Surjectivity of differential operators with constant coefficients I would like a proof or a reference (or a counter-example...) for the following fact. Let $P\in \mathbb{C}[x_1,\ldots ,x_n]$ and   $D\in \mathbb{C}[\frac{\partial }{\partial x_1} ,\ldots ,\frac{\partial }{\partial x_n}]$ be  nonzero homogeneous polynomials. Then there exists a homogeneous polynomial $Q\in \mathbb{C}[x_1,\ldots ,x_n]$ such that $D(Q)=P$. 
 A: Too long for a comment. I want to use a version of the Lojaciewicz theorem of division of distributions by an analytic function (in fact Hörmander's result of division by a polynomial). We may assume that $P(x) =x^\alpha=x_1^{\alpha_1}\dots x_n^{\alpha_n}$  a monomial homogeneous with degree $\vert \alpha\vert=\alpha_1+\dots +\alpha_n$. Let us replace your notation $D$ by an operator $$A(D)=\sum_{\vert \beta\vert =m}a_\beta D^\beta,\quad D=-i\nabla.$$
The question at hand is to find an homogeneous polynomial $Q$ such that
$
A(D) Q= x^\alpha.
$
By Fourier transformation, it is equivalent to solve
$$
A(\xi)\widehat Q(\xi)=i^{\vert \alpha\vert}\delta^{(\alpha)},
$$
which means divide a derivative of the Dirac mass by an homogeneous ($A$ is assumed to be non-zero) polynomial. It is indeed possible by the aforementioned results of division and we find that $\widehat Q$ is homogeneous with degree $-n-\vert \alpha\vert-m$, so that $Q$ is homogeneous with degree $\vert \alpha\vert+m$. Moreover $\widehat Q$ is supported at the origin, proving that $Q$ is a polynomial.
A: Here is another approach. Let $R$ be a non-zero homogeneous polynomial of degree $n$. We want to show that the mapping $Q\mapsto R(\partial)Q$ is surgective from $V_{m+n}$ to $V_m$ where $V_k$ is the space of homogeneous polynomials of degree $k$. Note that $\langle A,B\rangle=[A(\partial)\bar B](0)$ is a scalar product on $V_k$ (with monomials forming an orthogonal basis). If our mapping is not surjective, then there exists a non-zero polynomial $S\in V_m$ such that
$\langle S, R(\partial) Q\rangle=\langle S\bar R,Q\rangle=0$ for all $Q\in V_{n+m}$. But $S\bar R$ is a non-zero polynomial, so taking $Q=S\bar R$, we get a contradiction.
A: If $V_a$ is the    space of homogenous polynomials in $x_1,...,x_n$
of degree $a$ and $V_a^\vee$ its dual, then differentiation is a
$GL_n$-equivariant bilinear
map $V_{d+e}\times V_e^\vee\to V_d$.
Your question is whether, given non-zero $D\in V_e^\vee$, the linear map
$V_{d+e}\times\{D\}\to V_d$ is surjective.
For this you can specialise $D$ to define a closed $GL_n$-orbit in
$\mathbb P(V_e^\vee)$. That is,    you can take $D=\partial^e/\partial x_1^e$
and then the map is obviously surjective.
