2

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?

Darth Geek
  • 12,543

2 Answers2

1

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.

  • So ${(\pm t^4,t^5,t^3)\in\mathbb{C}^3\mid t\in \mathbb{C}} \subseteq V$, but how would one prove that $V\subseteq {(\pm t^4,t^5,t^3)\in\mathbb{C}^3\mid t\in \mathbb{C}}$? – Darth Geek Apr 19 '17 at 09:44
  • This is exactly what I wrote : I took $(x,y,z) \in V$. If $z = 0$ then $(x,y,z) = (0,0,0) \in T$. Else $(x,y,z) = (\pm t^4,t^5,t^3)$ but both are in $T$. –  Apr 19 '17 at 09:46
  • 1
    @DarthGeek : I was a bit tired this morning : my argument works perfectly without any assumptions on $z$. I edited my answer. –  Apr 19 '17 at 11:25
  • Why is $(-t^4,t^5,t^3)$ in $T$?if there is $u$ such that $(u^4,u^5,u^3) = (-t^4,t^5,t^3)$ doesn't that mean that $t = u = 0$? – Darth Geek Apr 19 '17 at 12:11
  • 1
    I was more tired than expected : you need two parametrization, $T$ and $T' = { (-t^4,t^5,t^3}$. We have $ V = T \cup T'$ and $T \cap T' = (0,0,0)$. –  Apr 19 '17 at 12:34
  • @DarthGeek : I have edited my answer. I think you can't have only one parametrization but I didn't do the computations (if you want I can try to do it but I don't know when I'll have time). –  Apr 19 '17 at 12:56
1

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 $(x,y,z) = (\pm t^4,t^5,t^3)$.

Darth Geek
  • 12,543