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Let $F$ be a field of prime order $p$. Suppose that $f\in F[x]$ is a non-zero polynomial of degree $\deg f<p$. If $f$ does not have multiple roots, then there exist polynomials $P,Q\in F[x]$ such that $$Pf+Qf'=1.$$ What is the smallest possible degree of $P$ in this representation, and how it depends on $f$?

Denoting this smallest possible degree by $\nu(f)$, some basic observations are:

  • $\nu(cf)=\nu(f)$ for any $c\in F^\times$;
  • if $g(x)=f(cx+b)$, then $\nu(g)=\nu(f)$ for any $b\in F$ and $c\in F^\times$;
  • $\nu(f)=0$ if and only if $f(x)=a(x-b)^d+c$ with $a,b\in F$ and $c\in F^\times$.

Is it possible to classify those polynomials $f$ with $\nu(f)<10$? With $\nu(f)<\deg f$? Does $\nu$ have any special properties allowing one to estimate or easily compute it?

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    $\begingroup$ The map (a,b)\mapsto af+bf’ is a linear map from \Sym^{n-1}\oplus \Sym^n\to \Sym^{n+\deg{f}-1} (\Sym^d is the space of polynomials of degree at most d). The RHS has dimension n+\deg{f}, the LHS has dimension 2n+1. Taking n = \deg{f}-1 both sides have the same dimensions and indeed the map is injective by virtue of f and f’ being coprime. Thus it is surjective. Thus you find (a,b) with \deg{a}\leq \deg{f}-2 and \deg{b}\leq \deg{f}-1 mapping to 1. [This is the usual definition of the resultant.] Anywho what this tells you is that you can compute (a,b) by inverting a matrix (in the coeffs of f). $\endgroup$
    – alpoge
    Commented May 31, 2019 at 16:29
  • $\begingroup$ Sorry —- I ran out of characters then was distracted! I meant to simply comment that therefore once one produces one f (out of laziness/busyness I haven’t tried) for which the corresponding a has degree exactly \deg{f} - 2, it follows that the generic f does too, since having lower degree is expressed as the vanishing of a polynomial. No comment on the other questions (besides that they’re expressed as the vanishing of various polynomials from the above, not that this gives you a good answer), and do let me know if I’ve overlooked something in characteristic p!!! $\endgroup$
    – alpoge
    Commented May 31, 2019 at 16:59
  • $\begingroup$ @alpoge: Thanks for the useful comment - and, could you elaborate on the second part of it? Do $f$ with $\deg a=\deg f-2$ exist at all? Why would this imply that one has $\deg a=\deg f-2$ for the generic $f$? $\endgroup$
    – W-t-P
    Commented May 31, 2019 at 20:25

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