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We know that in characteristic $0$, all algebraic series are differentiably finite. Is this true in positive characteristic? I look at the proof, indeed we need to the characteristic to be $0$ for the proof to work. If it is not true in positive characteristic, is there a counter-example?

Thank you in advance!

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    $\begingroup$ I don't see any mention of the field over which Comtet's proof is working. Have I missed something? $\endgroup$
    – Peter Taylor
    Commented Apr 17, 2022 at 13:39
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    $\begingroup$ Actually, it should have been stated in Theorem 6.4.6 that $\mathrm{char}\,K=0$ since the proof uses (6.12), for which it is assumed that $\mathrm{char}\,K=0$. $\endgroup$
    – Richard Stanley
    Commented Apr 17, 2022 at 15:03
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    $\begingroup$ @PeterTaylor First sentence in Comtet's article: "Soit y une fonction de la variable complexe x...". A part from this, in equation (4) from the proof, he divides by the derivative wrt y, which could be zero in positive characteristic. $\endgroup$
    – Jiu
    Commented Apr 17, 2022 at 15:24
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    $\begingroup$ I might be missing something, but over a field of characteristic $p$, isn't the $p$-th derivative always $0$ which implies that everything is $D$-finite? $\endgroup$
    – Random
    Commented Apr 17, 2022 at 15:39
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    $\begingroup$ @SamHopkins: I have done this, but for klein.mit.edu/~rstan/ec/ec2supp2.pdf. $\endgroup$
    – Richard Stanley
    Commented Apr 18, 2022 at 16:45

1 Answer 1

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Over a field of characteristic $p$, the $p$-th derivative of any power series is $0$, and so every power series over a field of finite characteristic is $D$-finite.

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