$X^2Y^2-Z^2(X^2+Y^2)=0$ is rational

Question

I'm trying to verify whether or not the curve $X^2Y^2-Z^2(X^2+Y^2)=0$ is rational.

I've verified that the only singularities are $(0:0:1),(0:1:0), (1:0:0)$, which are all ordinary with multiplicity $2$, so we can use the formula $g=\frac{(d-1)(d-2)}{2}-\sum_{P\in C}\frac{m_P(m_P-1)}{2}$. Then: $$g =\frac{(4-1)(4-2)}{2}-3\frac{(2)(2-1)}{2}=3-3=0$$

(hopefully this is correct)

So we've concluded the curve is rational. But I can't see any obvious way to find a birrational map between the curve and $\mathbb{P}^1$. Is there some kind of heuristic to find this map?

Answer 1

The other posts have given solution, but they have not explained a general method by which a problem of this kind can be solved. This is what I do here.

The method is to take the system of all a $2$nd or $3$rd degree curves (I take all $2$nd degree curves) and require that they pass through all the multiple points that you have given. So so take a quadratic curve $$dx^2+ey^2+fz^2+axy+bxz+cyz=0$$ and if this passes through the three given singular points then $d=e=f=0$ so we have the linear system of curves $$axy+bxz+cyz=0$$
Now a $2$nd degree curve and a $4$th degree curve intersect in $2\times 4=8$ points. The intersections at the singular points accounts for $6$ points of intersection, since they are all double points. This leaves $2$ more points of intersection. Now pick a further point, require that the quadratic curve passes through that point. Then there is one free point (because $g=0$), and by solving for this point you get a birational map. You can pick any point, but if you pick a finite point you get a lot of algebra (which all works out, amazingly enough, if you really want to test the theory you should do this) however the nicest results are obtained by chosing an infinite point, and as they are already taken I will say that the quadratic curve has an intersection of multiplicity $3$ at $[0,1,0]$. The tangents to the original curve at this point are given by $$x^2-z^2=(z+x)(x-z)$$ and I shall thus arbitrary say that the quadratic is tangent to $z+x$, this means that $z+x$ divides $ax+cz$, and thus we have $a=c$ so our linear system is given by

$$axy+bxz+ayz=0$$ or since multiplication by constant doesnt matter I write it as

$$xz=t(xy+yz)$$ so now this curve and the original $$x^2y^2=z^2x^2+z^2y^2$$
must have one further point in common for which I solve in affine form

$$x=t(xy+y)$$ $$x^2y^2=x^2+y^2$$

to get

$$x=\frac{1+t^2}{1-t^2}\ \ y=\frac{1+t^2}{2t}$$ where of course $$t=\frac{x}{y(1+x)}.$$

And this is our birational map, written in affine notation.

(Of course, your curve is just the image of the circle under the quadratic involuion $[x,y,z]\mapsto [yz,xz,xy]$)

Answer 2

There is an affine model $x^2+y^2=x^2y^2$, that is $$\left(\frac{x}{y}\right)^2=x^2-1.$$ Let $u=x/y$, then we get a birational equivalence to the conic $$u^2=x^2-1.$$

Answer 3

There is a map from $\Bbb P^1(\Bbb C)$ given by $[u : v] \mapsto [2uv(u²-v²) : -2iuv(u²+v²) : (u²-v²)(u²+v²)]$

The preimages of the nodes are the three pairs $([0:1],[1:0]),([1:1],[1:-1]),([1:i],[1:-i])$ and otherwise, the map is injective.