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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

Ryan Shesler
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    You’ve got the affine solutions (when $z=1$) correct, but the additional point(s) are when $z=0,$ which is one point, $(1,-1,0).$ (You are correct that $(0,0,0)$ is not a point in projective space.) – Thomas Andrews Jun 16 '20 at 04:04
  • In projective space, the points are actually equivalence classes. For example, $(1,2,3)\sim(2,4,6)\sim(3,6,9).$ So when $z\neq 0$ you get (x,y,z)\sim (x/z,y/z,1).$ We think of the points when $z=0$ as the “line at infinity” and the case where $z=1$ as the affine points. – Thomas Andrews Jun 16 '20 at 04:13
  • @ThomasAndrews thanks I remembered from reading after I commented that $(x,y,z) \sim (sx,sy,sz)$ in projective space. I should probably give myself a crash course in it before reading on in my book, thank you for explaining. – Ryan Shesler Jun 16 '20 at 04:15
  • Related. – Jyrki Lahtonen Jun 16 '20 at 10:21

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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.

Kenta S
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