The simplest way to find a parametrization of the plane projective curve $XZ-Y^2=0$.

Question

I have to explain to some first year math students that the projective algebraic set $\textbf{Z}(XZ-Y^2)\subset\mathbb P^2_k$ is $$V=\{(a^2_0:a_0a_1:a^2_1)\subset\mathbb P^2_k \,:\, \textrm{for}\; a_0,a_1\in k\}$$

It is evident that $V\subseteq \textbf{Z}(XZ-Y^2)$, but which is, in your opinion, the simplest way to show the other inclusion?

Answer 1

Suppose that $(x:y:z) \in \mathbf{Z}(XZ - Y^2)$. Then we know that $xz - y^2 = 0$, i.e. $$ y^2 = xz $$ Since $k$ is algebraically closed by hypothesis $\sqrt{x},\sqrt{z} \in k$, thus $$ (x:y:z) = (x : \sqrt{x} \sqrt{z} : z) \in V $$ so $\mathbf{Z}(XZ - Y^2) \subseteq V$ and we're done.

Answer 2

Just a note, lots of number theory of quadratic forms happens because all null vectors of your form, with rational integer entries, are integer multiples of $$ \left( u^2 , u v, v^2 \right) $$ with $\gcd(u,v)=1.$

I'm just sayin'