplus E. Define E=\bigoplus_{i=1}^\infty (A\oplus B\oplus C\oplus D). Hence \mathfrak{K}_0(R)=0.

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 G and H be any two groups. Show that there is some ring R so that R[G]\cong R[H]. (Here R[G] is the ring of finite R-linear combinations of elements of G, and multiplication is defined by the group multiplication of G.)

Solution: Let K=(G\times H)\times(G\times H)\times\cdots, and set R=\mathbb{Z}[K]. Then R[G]\cong R[H].

Lam makes the comment that, although consideration of the group rings \mathbb{Z}[G] and K[G] are very useful for determining properties of G (for instance, the modules over these rings are the objects of study in group cohomology and representation theory, respectively), the group ring R[G] for an arbitrary ring R might not give us much information about G.

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 R be a ring and M a right R-module. Define M^\ast=Hom_R(M,R) and S=Hom_R(M,M). Then M and M^\ast are S-R and R-S bimodules, respectively. Furthermore, we can define multiplications M^\ast M\subseteq R by m^\ast m=m^\ast(m) and MM\ast\subseteq S by mm^\ast(m')=m(m^\ast m'). We say that M is a progenerator if MM^\ast=S and M^\ast M=R.

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