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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 $n$.2012-05-04
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    Can you describe the field of functions on $X$?2012-05-04
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    @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.2012-05-04

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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?2012-05-04
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    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!2012-05-04
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    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! :)2012-05-04
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    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.2012-05-04
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    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.2012-05-04
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    Well in my case this is really a burden. I just cannot seem to be able to specialize. I often have grandiose ideas, but I cannot follow through to anything. I make projects, but so far with little success. That's why I try to post on math.stackexchange and mathoverflow, but all this moves very slowly in my head. And yes, algebraic geometry definitely is difficult, for relating machinery to examples, which is probably why there is a large mathoverflow community. There is a need to repeat folklore, spread things which are well-known but not worked out, because "obvious".2012-05-04
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    @GeorgesElencwajg I have a stupid question. I suppose that the projection is considering a point $x\in X$ and then to every $u\in X$ we consider the point of intersection of the line $ux $ with the hyperplane. If that is the projection , then let's take $ (1,1,0,0)\in X $ then the line is $ (1,1,0,0)+t(x-1,y-1,z,w) $ then intersecting with A , we have that $ t= \frac{1}{1-x} $ and then the projection is $ ( 0 , \frac{y-x}{1-x} , \frac{z}{1-x} , \frac{w}{1-x} )$ but it's different from your projection. I suppose that you considered $x=0 $ but why you put $y-x$ instead of $ y$ ? I'm confused :/2013-01-05
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    Dear @ Daniel: you are absolutely right. The algebra was correct but my geometric description of $f$ was false: I have now corrected it in an Edit. Thank you very much for catching that mistake.2013-01-05
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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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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 $X.$