Understanding how to calculate $f(\mathcal{O})$ for a given $f$

Question

I have an elliptic curve $E$ over $\mathbb{F}_7$ defined by $$y^2 = x^3 +2$$

I also have a function $f$ defined by $f(x,y)=\frac{4x-y+1}{5x-y-1}$

I need to calculate $f(\mathcal{O})$ where $\mathcal{O}$ is the point at infinity

In the paper I'm reading (specifically Example 2.4 on page 13), I am told to use projective coordinates: $$f(x:y:z)=\frac{4x-y+z}{5x-y-z}$$

and then $$f(\mathcal{O})=f(0:1:0)=1$$

I don't understand how they've worked this out - can anyone enlighten me please, so I can then use this technique on any definition of $f$

Answer 1

If we want to think of the curve $y^2=x^3+2$ as an equation over the projective space, we write it as $(\frac{y}{z})^2=(\frac{x}{z})^3+2$ and after multiplying by $z^3$ as $y^2z=x^3+2z^3$ so that a projective line $(x:y:z)$ with $z\neq 0$ satisfies this new equation iff $(\frac{x}{z},\frac{y}{z})$ satisfies $y^2=x^3+2$. The point at infinity is the line where $z=0$ and we cannot divide by it. If this is the case then $y^2\cdot 0=x^3 +2\cdot 0^3$ so that $x=0$ and the line is $(0:1:0)$.

Similarly, we can write $g(x,y,z)=\frac{4x-y+z}{5x-y-z}$ which is well defined as a function on projective space (i.e. $g(tx,ty,tz)=g(x,y,z)$ for $t\neq 0$). If $z\neq 0$, then we get that $g(x,y,z)=f(\frac{x}{z},\frac{y}{z})$ for $z\neq 0$. Thus, in a way $g$ is the generalization of $f$ where we are allowed to evaluate at "infinity" $(\frac{x}{0},\frac{y}{0})$. Since the elliptic curve has only one point at infinity, then evaluating $f$ at that point means to first construct $g$ using the relation $g(x,y,z)=f(\frac{x}{z},\frac{y}{z})$ and then evaluating it at $(0:1:0)$.