# Proving that if $X$ is the hypersurface $wx=yz$ in $\mathbb{A}^{4}$ then $X$ is rational.

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.

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We have to show that $X$ is a prevariety and it's function field $k[X] \cong \bar{k}(y_{1},\cdots,y_{n})$ for some $n$. –  Tomar May 4 '12 at 7:24
Can you describe the field of functions on $X$? –  Mariano Suárez-Alvarez May 4 '12 at 7:34
@MarianoSuárez-Alvarez: I think if $(x,y,w,z) \in \mathbb{A}^{4}$ then the set of all functions such that the product of the first and third co-ordinate = product of second and fourth co-ordinates. –  Tomar May 4 '12 at 7:54

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.

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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.

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How did you calculate the inverse? –  plm May 4 '12 at 8:48
Just by solving the system $y-x=\eta,z=\zeta,w=\omega, wx-yz=0$ for $x,y,z,w$. Try it, I am sure you can do it! –  Georges Elencwajg May 4 '12 at 8:54
Thanks. I wondered if there were a trick. But it's true too that I suck on that kind of exercise, so I am fearful of computations in general. This is a little depressing, I really try to fix that but my psychology really makes it hard. We'll see how it ends... Thanks alot for cheering me up! :) –  plm May 4 '12 at 10:37
Dear @plm, yes, it is perhaps a by-product of the tendency to teach ever more abstract and axiomatic mathematics that calculations are a bit neglected. This is especially true in algebraic geometry where we have to navigate between functors and explicit equations of varieties. In order to try to remedy this, I try to give computational and explicit prooofs here, even when I know more abstract and elegant arguments. So don't be depressed: almost all students have that difficulty with calculations, to a greater or smaller extent. –  Georges Elencwajg May 4 '12 at 11:26
Thinking about it. Can we obtain all rational hypersurfaces of projective space by projections? For irreducible equations of degree 3 in the plane singularity is equivalent to rationality, and I think projecting from the singular point always works. I guess this is really standard material (in elimination theory?) but I cannot find quickly a reference. There must be a moduli space of rational hypersurfaces, and one for those with a projection to a hyperplane. –  plm May 4 '12 at 11:39

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 $X.$

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