Exercise 11.11

Proof. Let $F$ be a finite field such that the characteristic of $F$ is not $2$. Here

Consider the maps

$\text{∙}$ The map $\phi$ is well defined: If $(x,y)\in C_1\setminus \{(0,0)\}$, then $y^2=x^3-\mathit{Dx}$, and $x\ne 0$, otherwise $y^2=x^3-\mathit{Dx}=0$, and then $(x,y)=(0,0)$.

Write $(u,v)=\left(2x-{\left(\frac{y}{x}\right)}^2,\frac{y}{x}\right)$, then $x=\frac{1}{2}(u+v^2)$ and $y=\frac{1}{2}v(u+v^2)$. The equality $y^2=x^3-\mathit{Dx}$ gives

so that $(u,v)=\left(2x-\left(\frac{y}{x}\right),\frac{y}{x}\right)\in C_2$.

$\text{∙}$ The map $\psi$ is well defined: if $(u,v)\in C_2$, then $u^2-v^4=4D$. Then $u+v^2\ne 0$, otherwise $4D=u^2-v^4=(u-v^2)(u+v^2)=0$, where $4D\ne 0$ ( $D\ne 0$ , and the characteristic is not $2$ by hypothesis).

Write $(x,y)=\left(\frac{1}{2}(u+v^2),\frac{1}{2}v(u+v^2)\right)$. Then $x=\frac{1}{2}(u+v^2)\ne 0$, and $(u,v)=\left(2x-{\left(\frac{y}{x}\right)}^2,\frac{y}{x}\right)$. The equality $u^2-v^4=4D$ gives

so that $(x,y)=\left(\frac{1}{2}(u+v^2),\frac{1}{2}v(u+v^2)\right)\in C_1$, and $(x,y)\ne (0,0)$.

Take any point $(x,y)\in C_1\setminus \{(0,0)\}$, then $x\ne 0$. Write $(u,v)=\phi (x,y)=\left(2x-\left(\frac{y}{x}\right),\frac{y}{x}\right)$. Then $(x,y)=\left(\frac{1}{2}(u+v^2),\frac{1}{2}v(u+v^2)\right)=\psi (u,v)=(\psi \circ \phi )(x,y)$. Thus $\psi \circ \phi =1_{C_1\setminus \{(0,0)\}}$. Similarly, take any point $(u,v)\in C_2$. Write $(x,y)=\psi (u,v)=\left(\frac{1}{2}(u+v^2),\frac{1}{2}v(u+v^2)\right)$. Then $(u,v)=\left(2x-{\left(\frac{y}{x}\right)}^2,\frac{y}{x}\right)=\phi (x,y)=(\phi \circ \psi )(u,v)$. Thus $\phi \circ \psi =1_{C_2}$.

This proves that $\phi$ and $\psi$ are bijections.

Therefore $|C_2|=|C_1\setminus \{(0,0)\}|=|C_1|-1$, and $|C_1|=|C_2|+1$.

To conclude, in any given finite field whose characteristic is not $2$, the number of finite points on $C_1$ is one more than the number of finite points on $C_2$. □