The set $$\{(a,b,c) \in \mathbb Z^3: \gcd(a,b,c)=1, a^2+b^2=c^2, ~\text{and}~z \neq 0 \}$$ of primitive Pythagorean triples is in bijection with the set $$P:=\{(u,v) \in \mathbb Q \times \mathbb Q: u^2+v^2=1 \}$$ of rational points on the unit circle. Then
$P \cong C(\mathbb Q)$, where $C(\mathbb Q)$ is the set of $\mathbb Q$-points on the projective variety $C:=V(X^2+Y^2=Z^2)$ given by the unit circle.
The map $f: \mathbb Q \to P$ defined by $f(t)=\left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2} \right)$ can be extended to the "rational map" $\varphi: \mathbb P^1 \to C$ given by $$\varphi([X,Y])=[Y^2-X^2,2XY, Y^2+X^2].$$
- To show $P \cong C(\mathbb Q)$, note that the unit circle $u^2+v^2=1$ defines the projective variety $C:=V(X^2+Y^2=Z^2)$ with homogeneous transform $u=X/Z, v=Y/Z$.
Now we have a map $\mathbb Q \hookrightarrow \mathbb P^1(\mathbb Q)$ by $t \mapsto [t,1]$.
In the same we way we can embedd $P \hookrightarrow \mathbb P^2(\mathbb Q)$ via $(u,v) \mapsto [u,v,1]$. So this map induces an injection between $P$ and $C(\mathbb Q)$, which would be surjective as well.
Thats all about the 1st part. Am I right ?
- What about the 2nd part ?
Note that $C$ or $C(\mathbb Q)$ is the projective subvariety in the prjective space $\mathbb P^2(\mathbb Q)$. Here $\varphi$ is homogeneous map and so $$\varphi(x_0,y_0,z_0)=\varphi(x_1,y_1,z_1)$$ where $(x_0,y_0,z_0) \sim (x_1,y_1,z_1)$ if $$(x_0,y_0,z_0)=(\lambda x_1, \lambda y_1, \lambda z_1)$$ for some nonzero $\lambda \in \mathbb Q$.
So the map $\varphi$ is well defined. Infact it is isomorphism since $\varphi^{-1}$ exists.
I would appreciate more clear/better understanding of the above questions.