Localizability of differential operators a la Grothendieck

Question

Hello,

Maybe this question is trivial, so sorry

Let $A$ be a (comm. with 1) $k$-algebra, where $k$ is a ring (comm. with 1).

Then we can define the module of differential operators $D^{\leq n} (A)$, a submodule of $End_k (A,A)$ (endomorphisms of vector spaces). $D^{\leq -1} = 0$, and then inductively $D^{\leq n} = \{ d | [d,a]\in D^{\leq n-1}\}$.

We have a lemma:

Lemma. Let $f \in A$. Then for every $d \in D^{\leq n}(A)$ we can find unique $e \in D^{\leq n}(A_f )$, such that $l\circ d = e \circ l$, where $l: A \to A_f$ is the localization map.

I think that I know how to prove the lemma, by induction on the order of diff. op. (just need to see how to apply operators to fractions). It gives us a map $D^{\leq n}(A) \to D^{\leq n}(A_f)$.

Question 1. Under which assumptions on $A/k$ this map $D^{\leq n}(A) \to D^{\leq n}(A_f)$ is a localization map (i.e. becomes an isomorphism after tensoring (say on the left, it does not matter) with $A_f$)?

Question 2 (my real question). If $A/k$ is finitely generated, or finitely presented, is this a localization map?

Somehow, I am having trouble with the "surjectivness" part. Maybe there is some reference?

Thank you, Sasha

Answer 1

Two observations (with $k$ a field of characteristic zero):

• If $A$ is a domain over a field $k$, then elements of $D(A)$ extend to elements of $D(A_S)$ for all multiplicatively closed sets $S$.

• Your questions become easier if you ask instead about the subalgebra $\Delta(A)$ of $D(A)$ generated by $A$ and derivations: then the answer is yes to your two questions. Now, if $A$ is finitely generated and regular, then $D(A)=\Delta(A)$, so in this case the answer is yes for $D(A)$ too.

You'll find this in the last chapter of McConnell and Robson's book on noetherian rings.

(Finally, $D^{\leq n}(A)$ is an $A$-bimodule which is not symmetric, so tensoring with $A_f$ on one side or the other is not the same thing)