Let $A$ be a commutative, regular, finitely generated $k$-algebra, where $k$ is a field of characteristic 0.

Then Grothendieck proved that in this case the algebra of differential operators on $A$ is simply the subalgebra of $End_k(A)$ generated by multiplication by $A$ and $Der_k(A)$, $k$-linear derivations. Denote this algebra by $D(A)$.

I have read that $D(A)\subseteq End_k(A)$ is 'strongly dense', i.e. given any finite-dimensional subspace $M\subseteq A$ and any linear operator $L\in End_k(A)$, there exists a differential operator $D\in D(A)$ such that $L|_M=D|_M$. Does anyone know that proof of this fact?

My thought process has been to try and work locally first: we know that $Spec(A)$ has a cover by affines $Spec(A_f)$ such that $D(A_f)\cong A_f[\frac{\partial}{\partial x_1},\dots,\frac{\partial}{\partial x_n}]$, where $x_1,\dots,x_n\in A_f$ are a system of parameters at all closed points in $Spec(A_f)$. Here the derivations $\frac{\partial}{\partial x_i}$ are only required to satisfy $x_j\mapsto \delta_{ij}$ (if I understand things correctly). So it's easy to find a differential operator which does anything you like to a subspace generated by monomials in the $x_i$, but I don't see what to do about other elements.


1 Answer 1


I think it may be easier if you do not choose the open cover in advance, so that the cover may depend on $M$. Equivalently, let us work with local rings first.

Here is a sketch of the argument. I am keeping your notation: $A$ is smooth over a characteristic zero field $k$, $D(A)$ is the algebra of differential operators, and $M\subset A$ is a finite-dimensional subspace. Consider the evaluation map $$ev:D(A)\to Hom_k(M,A).$$ We want to show that $ev$ is surjective.

Note that $ev$ is a morphism of $A$-modules, so it is enough to check its surjectivity locally. (Explicitly, this step involves looking at partitions of unity.) That is to say, for every point $x\in Spec(A)$, we consider the local ring $A_x$, an we need to prove surjectivity of the map $$ev_{A_x}:D(A_x)=A_x\otimes_A D(A)\to Hom_k(M,A_x).$$ Moreover, $Hom_k(M,A)$ is a finitely generated $A$-module (in fact, it is free of finite rank), so it suffices to consider closed points $x\in Spec(A)$.

Let now $x\in Spec(A)$ be a closed point, and we want to prove surjectivity of $ev_{A_x}$. By Nakayama's Lemma, it suffices to check that the composition $$D(A_x)\to Hom_k(M,A_x)\to Hom_k(M,k_x)$$ is surjective, here $k_x$ is the residue field of $x$. However, for this statement you can replace $A_x$ by its completion, which turns it into a claim about Taylor series.

P.S. It may be psychologically easier to extend scalars from $k$ to its algebraic closure at the very beginning... But if you resist the temptation, you get to think about things like $k$-linear differential operators on the ring of Taylor series with coefficients in a finite extension $k_x\supset k$...


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.