Show that a k-form can be expressed as wedge product

Question

I'm trying to show that given a $1$-form $\omega$ and a $k$-form $\alpha$ such that $\alpha \wedge \omega = 0$ then there exists a $(k-1)$-form $\beta$ such that $\alpha = \omega \wedge \beta$.

I'm struggling to show this even in the case of multi-linear algebra and forgetting forms. I've been trying to use the fact that if $v_i$ are a basis for $V$ then $\{v_{i_1}\wedge...\wedge v_{i_k} : 1 \le i_1 < ... < i_k \le n\}$ is a basis for $\bigwedge ^k V$.

I thought perhaps we could extend $\omega$ (I'm talking just in the multilinear algebra case here and not thinking about manifolds - I'll worry about that after!) to a basis of $V$ and then $\alpha \wedge \omega = 0$ kills any terms containing $\omega$ in the expansion of $\alpha$ but I'm struggling to make any further progress,

Thanks for any help

Answer 1

In terms of the multilinear algebra, it looks like you're on the right track. We extend $\omega$ to a basis $\omega, v_2, \dots, v_n$ and then write

$$\alpha = \sum a_I \omega \wedge v_{i_2} \wedge \dots \wedge v_{i_k} + $$ $$ \sum b_J v_{j_1} \wedge \dots \wedge v_{j_k} $$

Then as you've noted, wedging with $\omega$ kills everything in the first term. Now if you look at the second term of the above after wedging with $\omega$, what does this say about the constants $b_J$?

As for the case for forms, it's not literally true as stated, you probably need something like $\omega$ is never zero. For example take $\alpha = dx \wedge dy$ and $\omega = xdx$ on $\mathbb{R}^3$. Then $\alpha \wedge \omega = 0$, but $\alpha$ can never be written $\omega \wedge \beta$ for any $\beta$, since such a form will necessarily vanish anywhere $x=0$.

Edit: changed the example slightly