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Let $a$ be algebraic over $K:=\mathbb F_q\left(\!\left(\frac1T\right)\!\right)$. Can one extend continuously the hyperderivatives on $K(a)$?

Recal that the hyperderivative $D_h$ over $K$ is defined by $$D_h\left(\sum_{i\le m}\frac{a_i}{T^i}\right)=\sum_{i\le m}\binom{-i}h\frac{a_i}{T^{i+h}}$$ with $\binom{-i}h=\frac{-i(-i-1)\cdots(-i+h-1)}{h!}$

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  • $\begingroup$ You want to extend them as (a family of) Hasse-Schmidt derivations, right? $\endgroup$
    – darij grinberg
    Commented Aug 8, 2021 at 20:39
  • $\begingroup$ yes, you are right. $\endgroup$
    – joaopa
    Commented Aug 8, 2021 at 20:43
  • $\begingroup$ I suspect the answer is "yes", because a family of Hasse-Schmidt derivations $\left(D_0, D_1, D_2, \ldots\right)$ on a ring $A$ is equivalent to a ring homomorphism $\sum_{i\geq 0} D_i t^i : A \to A\left[\left[t\right]\right]$ (see, e.g., Theorem 2.10 in darijgrinberg.gitlab.io/algebra/va3.pdf ), and therefore all you need is to extend a ring homomorphism $K \to K\left[\left[t\right]\right]$ to a ring homomorphism $K\left(a\right) \to K\left(a\right)\left[\left[t\right]\right]$; but this should be easy "by tensoring". But I'll leave the details to you, particularly ... $\endgroup$
    – darij grinberg
    Commented Aug 8, 2021 at 20:44
  • $\begingroup$ ... making sure the maps are continuous. $\endgroup$
    – darij grinberg
    Commented Aug 8, 2021 at 20:44

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