Eilenberg swindle

Remarks

Example: every projective module $P$ is a direct summand of a free module $F$, so the swindle applies to it. Take $R = \Z/6\Z$ and $P = 3\Z/6\Z$, $Q = 2\Z/6\Z$; then $P \oplus Q \cong R$, and the swindle gives $P \oplus R^{(\infty)} \cong R^{(\infty)}$, although $P$ itself is not free.

Non-example: the regrouping needs infinitely many summands. With finitely many summands the argument breaks. Indeed, for $R$ commutative and $P \ne 0$, the isomorphism $P \oplus R^n \cong R^n$ fails. To see this, compare dimensions over the residue field at a prime where $P$ is nonzero.

For this reason the Grothendieck group $K_0(R)$ uses only finitely generated projective modules: with countably generated ones every class would vanish.