How to test if these two threefolds are birationally equivalent?

Question

Assume we have the projective plane $\mathbb{A}^2=Spec(\mathbb{C}[r,s])$. Now take the projective plane over this affine plane $\mathbb{P}^2_{\mathbb{A}^2}$ with homogenous coordinates $[u:v:w]$.

Define a threefold $Y$ by the vanishing of $u^2+rv^2+sw^2$ in $\mathbb{P}^2_{\mathbb{A}^2}$, i.e. $Y=V(u^2+rv^2+sw^2)$.

On the other hand there is the threefold defined in this question, where its singularities are studied. We have $X=(\mathbb{A}^2\times \mathbb{P}^1)/G$, where $\mathbb{A}^2=Spec(\mathbb{C}[x,y])$ and $G=\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$ with generators $g_1$ and $g_2$. These act via: $g_1.(x,y,[\lambda:\mu])=(-x,y,[\lambda:-\mu])$ and $g_2.(x,y,[\lambda:\mu])=(x,-y,[\mu:\lambda])$.

Are $X$ and $Y$ birationally equivalent? How to test such a thing? If they are, can we write down explicit rational maps $X --> Y$ and $Y --> X$?

Answer 1

These $\mathbb{C}$-varieties are both rational threefolds, so they are birationally equivalent. First, in $Y$ consider the open subset $$D_+(w) = \text{Spec} \ \mathbb{C}[u/w,v/w,r,s]/\langle (u/w)^2 + r(v/w)^2 + s \rangle.$$ The linear projection of $D_+(w)$ to $\mathbb{A}^3$ by $(u/w,v/w,r,s)\mapsto (u/w,v/w,r)$ is an isomorphism. Thus $Y$ is rational.

Next, in $X$, consider the open subset $U =X \setminus \text{Zero}(x,y)$. There is an action of $\mathbb{G}_m = \text{Spec}\ \mathbb{C}[t,t^{-1}]$ on $U$ by $t\cdot (x,y,[\lambda,\mu]) = (tx,ty,[\lambda,\mu])$. This action realizes $U$ as $\mathbb{G}_m$-torsor over the surface $\Sigma = A\times B \cong \mathbb{P}^1 \times \mathbb{P}^1$ with homogeneous coordinates $([\chi,\upsilon],[\lambda,\mu])$ via $(x,y,[\lambda,\mu]))\mapsto ([x,y],[\lambda,\mu])$. Inside $B$, let $B^o$ be the open subset $D_+(\lambda\mu(\lambda^2+\mu^2))$. Define $U^o$ to be the open inverse image of $B^o$ under projection. Consider the morphism $T:U^o\to \mathbb{G}_m$ by $T(x,y,[\lambda,\mu]) = \lambda \mu x/(\lambda^2+\mu^2)$. It is straightforward to check that $T$ is $G$-invariant, and thus factors as $U^o/G \to \mathbb{G}_m$. Because of this, $U^o/G$ is isomorphic to $\mathbb{G}_m\times (A\times B^o/G)$.

Since $A\times B^o/G$ is a unirational surface, it is rational by Lüroth's theorem (it is also not hard to find an explicit rational parameterization). Therefore, $X$ is also rational.