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Let $ X, Y $ be smooth affine varieties over $ \mathbb C $. Let $ T : X \rightarrow Y $ be a dominant quasi-finite morphism and let $ T^\# : \mathbb C[Y] \rightarrow \mathbb C[X] $ be the resulting map on coordinate rings.

($ T $ being quasi-finite means that it can be factored as an open embedding followed by a finite morphism. It also means that $T^\# $ is injective and every element of $\mathbb C[X] $ is algebraic over $ \mathbb C[Y] $.)

Let $ D(X) $ denote the ring of differential operators on $ X $. Is the following statement is true?

Let $ d \in D(X) $. Suppose that $ d(T^\#(f)) = 0 $ for all $ f \in \mathbb C[Y] $. Then $ d = 0 $.

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    $\begingroup$ I guess you want to add the condition that $T$ is dominant for $T^\#$ to be injective (otherwise a closed embedding is an example of a quasi-finite morphism). $\endgroup$
    – Sam Gunningham
    Commented Apr 13, 2018 at 17:26
  • $\begingroup$ I agree and made the edit. $\endgroup$
    – Joel Kamnitzer
    Commented Apr 13, 2018 at 20:00

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Edit: I think skipped a step in the original argument - it is not immediately clear to me that $d$ commutes with $\mathbb C[Y]$ in $D(X)$ unless $d$ is a derivation. I have added an inductive argument for this below.

I believe this is true. Here is a sketch of an argument.

We will prove this by induction on the degree of the differential operator $d$. It is certainly true for differential operators of degree 0.

Suppose $d\in D(X)$ is of degree $\leq k$, and $d(f)=0$ for all $f\in \mathbb C[Y]$ (which I identify with a subring of $\mathbb C[X]$).

Lemma: We have $[d,f]=0$ for all $f\in \mathbb C[Y]$.

Proof of lemma: The differential operator $[d,f]$ is of order $\leq k-1$, and $[d,f](g)=0$ for all $g\in \mathbb C[Y]$. The lemma now follows from the inductive hypothesis.

It remains to show that any $\mathbb C[Y]$-linear differential operator $d\in D(X)$ is zero.

Consider the finite field extension $$K := \mathbb{C}(Y) \subseteq L := \mathbb{C}(X)$$

Note that $D(X)$ injects in to $D(L/\mathbb C)$ and $d$ lands in the subring $D(L/K)$. In other words $d$ may be considered as an element of the ring $D(L/K)$ (here I am using Grothendieck's definition of differential operators for a $K$-algebra).

I claim that $D(L/K)=L$, and thus $d=0$. Indeed, $L$ is a smooth $K$-algebra of dimension $0$ (as we are in characteristic $0$), and thus:

  1. The ring of $D(L/K)$ is generated by derivations $Der_K(L,L)$ and $L$.
  2. $Der_K(L,L)=0$.
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  • $\begingroup$ This looks good to me. Thanks! Can you provide a reference for the last two facts? $\endgroup$
    – Joel Kamnitzer
    Commented Apr 13, 2018 at 20:18
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    $\begingroup$ According to a comment by Mariano Suarez-Alvarez in an answer to this question: math.stackexchange.com/questions/1974919/… , there is a proof of Fact 1 in the last chapter of McConner and Robson's book Noncommutative Noetherian Rings. However, I can't seem to get access to a copy at the moment to check. $\endgroup$
    – Sam Gunningham
    Commented Apr 13, 2018 at 20:53
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    $\begingroup$ Fact 2 is much more elementary. It is equivalent to the assertion that $\Omega^1_{L/K}=0$, which in turn is equivalent to the fact that $L/K$ is separable (immediate in characteristic $0$). The wikipedia page en.wikipedia.org/wiki/… indicates that there is a reference in Milne's book on etale cohomology, for example. $\endgroup$
    – Sam Gunningham
    Commented Apr 13, 2018 at 20:55
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    $\begingroup$ Thanks! Yes, I think that I know how to prove Fact 2. $\endgroup$
    – Joel Kamnitzer
    Commented Apr 13, 2018 at 22:48

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