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Let $k$ be a field. If $f \in k[x]$ is a polynomial, and $d/dx\ f = 0$, then either

  1. $f$ is constant, or

  2. $char\ k = p$ and $f \in k[x^p]$.

So "annihilated by all derivations" is perhaps not the right thing to ask for in characteristic $p$ (though that's what I asked for in Is the singular locus ideal preserved by all derivations? ).

What is the right thing to ask for?

I would like an invariance condition one could state of a subscheme $Y$ of $X$, that holds for the singular locus, but doesn't hold for (say) any regular closed point on a rational variety.

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    $\begingroup$ Hasse derivatives, $f$ is constant if and only if all Hasse derivatives of $f$ are zero. (and @Qiaochu: no.) $\endgroup$
    – Felipe Voloch
    Commented Jul 27, 2011 at 19:09
  • $\begingroup$ And, Hasse derivatives are just generators of the algebra of differential operators on your field/ring. So, an element of the ring is constant(from the base i.e. $k$) iff it is killed by all differential operators with zero constant terms. $\endgroup$
    – P. Grabowski
    Commented Feb 7, 2023 at 21:16

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In principle, there are two possible approaches. One is based on the Hasse derivatives (also called hyperdifferentiations). See http://math.fontein.de/2009/08/12/the-hasse-derivative/ for elementary definitions and properties, and the paper

P. Vojta, Jets via Hasse-Schmidt derivations, ArXiv: math/0407113,

for the use of Hasse derivatives in algebraic geometry.

On the analysis level, there are also the Carlitz derivatives, special difference operators working efficiently just on functions annihilated by usual derivatives, with a rich theory of "differential equations", Fourier series, special functions etc. See

A. N. Kochubei, Analysis in Positive Characteristic, Cambridge University Press, 2009.

However there is no geometry around this approach so far.

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  • $\begingroup$ I'd rather deal with algebra, and I imagine this Hasse approach is what I'll have to come to terms with. Thanks! Is it fair to view it in the following way? In any characteristic, all the derivations on $k[x]$ look like $f\ d/dx$. In char 0, the only things deserving the name "higher-order differential operator" come from composing $d/dx$. But in char p, there are more things deserving that name. $\endgroup$
    – Allen Knutson
    Commented Jul 29, 2011 at 6:39
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    $\begingroup$ That is right. However, so far virtually all interesting examples of special polynomials or functions known for the char p situations are connected with the Carlitz derivatives. Perhaps this subject is still waiting for a researcher with algebraic backgound. $\endgroup$
    – Anatoly Kochubei
    Commented Jul 29, 2011 at 7:30

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