I would like a proof or a reference (or a counterexample...) 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 $D(Q)=P$.
3 Answers
Here is another approach. Let $R$ be a nonzero 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 nonzero 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 nonzero polynomial, so taking $Q=S\bar R$, we get a contradiction.

1$\begingroup$ Nice, thanks! In fact that works over any field, you just need to view $R(\partial )$ as an element of $V_n^{*}$. $\endgroup$– abxNov 21, 2018 at 17:27

3$\begingroup$ @abx I think the argument needs a field of characteristic 0, or at least characteristic not dividing $(m+n)!$ (here $n$ is not the number of variables) $\endgroup$– MizarNov 21, 2018 at 22:07

$\begingroup$ Right, thanks. This is needed for the duality between symmetric powers. $\endgroup$– abxNov 22, 2018 at 7:12

$\begingroup$ @abx more specifically to this argument, monomials are orthogonal but not orthonormal, and for a monomial $M$, the pairing $\langle M,M\rangle$ vanishes in characteristic $p$ if $p$ is less than or equal to one of the exponents of $M$. $\endgroup$ Nov 23, 2018 at 12:18
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 nonzero) polynomial. It is indeed possible by the aforementioned results of division and we find that $\widehat Q$ is homogeneous with degree $n\vert \alpha\vertm$, 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.

1$\begingroup$ Interesting, thanks! Though a bit of an overkill... $\endgroup$– abxNov 22, 2018 at 7:14
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 nonzero $D\in V_e^\vee$, 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.

1$\begingroup$ I'm not sure I understand the last two sentences, What is obvious (to me) from what was written before is that it is enough to get $x_1^d$ in the image, but that's not what you are saying. Can you elaborate a bit? $\endgroup$– fedjaNov 21, 2018 at 14:34

$\begingroup$ Same as @fedja: why is it enough to check surjectivity for one particular orbit? $\endgroup$– abxNov 21, 2018 at 14:38

$\begingroup$ In the projective space $\mathbb P(V_e^\vee)$ there is a unique closed orbit, the orbit of the highest weight vector $\partial^e/\partial x_1^e$. This is contained in the closure of every orbit. Let $U$ be the locus of points $D$ in $\mathbb P(V_e^\vee)$ for which $V_{d+e}\times\{D\}$ is surjective. Then $U$ is Zariski open and $GL_n$invariant, so it is enough to show that $U$ contains $\partial^e/\partial x_1^e$. For this the map is obviously surjective. $\endgroup$– inkspotNov 21, 2018 at 17:16