Let $\mathbb{F}_p$ be a finite field with $p$ elements, and let $$x+y+z=0$$ be a projective line with $x,y,z \in \mathbb{F}_p$. In a book I am currently reading about elliptic curves, it uses the fact that this projective line obviously has $p+1$ solutions to prove a theorem of Gauss, but doesn't explain (probably because it assumes general background on projective geometry). I have barely touched on projective geometry, so I was hoping someone could explain why there are obviously $p+1$ solutions.
The only thing I can think of is $x+y = -z$ corresponds to the equation $x^{\prime} + y^{\prime} = 1$ in affine space by $\frac{-x}{z} + \frac{-y}{z} = 1$ when $z \neq 0$ with $x^{\prime}, y^{\prime} \in \mathbb{F}_p$. Then if $x^{\prime} = s$, we have $y^{\prime} = 1-s$ and there are $p$ choices for $s$. So we have $p+1$ solutions, the $p$ mentioned and $(0,0,0)$. The only problem is I don't know if this is right and I thought $(0,0,0)$ wasn't a point in projective space. If not, are we assuming the extra solution is $\mathcal{O}$ in the context of elliptic curves? Thank you
Just think about it in cases. When $x=0$, you must have $y=-z$, so there is essentially only one solution, $(x,y,z)=(0,1,-1)$.
Otherwise, you may assume $x=1$. There are $p$ possibilities for $y$, and all of them give a distinct point in projective space. These give the solutions $(x,y,z)=(1,y,-1-y)$.
Thus, in total, there are $p+1$ solutions.
Here is another way to think about this: (this method generalizes easier)
First, we consider solutions in $\mathbb F_p^3$. Any choice of $x$ and $y$ work, so there are $p^2$ of these. Thus, there are $p^2-1$ solutions in $\mathbb F_p^3\setminus\{(0,0,0)\}$. Finally, since each equivalence class in projective space has $p-1$ points, this gives $(p^2-1)/(p-1)=p+1$ solutions.