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 $M^\ast M=R$.