If $M\oplus M$ is free, is $M$ free? If $M$ is a module over a commutative ring $R$ with $1$, does $M\oplus M$ free, imply $M$ is free? I thought this should be true but I can't remember why, and I haven't managed to come up with a counterexample.
I apologize if this has already been answered elsewhere.

This would mean that there is no element of order 2 in a $K$-group, which is clearly not correct. To find an example, you can try to find a ring $R$ such that $K_0(R)$ has an element of order 2. As an example related to my research interest, I can say $K_0(C(\mathbb{RP}^2))= \mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}$, see Karoubi 1978, IV.6.47.
