how to prove that exist one scalar function $f$ such that $\omega=f \eta\wedge \theta$ How to prove that exist one scalar function $f$ such that $\omega=f (\eta\wedge \theta)$. The question is: let $\eta,\theta\in \Omega^1(A)$ where $A\subset \mathbb R^3$ open set, where $\eta\wedge \theta(x)\neq 0 $ for all $x\in A$. If $\omega$ is two form  that satisfy $\omega\wedge \eta= \omega\wedge \theta=0$. Show that exist $f:A\to \mathbb R$ such that $\omega=f(\eta\wedge\theta)$.
We try the next:  choose $B_i=\{x\in \mathbb R^3: \eta_i\wedge\theta_i(x)\neq 0\}$. Hence $\eta\wedge \theta(x)\neq 0 $ then $R^3= \cup_{i=1}^3 B_i$ so if I will defined  $f_i= \frac{\omega_i}{\eta_i\wedge\theta_i}$ and maybe this work? but i dont use the hypothesis $\omega\wedge \eta= \omega\wedge \theta=0$.
the other side i thought too in to do this:  like $\omega\wedge \eta= \omega\wedge \theta=0$ multiply with $wedge$ product in both side in the above equality then $\omega\wedge \eta\wedge \theta= \omega\wedge \theta\wedge \theta=0$ so when I will applicated $x\in A$ we have
$\omega \wedge (\eta\wedge\theta)(x)=0$ but $\eta\wedge\theta(x)\neq 0$ so using properties of the wedge product we have $\omega =c (\eta\wedge\theta)$. In this case maybe $f=c$. Is possible? I have two ideas but I not sure if both are right, or maybe just one, or maybe need other side, somebody can help me please? Thank you so much.
 A: This is true more generally than in $\Bbb R^3$. An approach you should consider is to start with the linear algebra, thinking just at a point. Then you can make this argument work locally on a manifold (or on an open subset of $\Bbb R^n$) and check smoothness.
If $\eta(p)$ and $\theta(p)$ are linearly independent elements of $\Bbb R^{n*}$ (so, $1$-forms at a point), then you can extend to a basis. Then, knowing what a basis for $\Lambda^2(\Bbb R^{n*})$ is, you write $\omega(p)$ as a linear combination of basis elements and deduce from $\omega(p)\wedge\eta(p)\wedge\theta(p)=0$ that $\omega(p)$ must be a scalar multiple of $\eta(p)\wedge\theta(p)$.
(Notationally, if will probably be easier to write $\eta=\theta_1$ and $\theta=\theta_2$, extend to a basis $\theta_1,\dots,\theta_n$, and then recall that $\theta_i\wedge\theta_j$, with $i<j$, form a basis for $\Lambda^2(\Bbb R^{n*})$. Then you write $\omega = \sum\limits_{i<j}c_{ij}\theta_i\wedge\theta_j$. Then show that only $c_{12}\ne 0$; for example, to see $c_{13}=0$, consider $0=\omega\wedge\theta_2 = -c_{13}\theta_1\wedge\theta_2\wedge\theta_3 + \dots$, and use linear independence of $\theta_i\wedge\theta_j\wedge\theta_k$ for $i<j<k$.)
