Decomposite a vector field into two parts Let A be a region in $\mathbb R^3$, and suppose $ \vec {\mathbf F}$ is a smooth vector field on A. I was asked to show that I can write $\vec {\mathbf F}=\vec {\mathbf F_1}+\vec {\mathbf F_2}$, s.t. $\operatorname{rot}(\vec {\mathbf F_1})=0, \operatorname{div}(\vec {\mathbf F_2})=0$. How can I show this? I think this has something to do with physics? 
 A: I think this is right--I hope it's right, but I'm digging into memory with no notes or books; anyway, here goes:  Consider $\nabla \cdot F$; it is a scalar function on $A$.  Consider the Poisson equation on $A$ $\nabla^2 \phi = \nabla \cdot F$.  Find (or, better in the present context, assume the existence of) such a $\phi$.  Then $\nabla \cdot (F - \nabla \phi) = 0$ on $A$.  Invoke the classical result that a divergence-free field is a curl (a "rot" in the OP's terminology), thus there exists a vector field $C$ on $A$ such that $\nabla \times C = F - \nabla \phi$; then $F =  \nabla \times C + \nabla \phi$, with $\nabla \cdot \nabla \times C = 0$ (since it's a curl--a "rot", if you will) and, of course, $\nabla \times \nabla \phi = 0$, since $\nabla \phi$ is a gradient.  Whew!  This stuff in fact has everything to do with physics; check out electromagntism and the equations of J. C. Maxwell; see Feynman's Lectures, vol. II.
This question and answer are deeply related to this one.
Best of success in these endeavors.  Cheers.
