# polynomial over finite field, roots forming additive subgroup

Let $q=2^h$ and $t=2^r$ for some $h\ge r$ and denote by $\mathbb{F}_q$ the finite field of order $q$.

(since the previous, simple version was wrong, I'm posting here a new version)

Let $f$ be a monic polynomial with $f(0)=0$, $f(1)=0$, such that the equation $f(z)=c$ has either $0$ or $t$ solutions in $\mathbb{F}_q$. Let $F_a(z)=\frac{f(z)+f(a)}{z+a}$, and we are given that the restriction of $F_a(z)$ to its non-roots (and $\neq a$) is injective, and there are constants $w_1,...,w_{t-1}$ such that none of the polynomials $w_i^{-1}+F_a(z)$ has a root in $\mathbb{F}_q\setminus\{a\}$.

Computationally, for $q\le 32$, I turns out that $\{0,w_1,\ldots,w_{t-1}\}$ is always an additive subgroup in $\mathbb{F}_q$, and that the $\{z|f(z)=c\}$ are also cosets of an additive subgroup. I wonder if any of this could be proven in general.

• This seems interesting. What is the motivation? – Slade Oct 16 '14 at 12:51
• Are the $w_i$ supposed to be distinct, or correspond to $a$ in some way? – Slade Oct 16 '14 at 13:39
• The $\{w_i | 1\le i\le t-1\}$ are a fixed set for which a polynomial $f$ with said properties exist. If it helps, it can be derived from the above properties that when writing $f(z)=\sum_{i=0}^{q-1} a_i z^i$ then $a_1=\sum_{i=1}^{t-1} w_i^{-1}$ and that $a_{2s+1}=0$ for all $s\ge 1$, and also that $a_{q-2}=0$. – user1111929 Oct 16 '14 at 13:53
• It does look like you are seeing linearized polynomials. I don't see right away why that should be the case though. – Jyrki Lahtonen Oct 17 '14 at 18:02
• The motivation by the way is proving a property of certain point sets in projective geometry over finite fields. A proof of this (even just the {0,w1,...,wt} part) can be completed by me to a good research article (web of science level). I will gladly offer coauthorship to whoever provides me with a proof. – user1111929 Oct 17 '14 at 19:43

If $t\neq 1,2,q$, then there exist subsets of $\mathbb{F}_q$ of size $t$ that include $0$, but that are not additive subgroups. For example, take $S = (\mathbb{F}_t\setminus \{1\}) \cup \{a\}$, where $a\notin \mathbb{F}_t$.
Since any function from $\mathbb{F}_q$ to itself can be interpolated with a polynomial, this is quite false as stated—just let $f$ be any $t$-to-$1$ function with $f(S) = 0$—though I'm still curious what the idea is, and whether there is a less trivial statement of the problem.
• I updated the problem statement even more. These $q/t+1$ sets all appear to be cosets of additive subgroups of size $t$ of $\mathbb{F}_q$. – user1111929 Oct 16 '14 at 13:33