Denote a point in $\mathbb{P}^3$ by $[x,y,z,t]$. I want to show that $\mathbb{P}^2$ is birational to $Q = V(xt-yz)$. Here is what I got so far:
Let $f:Q \dashrightarrow \mathbb{P}^2, [x,y,z,t] \mapsto [x,y,z]$, then $f$ is a regular map on the open and dense set $Q \setminus \{[0,0,0,1]\}$, and im$(f) = \mathbb{P}^2$, so $f$ is dominant. Let $g:\mathbb{P}^2 \dashrightarrow Q, [x,y,z] \mapsto [x^2, xy, xz, yz]$, this map is well-defined on $U_0 \cup (U_1 \cap U_2)$, where $U_i$ means that the $i$-th homogeneous coordinate is not zero. Since the $U_i$ are open and dense, $U_0 \cup (U_1 \cap U_2)$ is open and dense as well. So $g$ is regular on that set. Now I am a little bit unsure. I would like to show that the image of $g$ is dense, but I don't quite see it. Is it even the case here?
If yes, then I would argue that $f$ and $g$ are inverse rational maps, in the sense that there exist open and dense subsets $U$ and $V$, such that $f \circ g = \text{Id}_{\mathbb{P}^2}$ on $U$ and $g \circ f = \text{Id}_{Q}$ on $V$. For example, one could choose $U = U_1 \cap U_2 \cap U_3$ and $V = U_1 \cap U_2 \cap U_3 \cap U_4$, so basically all the entries are non-zero. These sets should be open and dense, and if $xt-yz = 0$, then $t = \frac{yz}{x}$, and $[x^2, xy, xz, yz] = [x, y, z, \frac{yz}{x}]$, so we can see that $f$ and $g$ are indeed inverse to each other.
Is this proof okay?