# Show that the variety C is rational

I have to show that the variety $$C:=\{(x_0 :x_1:x_2)\in \mathbb{P}_k^2\ |\ x_0^2x_1^2 + x_0^2x_2^2+x_1^2x_2^2 = 0\}\subset \mathbb{P}_k^2$$ is rational.

I already found a solution on this but I don't understand it completely. The solution says that we can use the Cremona-transformation $$\phi: \mathbb{P}_k^2 \to \mathbb{P}_k^2,\ (x_0:x_1:x_2)\mapsto (\frac{1}{x_0}:\frac{1}{x_1}:\frac{1}{x_2})=(x_1x_2:x_0x_2:x_1x_2),$$ which is birational with $$\phi=\phi^{-1}.$$ Therefore, $$\phi^*: k(\mathbb{P_k^2)}\to k(\mathbb{P_k^2})$$ is an isomorphism between the function-fields.

It is $$\phi^*(x_0^2x_1^2 + x_0^2x_2^2+x_1^2x_2^2) = x_2^2 + x_0^2 + x_1^2$$, now the solution says it holds that $$C\cap\{x_0x_1x_2 \neq0\}$$ is isomorph to $$Q\cap\{x_0x_1x_2 \neq0\}$$, where $$Q=\{x_0^2+x_1^2+x_2^2=0\}$$ is a smooth conic section, which is known to be isomorph to $$\mathbb{P_k^1}$$.

I understand that $$\phi^*$$ is an isomorphism and that $$\phi^*(x_0^2x_1^2 + x_0^2x_2^2+x_1^2x_2^2) = x_2^2 + x_0^2 + x_1^2$$. But why can I follow that the two varietys, induced by those polynomials, are isomorph?

• Thanks for your answer, I now see that $\phi$ is an isomorphism between $C_1$ and $Q_1$. Why does this imply that $C$ is birational to $Q$? Dec 13, 2021 at 22:57

Answer: You have constructed a map $$ϕ:C_1:=C−D(x_0x_1x_2)→Q_1:=Q−D(x_0x_1x_2)$$ with an inverse map $$ϕ^{−1}$$. Hence $$C_1≅Q_1$$ are isomorphic. Hence $$C$$ is birational to $$\mathbb{P}^1_k\cong Q_1$$ , and is therefore rational.
Note: By definition two (irreducible) varieties $$X,Y$$ are birational iff there are open subsets $$U\subseteq X, V\subseteq Y$$ and an isomorphism $$U \cong V$$.