In $A$-Mod, $M\oplus A\cong A\oplus A$ implies $M\cong A$

Question

(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 $A$?

Answer 1

How do you use the splitness of your exact sequence? Once done, it is easy. So, let me call the middle term with basis $e_1,e_2$ and the map on the left by $i$. Then $i(1)=ae_1+be_2$. This splits says we have a map $j:Ae_1\oplus Ae_2\to A$ with $j\circ i$ identity. If $j(e_i)=c_i$, we see that $ac_1+bc_2=1$. Let $v_1=i(1), v_2=-c_2e_1+c_1e_2$. Then using the above equation, we immediately see that $Ae_1\oplus Ae_2=Av_1\oplus Av_2$. But, now $i(1)=v_1$ and the rest is clear.