# zeros of a polynomial in a finite field

Hi everyone. While looking for examples of asymptotically bad towers of function fields over a finite field $F_q$ defined by a Kummer equation of the form $y^m=f(x)$ with $p\equiv 1\mod m$ where $p=Char(F_q)$ the following situation came up: let $$a, b, c, d\in F_q$$ such that $a$, $c$ and $ad-bc$ are non zero in $F_q$. Now consider the rational function $$g(x)=(a/c)^m-\left(\frac{ax+b}{cx+d}\right)^m$$ and let $f(x)$ be the denominator of $g(x)$. Clearly $f(x)$ is a polynomial of degree $m-1$ with coefficients in $F_q$. To my surprise, after many numerical experiments, $f(x)$ seems to be a separable polynomial splitting into linear factors in $F_q[x]$. I could prove this in one particular situation. Maybe someone already knows this is true or, equally better, someone already found a counterexample. Thanks!

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The tags for your question are probably not what you intended – Trevor Wilson Jul 19 '12 at 22:37
To (possibly) clarify what Wanax says: since $p\equiv 1\pmod{m}$, the polynomial $h(y)=y^m-1$ has $m$ distinct roots in $\mathbb{F}_p$ (the multiplicative group of a finite field is cyclic). You have $g(x)=-\frac{a^m}{c^m}h(\frac{x+b/a}{x+d/c})$, so roots of $g(x)$ are now very easy to figure out. – Vladimir Dotsenko Jul 20 '12 at 9:38
oops...how could i miss that??? thanks a lot wanax and vladimir. – R. Toledano Jul 27 '12 at 18:09