$df\wedge \alpha=0$ implies $\alpha=g\,df$ for some $g$

Question

Let $A \subset \mathbb{C}$ be an open set and $f\in {C}^{\infty}(A)$ with $df\neq0$.

I consider $\alpha$ a $1$-form such that $df\wedge \alpha=0$ and I want to prove that exists a function $g$ such that $\alpha=g\:df$.

I have that $df=f_x\,dx+{f}_y\,dy$ and $\alpha=P\,dx+Q\,dy $ for some $P,Q\in C^\infty(A) $.

Imposing

\begin{align} 0 & =df\wedge \alpha=(f_x\,dx+f_y\,dy) \wedge(P\,dx+Q\,dy) \\[6pt] & =f_x\,dx\wedge Q\,dy+f_y\,dy\wedge P\,dx \\[6pt] & =(f_x Q-P f_y)\,dx\wedge dy \end{align}

I obtain $f_x Q=P f_y$.

How can I prove now that exists a $g$ such that $P=gf_x$ and $Q=gf_y$?

Thanks for the help!

Answer 1

Since $df$ is nowhere $0$, you get a basis for the $1$-forms by taking $df, \star df$. (If you aren't familiar with $\star$, it's defined by $\star (a dx + b dy) = a dy - b dx$. You can check that $df\wedge\star df\ne 0$; hence the two $1$-forms are linearly independent.) So write $\alpha = s df + t{\star} df$ for some functions $s$ and $t$. Can you finish?