Algebraic construction of a $C_4$ free graph I'm reading the proof of theorem 7 from page 5, here. I am not sure about two parts of the proof, it would be great if someone could explain it to me:

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*Why do we need $\mathbb{F}_q^2$ at all? Why not consider something like $\mathbb{Z}/n\mathbb{Z}$ if we only need a finite number of points?

*What does "all lines in $\mathbb{F}_q^2$" mean? Does it mean all lines $y=mx+c$ where both $m, c$ are in $\mathbb{F}_q$? Why does every line contain $q$ points?


We can take $\mathbb{Z}/q\mathbb{Z}$ directly, because it will be a finite field ($q$ is prime), but we can take any other finite field of order $q$. A line have $q$ points because for any $a$: $mx + c = a \iff x = m^{-1}(a-c)$. Two lines can intersect at only one point because otherwise we would have with $x\neq y$:
$mx+c = m'x+c'$, $my+c = m'y+c'$. By substitution, lines intersect at $m(x-y) = m'(x-y) \implies m = m'$, and so $c=c'$, so the lines are the same.
