Wedge product on free module

Question

Let $R$ be a unital commutative ring, let $M$ be a free module of rank $n$ over $R$ with a basis $e_1,\dots,e_n$. Suppose that $m_1,\cdots,m_n$ are elements of $M$ such that $$m_1\wedge\cdots\wedge m_n=ce_1\wedge\cdots\wedge e_n$$ for some invertible $c\in R^{\times}$. How to show that $m_1,\cdots,m_n$ is also a basis for $M$? I can show that $m_1,\dots,m_n$ are linear independent, but I don't know show that they span $M$.

Answer 1

Maybe there is a direct exterior algebra argument, but otherwise you can write $M\in M_n(R)$ the matrix of $(m_i)$ in the basis $(e_i)$, and notice that $c = \det(M)$.

Then it is well-know that a matrix with invertible determinant is invertible (you just have to use the usual formula with the comatrix which holds over any commutative ring), so each $e_j$ can be expressed as a linear combination of the $m_i$.