Let $R$ be a commutative ring and $(e_1, \dotsc, e_n)$ be a basis for $R^n$, the free $R$-module of rank n. Let $A$ be an $n \times n$ matrix with entries in $R$. Let $f_i = \sum \limits_{j=1}^n a_{ij}e_j, 1\leq i \leq n$. Show that $(f_1, \dotsc, f_n)$ form a basis for a free sub-module $K$ of $R^n$ if and only if the determinant of A is not a zero-divisor in the $R$.
Thank you for your help.
My thought: In class, we proved that if D is a principle ideal domain, sub-modules of finitely generated free D-module are also free. I'm still thinking about the relationship between det(A) and the divisor of R