# Parameterizing a rational curve

I'm having trouble finding a parameterization for the following curve: $x^4 - 2x^2yz + y^2z^2 - y^3z = 0$ taken to be a curve in $\mathbb{C}\mathbb{P}^2$. I followed the example on Wikipedia where they parameterized a circle's equaton, ie., I demohogenized the polynomial at $z$, so I reduce to the curve $x^4 - 2x^2y + y^2 - y^3 = 0$. Then I observed that the curve contains the point $(0,1)$, so I considered the line through $(0,1)$ with slope $t$, ie. $y = tx + 1$ and I plugged this into my equation.

I ended up with an ugly quartic polynomial in $x$ with coefficients in $t$...Specifically, I got $x^4 -2tx^3 + (2+t^2)x^2 + (-t-2t)x + 2 = 0$ and hopefully I didn't make any mistakes. Solving for $x$ doesn't seem easy. Is this the right approach to doing this?

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This method of parametrizing a curve typically only works if it's a degree-2 curve; that's because you end up with a quadratic with one known root, so the other root is well defined. Here it's better if you dehomogenize by setting $x=1$, since the resulting curve is quadratic in $z$ - but it's still cubic in $y$, so this method still doesn't seem to apply. –  Greg Martin Apr 3 '12 at 6:32

Let us de-homogenize the equation by setting $X=x/y$, $Z=z/y$ (which is the same as putting $y=1$): $$Z^2-(1+2X^2)Z+X^4=0.\qquad(+)$$ Let us project to the $X$-axis and see where the branch points are: those are the $X\in\mathbb{C}$ where the equation for $Z$ has a double root, i.e. where $1+4X^2=0$. We thus have a double cover of the $X$-sphere with the branch points $\pm i/2$, hence the curve is indeed rational. To get a rational parametrization, we just need to find a rational function $\mathbb{CP}^1\to\mathbb{CP}^1$, $T\mapsto X(T)$, which gives the same branched double cover of $\mathbb{CP}^1$. We can take e.g. $(X+i/2)/(X-i/2)=T^2$, i.e. $$X(T)=-\frac{i}{2}\frac{1+T^2}{1-T^2}.$$ We can now compute $Z(T)$ out of $(+)$ and get the following: $$Z(T)=\frac{T^8-6T^6+14T^4-8T^2+1+4iT(T^2+1)^3}{4(1-T^2)^4}$$ and we have a rational parametrization. If you prefer, in homogeneous coordinates, setting $T=t/s$, we get $$z=\frac{t^8-6t^6s^2+14t^4s^4-8t^2s^6+s^8+4its(t^2+s^2)^3}{4}$$ $$x=\frac{i}{2}(s^2+t^2)(s^2-t^2)^3$$ $$y=(s^2-t^2)^4.$$