Parametrizing algebraic variety. I am to find a parametrization of the variety $V:=V(x^2-yz,y^3-z^5) \subseteq \mathbb{C}^3$.
I've come up with $\{(t^4,t^5,t^3)\in \mathbb{C}^3\mid t\in \mathbb{C}^3\}=:T$.
It is clear that $T\subseteq V$ but I'm not so sure about the other inclusion.
I've tried to prove that $(y/x)^3 = z$ using some substitutions valid in the Variety, but to no avail.
Is the found parametrization correct?
 A: Let $(x,y,z) \in V$. We have $y^3 = z^5$ and so we can write $(x,y,z)$ as $(x,t^5,t^3)$ for some $t \in \mathbb C$. But since $x^2 = zy$ we can say $x = \pm t^4$. So $(x,y,z) = (\pm t^4, t^5, t^3)$. This means you need two parametrizations for your curve, defined by $\phi_{\pm}(t) = (\pm t^4, t^5, t^3)$. 
I don't think you can find a unique parametrization. An idea of proof could be the following : such parametrization gives you a birational morphism between the closure of $V$ in $\mathbb P^3$ and $\mathbb P^1$. In particular, the genus of the curve would be $0$. But your curve is a smooth intersection of two surfaces of degree $2$ and $5$, so if your curve was smooth its genus would be $16$. But your curve is singular, so one needs to study the singularity for finding the genus, if it's still nonzero then single parametrization cannot exist. Curves with a parametrization "almost everywhere" (i.e an isomorphism $ U \to V$  where $ U,V$ are Zariski open in $C$, resp. $\mathbb P^1$) are called unicursal. 
A: Let $(x,y,z)\in V$ and let $t = \dfrac{y^2}{z^3}$. Then
$$t^3 = \frac{y^6}{z^9} = \dfrac{z^{10}}{z^9} = z, \qquad t^5 = t^2z = \frac{y^4z}{z^6} = \frac{y^4}{z^5} = \frac{y^4}{y^3} = y.$$
Also, $x^2 = t^8$ so $x = \pm t^4$ so $(x,y,z) = (\pm t^4,t^5,t^3)$.
