(Exercise from an introductory course in homological algebra)
Whenever $A$ is a commutative ring with unit and $M$ an $A$-module, the following holds: $$M\oplus A\cong A\oplus A \Rightarrow M\cong A.$$
There's already an answer to this question ($M\oplus A \cong A\oplus A$ implies $M\cong A$?) yet I need a proof which does not involve tensor algebras, and the other answer using determinants doesn't seem 'natural' to me...
What I have concluded up 'til now is that $M$ is projective (hence torsion-free) and the s.e.s.
$$0\rightarrow A\rightarrow A\oplus A\rightarrow M\rightarrow 0 $$
splits (the first arrow however needn't be a canonical inclusion). $M$ is generated by two elements because of the epi from $A\oplus A$ but they are linearly dependent: does this imply that $M$ is generated by one element? Adding this to the torsion-free property, wouldn't it suffice to show that $M$ is in fact isomorphic to $A$?