Finitely generated projective module $M$ over integral domain $R$ such that $M \oplus R \cong R^2$ and $M\otimes N \cong R$ for some module $N$

Question

Let $M$ be a finitely generated projective module over an integral domain $R$. Also assume that $M \oplus R\cong R^2$ and that $M$ is locally free of rank $1$ i.e. $M \otimes N \cong R$ for some $R$-module $N$. Then is it true that $M$ is free $R$-module ?

Answer 1

The statement $M\oplus R\cong R^2$ implies we have a presentation $0\to R\to R^2\to M\to 0$, and this splits. The map $R\to R^2$ is given by a vector $(a,b)$, $a,b\in R$ or if you fix this basis $e_1,e_2$ for $R^2$, the map is given by $1\mapsto ae_1+be_2$. This splits implies you have $c,d\in R$ such that $ad-bc=1$. Then, it is immediate that $ae_1+be_2, ce_1+de_2$ also form a basis for $R^2$. Thus, $M$ is generated by the single element, the image of $ce_1+de_2$.

Answer 2

$R=\bigwedge_R^2R^2=\bigwedge _R^2(M\oplus R)=\bigwedge^2_RM\oplus ( M\otimes_R R) \oplus \bigwedge_R^2R=0\oplus M\oplus 0=M$