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.