Stably free and the Eilenberg swindle

Posted by Jason Polak on 29. March 2018 · 2 comments · Categories: modules · Tags:

I already mentioned the idea of stably isomorphic for a ring $R$: two $R$-modules $A$ and $B$ are stably isomorphic if there exists a natural number $n$ such that $A\oplus R^n\cong B\oplus R^n$.

Let's examine a specific case: if $A$ is stably isomorphic to a free module, then let's call it stably free.

So, to reiterate: a module $A$ is called stably free if there exists a natural number $n$ such that $A\oplus R^n$ is free. We already saw an example of a stably free module, where $R = \mathbb{H}[x,y]$, the two variable polynomial ring over the quaternions.

One might wonder: why don't we allow infinite $n$ in this definition? It's because of the Eilenberg swindle, named after mathematician Samuel Eilenberg.

The Eilenberg swindle goes like this: suppose $P$ is a projective $R$-module. Then, there exists a module $Q$ such that $P\oplus Q \cong F$ where $F$ is a free module. Now, let $E = \oplus_{i=1}^\infty F$.

Then:
$$P\oplus E \cong P\oplus (P\oplus Q)\oplus (P\oplus Q)\cdots\\ P\oplus (Q\oplus P)\oplus (Q\oplus P)\oplus\\ F\oplus F\oplus F\oplus\cdots$$
Therefore, $P\oplus F$ is free. Hence, if we allowed infinite $n$ in the definition of stably free, every projective module would be stably free and there wouldn't be much point in the definition 'stably free'.

Here is an exercise for you:

Show that if $A$ is a stably free $R$ module that is not finitely generated, then $A$ is free.