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 .

### Like this:

Like Loading...

## 5 comments

Comments feed for this article

Friday, March 20, 2009 at 1:57 pm

ZygmundInteresting, I was just reading about this in a book on K-theory.

Friday, March 20, 2009 at 2:01 pm

complexzetaI read about this first in Rosenberg’s book on algebraic K-theory.

Friday, March 20, 2009 at 2:23 pm

ZygmundActually that’s the same book I am currently studying. It was an exercise though. I think Weibel probably mentioned the trick somewhere in his book on homological algebra, so I realized the connection.

Saturday, July 11, 2009 at 4:16 pm

Grothendieck Groups and the Eilenberg Swindle « Delta Epsilons[...] The Eilenberg swindle can also be stated in a slightly different form. [...]

Tuesday, December 22, 2009 at 8:24 am

Svadobne ucesyGreat post, thank you.