Let $R$ be a commutative ring with $1\in R$. Let $M$ be an $R$-module. The exterior product of degree 2 is defined as $$M\wedge M:=\frac{M\otimes_RM}{\langle m\otimes m:m\in M\rangle}.$$
Let $x,y\in M$ be two elements. If there exists $m\in\mathbb N_+$ and $e_1,\cdots,e_m\in M$ and $r_1,\cdots,r_m,s_1,\cdots,s_m\in R$ such that
- $x=\sum_{i=1}^mr_ie_i$ and $y=\sum_{i=1}^ms_ie_i$;
- any $2\times 2$-minor of the $2\times m$-matrix $\begin{pmatrix}r_1&\cdots&r_m\\s_1&\cdots&s_m\end{pmatrix}$ is zero
then we have $x\wedge y=0$ in $M\wedge M$.
My question is about the converse.
If $x,y\in M$ and $x\wedge y=0$, can we find $m\in\mathbb N_+$, $e_1,\cdots,e_m\in M$ and $r_1,\cdots,r_m,s_1,\cdots,s_m\in R$ such that $(1),(2)$ hold?
When $M$ is a free module, the converse is true. Indeed when $M$ is locally free, the converse is also true. I wonder if this is true in other situations
- $M$ is projective;
- $M$ is flat.
