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Let $q$ be a power of a prime $p$, and $m,l$ positive integers with gcd$(l,q^m-1)=1$. Denote $Tr$ to be the trace of $GF(q^m)$ over $GF(q)$.

Suppose that there exists a nonzero $\gamma \in GF(q^m)$ such that $$ Tr(x)=0 \Leftrightarrow Tr(\gamma x^l)=0, \;\text{ for all } x\in GF(q^m). \;\;\;\;\;\;\text{(1)}$$

I'm pretty sure that Equation $(1)$ implies that $\gamma \in GF(q)$ and $l$ is a power of $p$ (the converse is clearly true).

Am I right? If so, how can I prove it?


So far I have unsuccesfully tried to prove it, using the following in one way or the other:

  • $\gamma x^l$ is a permutation of $GF(q^m)$
  • The roots of $Tr$ are exactly the elements $a^{q}-a$ for $a\in GF(q^m)$
  • The formal derivative of $Tr(\gamma x^l)$ is $l x^{l-1}$. Showing that it is zero implies that $p|l$.
  • Try to show that $Tr(\gamma x^l)=\gamma(Tr(x))^l$

I feel that the last point is the key; if this is shown, then one can compare coefficients and take it from there. I also feel that the proof should be very simple, but I'm stuck.


Another thing I just read about, from Lidl & Niederreiter's book "Finite Fields", and looks highly relevant:

Theorem. A polynomial $f\in GF(q)$ is a permutation polynomial of all finite extensions of $GF(q)$ iff it is of the form $f(x)=ax^{p^h}+b$, where $a\neq 0$, $p$ is the characteristic of $GF(q)$, and $h$ is a nonnegative integer.

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  • $\begingroup$ Ok. This is a cool question. Unfortunately it's late here, and I'm a bit drunk. I don't see it right away, but there might be a trick... Anyway, I would be at least a bit surprised if this is not true. $\endgroup$ Commented Mar 15, 2015 at 23:06
  • $\begingroup$ I know a couple of people who were claiming that they could do math better after a couple of beers :) Anyway, I think I have an idea, but I have to double think before posting anything, I'm not sure yet. $\endgroup$
    – geo909
    Commented Mar 16, 2015 at 15:27
  • $\begingroup$ If $q=2$ then the claim follows because the assumption implies that $Tr(x)=Tr(\gamma x^\ell)$ for all $x\in GF(2^m)$. The rest is along the lines of this argument. For the difference of the two traces of monomials to vanish universally it is necessary that their exponents belong to the same cyclotomic coset et cetera. $\endgroup$ Commented Mar 16, 2015 at 20:02
  • $\begingroup$ Many thanks Jyrki, I'll need to study this a bit. I'll update later. $\endgroup$
    – geo909
    Commented Mar 16, 2015 at 20:18

1 Answer 1

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I Think is kinda late, but i was looking for something like that i would like to share it: I think the easy way is to see that $|Ker(Tr)|=q^{m-1}$ (if you see it as a linear transformation $Ker : GF(q^{m})\rightarrow GF(q)$), and then count all the elements $\gamma x^{l}$

Or you could also use that $Tr(x)=0 \leftrightarrow x=y^{q}-y $ for some $y\in GF(q^{m})$

Plus i dont think this is relevant, but your question can traslate to Coding Theory, and you´ll be asking what are all the linear transformations of a code (In this case a Trace code) that preserve the Hamming weight, which would be all the multiplication for an element of the base field and a permutation of the coordinates, which would traslate to the "trasformations" $\gamma x^{l}$ in this context.

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