Finding the number of roots of u(X,Y) under certain restrictions

Question

Suppose $q$ is a prime power, gcd($s$,$q$)$=1$ and gcd($s$,$t$)$=1$ and $f(X)$ is a polynomial of degree $t$ over $GF(q)$. Also $u(X,Y)=\sum_{i,j} u_{ij}X^iY^j$, where $u_{ij}\in GF(q)$, $0\leqslant i$, $j<s$ and $si+tj\leqslant m$, $m \in\mathbb{N}$. Suppose $P=\{(x,y)|x,y \in GF(q) \text{ and } y^s=f(x) \}$ and $Z=\{(x,y)|x,y \in GF(q) \text{ and } u(x,y)=0 \}$. We want to show that $P \cap Z$ has no more than $m$ elements. Here's how they do it in the text I'm reading. Since gcd($s$,$q$)$=1$, a field extension $GF(q^l)$ of $GF(q)$ will have a primitive $s$-th root of unity $\zeta$. They claim that if $P'=\{(x,y)|x,y \in GF(q^l) \text{ and } y^s=f(x) \}$ and $Z'=\{(x,y)|x,y \in GF(q^l) \text{ and } u(x,y)=0 \}$, then even $P' \cap Z'$ will have at most $m$ elements. Now, $P'$ can be divided into classes $\{ (x,0)\}$ if $f(x)=0$ and $\{ (x,y),(x,\zeta y),\ldots (x,\zeta^{s-1}y)\}$ in the other case. Consider the classes in which the zeros of $u(X,Y)$ are contained. Apparently, one finds the same classes when one looks at the zeros of $$u^*(X,Y)=u(X,Y)u(X,\zeta Y)\ldots u(X,\zeta^{s-1}Y)$$ (how is this even relevant?). One can show that (I get that now, subject of another question) $u^*(X,Y)=V(X)$ with the degree of $V(X)$ at most $ m$. Now we are supposed to be ready to finalise: The zeros of $u^*(X,Y)$ can thus be grouped into at most $m$ classes. A class of the form $\{ (x,0)\}$ trivially contains at most one zero of $u(X,Y)$. But now for the mysterious part. I quote: A class with $s$ elements contributes at most one zero of $u(X,Y)$ (and at most one zero to each of the other $u(X,\zeta^kY)$). I really can't see why that is. What am I overlooking?

Answer 1

In fact, the proof is wrong, which is a pitty, since this was part of an example of an algebraic geometry (Goppa) code, without going into algebraic geometry (Coding theory, Van Tilborg). Bad Math books (seems to be the norm rather than the exception) take all the fun out of doing maths. Here is a counterexample. If, for example $u(X,Y)= X+XY+X^2+XY^2$ and $f(X)=X^t +1$ with $m\geqslant s + 2t$, then $(0,1)$ and $(0,\zeta)$ belong to the same class of $P'$ and are two different zero's of $u(X,Y)$ nonetheless.