How to show two varieties are NOT birationally equivalent?

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This is an exercise from Ideals, Varieties and Algorithms by Cox et al.

Some backgroud

It comes from a problem showing that $Q=V(x^2+y^2-z^2-1)$, a hyperboloid, and $W=V(x+1)$, a plane, are birationally equivalent by two rational maps that define a one-to-one correspondence between $Q-V_Q(x+1)=Q-V$ and $W-V_W(y^2-z^2+4)=W-H$. The two rational maps are $$\phi: Q-V \rightarrow W-H, \quad (x,y,z)\mapsto (-1, \frac{-2y}{x-1}, \frac{-2z}{x-1})\\ \psi: W-H \rightarrow Q-V, \quad (-1, a,b)\mapsto (\frac{a^2-b^2-4}{z^2-b^2+4},\frac{4a}{a^2-b^2+4},\frac{4b}{a^2-b^2+4})$$

Then it asks to show that although $Q$ and $W$ are birationally equivalent, the excluded part from the map

$V$ and $H$, as varieties, are neither isomorphic nor birationally equivalent.

My attempt

It is equivalent to show that the two varieties $V=\{(x,y)\in \mathbb{R}^2| x=\pm y\}$ and $H=\{(a,b)\in \mathbb{R}^2 | a^2-b^2+4=0\}$ are not birationally equivalent.

Suppose there exists rationally mappings $\phi: H\rightarrow V$ and $\psi: V\rightarrow H$ such that they are inverse of each other.

My question:

I then defined them and set up the equations by the definition of the varieties. It makes it more complicated and I don't know where to go.

I also noticed one of them is two lines intersecting with each other, the other is two branches of a hyperbola. I wonder if that tells anything, because the author didn't mention how to identify equivalence using geometry.

Any help would be appreciated!