Showing that $\mathbb{P}_k^2$ is birational to $\mathbb{P}_k^1 \times \mathbb{P}_k^1$

Question

I wanted to check that there was nothing (roughly) wrong with my reasoning in showing that $\mathbb{P}_k^2$ is birational to $\mathbb{P}_k^1 \times \mathbb{P}_k^1$.

First of all, I know that for two irreducible quasi-projective varieities $X$ and $Y$, $X$ is birational to $Y$ if and only if the function fields $k(X) \simeq k(Y)$.

Now, if we consider $\mathbb{P}^2_k \cap U_z \simeq \mathbb{A}^2_k$, then the function field of $\mathbb{P}^2_k$ is that of $\mathbb{A}^2_k$, so is $k(x,y)$.

On the other hand, as a variety, $\mathbb{P}_k^1 \times \mathbb{P}_k^1 \simeq \mathbb{V}(xw-yz) \subset \mathbb{P}_k^3$ (using the Segre morphism). Then taking $\mathbb{V}(xw-yz) \cap $ $ U_w \simeq \mathbb{V}(x-yz) \subset \mathbb{A}^3_k$, we have that $\mathbb{V}(xw-yz) \cap U_w $ $= k[x,y,z]/(x-yz) $ $ \simeq k[y,z]$. So then $k(\mathbb{P}_k^1 \times \mathbb{P}_k^1) = k(y,z) \simeq k(\mathbb{P}_k^2)$.

So then $\mathbb{P}_k^2$ is birational to $\mathbb{P}_k^1 \times \mathbb{P}_k^1$.

Does this look essentially right?

Answer 1

This looks solid.

Alternatively you can construct mutual inverse rational maps $\phi:\Bbb P^2\dashrightarrow\Bbb P^1\times\Bbb P^1$ and $\psi:\Bbb P^1\times\Bbb P^1\dashrightarrow\Bbb P^2$ using the homogeneous coordinates on $\Bbb P^2$ and $\Bbb P^1\times\Bbb P^1$. These homogeneous coordinates are described by \begin{align*} (t_0,t_1,t_2)\sim(\lambda\cdot t_0,\lambda\cdot t_1,\lambda\cdot t_2) && (s_0,s_1,t_0,t_1)\sim(\lambda\cdot s_0,\lambda\cdot s_1,\mu\cdot t_0,\mu\cdot t_1) \end{align*} for $\lambda\neq 0$ and $\mu\neq 0$.

There are open sets $U\subset \Bbb P^2$ and $V\subset\Bbb P^1\times\Bbb P^2$ described by \begin{align*} U&=\{(s:t:1)\in\Bbb P^2:t_0,t_1\in k\} & V &=\{(s:1:t:1)\in\Bbb P^1\times\Bbb P^1:s_0,t_0\in k\} \end{align*} Defining \begin{array}{ccccccc} U & \xrightarrow{\phi} & V & & V & \xrightarrow{\psi} & U \\ (s:t:1) & \mapsto & (s:1:t:1) & & (s:1:t:1) & \mapsto & (s:t:1) \end{array} gives us our mutually inverse rational maps.

To see that $\Bbb P^2$ and $\Bbb P^1\times\Bbb P^1$ are not isomorphic, recall that any two lines in $\Bbb P^2$ intersect at a unique point (can you prove this?). Is the same true for $\Bbb P^1\times\Bbb P^1$?

Answer 2

Alternatively, the open subsets $\mathbb A^1\times\mathbb A^1\subset\mathbb P^1\times\mathbb P^1$ and $\mathbb A^2\subset \mathbb P^2$ are isomorphic, so $\mathbb P^1\times\mathbb P^1$ and $\mathbb P^2$ are birational.