How do i prove that :
$X$ is the hypersurface $wx=yz$ in $\mathbb{A}^{4}$ then $X$ is rational.
I do know the definition of $X$ being rational, but don't know how to apply that prove the above result.
How do i prove that :
$X$ is the hypersurface $wx=yz$ in $\mathbb{A}^{4}$ then $X$ is rational.
I do know the definition of $X$ being rational, but don't know how to apply that prove the above result.
Let $H\subset \mathbb A^4$ be the hyperplane $z=w$ and $U=X\setminus H$ the complementary open subset .
Let $A\subset \mathbb A^4$ be the hyperplane $x=0$ , isomorphic to $\mathbb A^3$, and consider the open dense subset $V=A\setminus H \subset A$ .
Birationality of $X$ will be proved by exhibiting an isomorphism between $U$ and $V$ .
That isomorphism is $f:U\stackrel {\cong}{\to} V: (x,y,z,w) \mapsto (0,y-x,z,w)$
Its inverse is $f^{-1}:V\stackrel {\cong}{\to} U: (0,\eta,\zeta,\omega) \mapsto (\frac {\eta\zeta}{\omega-\zeta},\frac {\omega\eta}{\omega-\zeta},\zeta,\omega)$
Edit: The secret revealed
Here is how $f$ is obtained.
Let $v$ be the vector $v=(1,1,0,0)$. For every point $q=(x,y,z,w)\in X$ consider the line given parametrically by $q+tv$ .
Its point of intersection with the hyperplane $A$ corresponds to $t=-x$ and is the point $\hat f(q)=(0,y-x,z,w)$.
Notice that $\hat f$ is not injective on $X$: it collapses the planes $z=w=0$ and $x=y,z=w$ (whose union constitute the intersection $X\cap A$) respectively to the lines $x=z=w=0$ and $x=y=0, z=w$ of the hyperplane $A$.
However the restriction $f$ of $\hat f$ to $U$ is injective and is even an isomorphism $f=\hat f\mid U:U\stackrel {\cong}{\to} V$, as already mentioned.
There is a map $f:\mathbb C[x,y,z,w]/(xw-yz)\to \mathbb C(X,Y,Z)$ such that $f(x)=X$, $f(y)=Y$, $f(z)=Z$ and $f(w)=YZ/X$. This extends to a map $\bar f:\operatorname{Frac}\bigl(\mathbb C[x,y,z,w]/(xw-yz)\bigr)\to \mathbb C(X,Y,Z)$ which is clearly surjective. Since it is injective because its domain is a field, it must be an isomorphism.
I hope it works! (maybe not much different from dear Georges Elencwajg’s solution.)
Let $U=X \setminus \{0\}$ and let $X':=\pi(U)$ where $\pi : \mathbb{A}^4 \setminus \{0\} \to \mathbb{P}^3$ be the usual projection. In fact, $X'$ can be identified with the image of the Segre embedding $\mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^3$ defined by $t_0t_3=t_1t_2$ where $[t_0:t_1:t_2:t_3]$ is the homogeneous coordinates of $\mathbb{P}^3$ and $k(X)\cong k(X').$
Let $U_0=\{t_0 \neq 0\}$ be the affine open subset of $\mathbb{P}^3,$ then, $U_0 \cap Z(t_0t_3=t_1t_2)$ is an open subset of $X’.$ Let $V_0= \{u_0 \neq 0\}$ be an affine open subset of $\mathbb{P}^2$ where $[u_0:u_1:u_2]$ is the homogeneous coordinates of $\mathbb{P}^2.$ Now, $U_0 \cap X’ \cong V_0$ since their coordinate rings are isomorphic to $k[x,y].$ Hence, $X'$ is birational to $\mathbb{P}^2$ and so is $X.$