Solutions to a linear equation over a subfield of a finite field

Question

I know that if we have a line defined over a finite field $\mathbb{F}_p=K$ then this line has p+1 solutions in $P^2(K)$, $p^2+1$ in $P^2(\mathbb{F}_{p^2})$ and so on. But what about the opposite? If the line is defined over $\mathbb{F}_{p^2}$ how many solutions does it have in $P^2(K)$?

Answer 1

A line in $P^2$ is defined by an equation $ax+by+cz=0$ where $a,b,c$ are not all $0$. The set of solutions $(x,y,z)\in K^3$ to such an equation is just a vector subspace of $K^3$ whose dimension is $3-d$ where $d$ is the dimension of the span of $\{a,b,c\}$ as a vector space over $K$. The solutions in $P^2(K)$ are the just the projectivization of this vector subspace, which has $\frac{p^{3-d}-1}{p-1}$ elements. In the case where $a,b,c\in\mathbb{F}_{p^2}$, this $d$ can only be $1$ or $2$. In the case $d=1$, you get $p+1$ elements in $P^2(K)$ and the line is actually defined over $K$ (any two of $a,b,c$ are linearly dependent over $K$, so you can rescale to make one of them $1$ and then the rest are also in $K$). In the case $d=2$, you get just $1$ element on $P^2(K)$.