Let $R$ be a commutative ring, $M$ some $R$-module, and $m,n \in M$. Is there some criterion when $m \wedge n = 0$ in $\Lambda^2(M)$? There are some sufficient criterions, for example that $m \in \langle n \rangle$ (or $n \in \langle m \rangle$), or more generally that $a \cdot m + b \cdot n = 0$ for some $a,b \in R$ such that $(a,b) \cap (1+\mathrm{Ann}(m) + \mathrm{Ann}(n)) \neq \emptyset$. (Proof: $a (m \wedge n) = 0$ and $b (m \wedge n)=0$, hence $(a,b) (m \wedge n)=0$. If $ua+vb=1+x$ with $x \in \mathrm{Ann}(m) + \mathrm{Ann}(n)$, then $x (m \wedge n)=0$, hence $0 = (ua+vb)(m \wedge n) = m \wedge n$.)
I would like to know necessary-sufficient criterions. There is such a criterion for tensor products (Bourbaki, Algèbre commutative, ch. I, §2, no. 11, lemma 10).