Show that the variety C is rational

Question

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?

Answer 1

Question: "But why can I follow that the two varietys, induced by those polynomials, are isomorph?"

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 $U \cong V$.