Exercise 5.7.17* ($(\vartheta \circ \varphi)(P) = 2P$)

Proof. Recall that $O=(0:1:0)$ is the point at infinity of ${𝒞}_f(ℝ)$ and ${𝒞}_g(ℝ)$. We define $\phi$ and $𝜗$ without ambiguity by

$$\begin{array}{lll}\hfill \phi \{\begin{array}{ccc}\hfill {𝒞}_f(ℝ)\hfill & \hfill \to \hfill & {𝒞}_g(ℝ)\hfill \\ \hfill P=(x,y)\hfill & \hfill \mapsto \hfill & (u,v)=\left(\frac{y^2}{x^2},y-\frac{4y}{x^2}\right)\text{ if }P\ne O\text{ and }P\ne T=(0,0),\hfill \\ \hfill T=(0,0)\hfill & \hfill \mapsto \hfill & O\hfill \\ \hfill O\hfill & \hfill \mapsto \hfill & O,\hfill \end{array}& & \hfill \end{array}$$

$$\begin{array}{lll}\hfill 𝜗\{\begin{array}{ccc}\hfill {𝒞}_g(ℝ)\hfill & \hfill \to \hfill & {𝒞}_f(ℝ)\hfill \\ \hfill Q=(u,v)\hfill & \hfill \mapsto \hfill & (x,y)=\left(\frac{v^2}{4u^2},\left(1-\frac{20}{u^2}\right)\frac{v}{8}\right)\text{ if }Q\ne O\text{ and }Q\ne T=(0,0),\hfill \\ \hfill T=(0,0)\hfill & \hfill \mapsto \hfill & O\hfill \\ \hfill O\hfill & \hfill \mapsto \hfill & O.\hfill \end{array}& & \hfill \end{array}$$ These definitions make sense if $P\in {𝒞}_f(ℝ)$ implies $\phi (P)\in {𝒞}_g(ℝ)$, and if $Q\in {𝒞}_g(ℝ)$ implies $𝜗(Q)\in {𝒞}_f(ℝ)$. This is the case if $P=T$ or $P=O$. Let $P=(x,y)\in {𝒞}_f(ℝ)$, and suppose that $P\ne O$ and $P\ne T$. Then $$\begin{array}{lll}\hfill y^2=x^3+6x^2+4x,x\ne 0.& & \hfill \text{(1)}\end{array}$$

Now, if $(u,v)=\left(\frac{y^2}{x^2},y-\frac{4y}{x^2}\right)\in {𝒞}_g(ℝ)$, then, using (1),

$$\begin{array}{llll}\hfill g(u,v)& =v^2-u^3+12u^2-20u& \hfill & \\ \hfill & ={\left(y-\frac{4y}{x^2}\right)}^2-\frac{y^6}{x^6}+12\frac{y^4}{x^4}-20\frac{y^2}{x^2}& \hfill & \\ \hfill & =\frac{1}{x^6}\left[{(x^{3}y-4\mathit{xy})}^2-y^6+12x^2{y}^4-20x^4{y}^2\right]& \hfill & \\ \hfill & =\frac{y^2}{x^6}\left[{(x^2-4)}^2{x}^2-y^4+12x^2{y}^2-20x^4\right]& \hfill & \\ \hfill & =\frac{y^2}{x^6}\left[{(x^2-4)}^2{x}^2-{(x^3+6x^2+4x)}^2+12x^2(x^3+6x^2+4x)-20x^4\right]& \hfill & \\ \hfill & =0.& \hfill & \end{array}$$

Therefore $(u,v)=f(x,y)\in {𝒞}_g(ℝ)$.

Similarly, let $Q=(u,v)\in {𝒞}_g(ℝ)$, and suppose that $Q\ne O$ and $Q\ne T$. Then

$$\begin{array}{lll}\hfill v^2=u^3-12u^2+20u,u\ne 0.& & \hfill \text{(2)}\end{array}$$

If $(x,y)=\left(\frac{v^2}{4u^2},\left(1-\frac{20}{u^2}\right)\frac{v}{8}\right)$, then, using (2),

$$\begin{array}{llll}\hfill f(x,y)& =y^2-x^3-6x^2-4x& \hfill & \\ \hfill & ={\left(\left(1-\frac{20}{u^2}\right)\frac{v}{8}\right)}^2-{\left(\frac{v^2}{4u^2}\right)}^3-6{\left(\frac{v^2}{4u^2}\right)}^2-4\left(\frac{v^2}{4u^2}\right)& \hfill & \\ \hfill & =\frac{1}{64u^6}\left[u^2{(u^2-20)}^2{v}^2-v^6-24u^2{v}^4-64u^4{v}^2\right]& \hfill & \\ \hfill & =\frac{v^2}{64u^6}\left[u^2{(u^2-20)}^2-v^4-24u^2{v}^2-64u^4\right]& \hfill & \\ \hfill & =\frac{v^2}{64u^6}\left[u^2{(u^2-20)}^2-{(u^3-12u^2+20u)}^2-24u^2(u^3-12u^2+20u)-64u^4\right]& \hfill & \\ \hfill & =0.& \hfill & \end{array}$$

Therefore $(x,y)=g(u,v)\in {𝒞}_f(ℝ)$. Now take $P=(x,y)=(-1,1)$. Then $f(-1,1)=1+1-6+4=0$, so $P\in {𝒞}_f(ℝ)$. Moreover $(u,v)=(1,1-4)=(1,-3)$, and $g(1,-3)=9-1+12-20=0$, so $\phi (P)\in {𝒞}_g(ℝ)$ (this is also a consequence of our preceding proof).

Then

$$\begin{array}{llll}\hfill (𝜗\circ \phi )(P)& =𝜗(1,-3)& \hfill & \\ \hfill & =\left(\frac{9}{4},(1-20)\frac{-3}{8}\right)& \hfill & \\ \hfill & =\left(\frac{9}{4},\frac{57}{8}\right).& \hfill & \end{array}$$ □