Let $x, y, z$ be the usual coordinates on $\mathbb{R}^{3}$. Consider the 1-form on $\mathbb{R}^{3}$ given by $\phi = dx+ydz$. Do there exist smooth functions $u$ and $v$ such that $\phi=u\ dv$? Why?
Attempt at answer: Suppose there existed functions $u=u(x,y,z), v=v(x,y,z)$ with the desired properties. Then expanding out we would have $dx+ydz=u\frac{\partial v}{\partial x}dx+u\frac{\partial v}{\partial y}dy+u\frac{\partial v}{\partial z}dz. $ Since $u$ is not the zero function, then $0=\frac{\partial v}{\partial y}$, hence $v=v(x,z)$ is independent of y. Then, $y=u\frac{\partial v}{\partial z}$ implies that $u$ does indeed depend on $y$ (since the partial derivative does not). However, $1=u\frac{\partial v}{\partial x}$ does not depend on y, a contradiction.
