Cardinality of a conic and quadric over the field $\mathbb{F}_9$ I have the following question in an assignment: 
Over the field $\mathbb{F}_9$ of nine elements, find the cardinality of
a) the conic $x_0^2 + x_1^2 + x_2^2 = 0$ in $\mathbb{P}_2$
b) the quadric $x_0^2 + x_1^2 + x_2^2 + x_3^2 = 0$ in $\mathbb{P}_3$
Any hints on how to proceed? One way would be to just substitute the points of $\mathbb{F}_9=\mathbb{Z_3}[x]/\langle x^2+1\rangle$. But that's way too much computation. 
 A: Your two degree $2$ varieties  are smooth and contain at least a point over $\mathbb F_9$, say $(1:1:1)$ on the conic and $P=(1:1:1:0)$ on the quadric $\mathcal Q$.
Projecting from that point on the line $x_0=0$ of $\mathbb P^2(\mathbb F_9)$, respectively on the plane $\Pi:x_0=0$ of $\mathbb P^3(\mathbb F_9)$, you will find that over $\mathbb F_9$ your conic has $10$ points and your quadric $100$ points.  
Edit: a few details  for the more difficult case of the quadric 
The tangent plane $T_P$ to the quadric $\mathcal Q$ at $P=(1:1:1:0)$  has equation $T_P: x_0+x_1+x_2=0$.
It cuts the the quadric $\mathcal Q: x_0^2 + x_1^2 + x_2^2 + x_3^2 = 0$ along the degenerate conic $$x_0+x_1+x_2=0,\quad x_1^2+x_2^2+x_1x_2-x_3^2=(x_1-x_2+x_3)(x_1-x_2-x_3)=0$$ consisting of the union of the lines $$L_1:x_0+x_1+x_2=0,x_1-x_2+x_3=0 \\L_2:x_0+x_1+x_2=0,x_1-x_2-x_3=0$$ 
The projection of our quadric $\mathcal Q$ from its point $P$ to the plane $\Pi: x_0=0$ of $\mathbb P^3$ induces a bijection from $\mathcal Q\setminus (L_1\cup L_2)$ onto the affine plane $\Pi\setminus T_P\cong \mathbb A^2(\mathbb F_9)$.
This accounts for $\vert \Pi\setminus T_P\vert=\vert \mathbb A^2(\mathbb F_9)\vert=9^2=81$ points of the quadric.
To obtain the complete  quadric we have to add the points of $L_1\cup L_2$.
Since there are $19$ of them (union of two $10$-point projective lines meeting in $1$ point) , we see that $\mathcal Q$ contains a total of $$81+19=100$$points over $\mathbb F_9$
The exact same reasoning proves that the quadric $\mathcal Q: x_0^2 + x_1^2 + x_2^2 + x_3^2 = 0$ in $\mathbb P^3(\mathbb F_q)$ has $q^2+2q+1=(q+1)^2$ points over an arbitrary finite field with $q$ elements $\mathbb F_q$.
