I have been given the following problem:
Let $\alpha$ be a nowhere-zero 1-form. Prove that for a (p+1)-form $\beta$ $(p\geq0)$, one has $\alpha \wedge \beta = 0$ if and only if $\beta = \alpha \wedge \gamma$ for some p-form $\gamma$. [Hint: You might like to do it on $\mathbb{R}^n$ fist. Partition of unity is useful in the general case].
It is true that $\alpha \wedge \alpha = 0$ for 1-forms (and $k$-forms whenever $k$ is odd I believe), so the if statement is easy. However, for the converse, I couldn't find an argument even on $\mathbb{R}^n$. I tried to do this locally first by writing $\alpha = \sum_i \alpha_i(x) dx_i$, $\beta = \sum_J \beta_J(x) dx_J$ where $J = j_1j_2...j_{p+1}$ and $1 \leq j_1 \leq j_2 \leq...\leq j_{p+1} \leq dim(M)$ for whatever the dimension of M might be. Then
$\alpha \wedge \beta = \sum_{i,J} \alpha_i(x) \beta_J(x) dx_i\wedge dx_J$.
It is given that $\alpha$ is nonvanishing, but nothing is known about $\beta$. The sum seems to be a combinatorial thing if you want to arrange it into a form with $dx_i\wedge dx_J$ all distinct. Anyhow, I cannot see how to rewrite this expression in such a form that it becomes clear that one can sort of factor out $\alpha$.