I'm reading Serre's A Course in Arithmetic and I can't understand the following statement in the proof of Ch. 4, 1.7 prop 4. It says
(here, $q=p^f$ is a power of a prime $p\neq 2$ and $F_q$ is a field with $q$ elements.)
Let $A$ (resp. $B$) be the set of elements of $F_q$ of the form $ax^2$(resp. of the form $c-by^2$) with $x \in F_q$ (resp. with $y\in F_q$). One sees easily that $A$ and $B$ each have $(q+1)/2$ elements.
I can easily see why this is true when $q=p$ but I'm so lost when $q=p^f$ for some $f>1$. For example, if we take $q=3^2$, then I can only find 4 elements of the from $x^2$, $(0,0), (0,1), (1,0)$ and $(1,1)$ - which makes sense to me since $F_q \cong Z_3\oplus Z_3$.
I bet I'm making a really dumb mistake but I spent more than 3 hours and still have no clue. Can anyone shine me some light please?
Your mistake (I won't call it "dumb"—we all make mistakes—and this is a somewhat subtle point!) is that as rings, $\Bbb F_9$ is not isomorphic to $\Bbb Z_3\times\Bbb Z_3$. (Here, to be clear, $\Bbb Z_n$ is a shorthand for $\Bbb Z/3\Bbb Z$.) Indeed, it's an even more common mistake to assert that $\Bbb F_9$ is isomorphic to $\Bbb Z_9$, but that's also incorrect. For example, note that $\Bbb F_9$ has no zero-divisors (since it's a field). But $\Bbb Z_3\times\Bbb Z_3$ has four zero-divisors, namely $(0,1)$, $(0,2)$, $(1,0)$, and $(2,0)$. And $\Bbb Z_9$ has two zero-divisors, namely $3$ and $6$.
It is true that $\Bbb F_9$ is a vector space of dimension 2 over $\Bbb F_3$; in particular, $\Bbb F_9$ is isomorphic to $\Bbb Z_3\times\Bbb Z_3$ as additive groups. But rings have multiplications as well; and if we think of $\Bbb Z_3\times\Bbb Z_3$ as a product ring with its component-wise multiplication, then that is not the same as the multiplication in $\Bbb F_9$. As Gerhard "Too Many Middle Names to Count" Paseman comments, representing $\Bbb F_9$ as $\Bbb F_3[x]/\langle f(x)\rangle$ for an irreducible quadratic polynomial $f(x)\in\Bbb F_3[x]$ is probably the best way to keep track of the multiplication.
(In general, the rings $\Bbb F_p \cong \Bbb Z_p$ when $p$ is a prime, but $\Bbb F_{p^m} \not\cong \Bbb Z_{p^m}$ when $m\ge2$; indeed, $\Bbb F_{p^m}$ and $\Bbb Z_{p^m}$ are not even isomorphic as additive groups.)