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I want to prove that an irreducible quadric $ X \subset \Bbb A^n$ defined by a quadratic equation $ F(T_1,\ldots,T_n)=0$ is rational (i.e birational to some affine space $\Bbb A^m$ ).

I'm not sure how to do this, the book "Algebraic Geometry of Shafarevich" says that it's the same as in the case of $ \Bbb A^2$ , in this case we take a point $(a,b)\in X$ (in the curve) where $X$ is defined by $ F(x,y)=0 $ and $F$ has degree $2$. We consider $ y = t(x-a)+b$ and then $ F( x , t(x-a)+b ) $ is a polynomial of degree $2$ on $x$ , one solution is $(x-a)$ so we can divide and have a polynomial $ H(x,t) $ $f$ degree $1$ on $x$ , so we parametrized $x$ and $y$ . Shafarevich says that the idea in $\Bbb A^n$ is the same, but I don't understand how, the hint is to consider a non singular point of the curve, and following the same idea. Please help me with this )=

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See here for a similar result. – Georges Elencwajg Jan 5 '13 at 12:30

First, you have to assume that the ground field $k$ is algebraically closed: else you have counterexamples, like the conic $x^2+y^2=3$ in $\mathbb A^2_\mathbb Q$, which is certainly not rational over $\mathbb Q$, since it has no points with coordinates in $\mathbb Q$.

If $k$ is algebraically closed, take a smooth point $p\in X$ on your quadric and project from it to some affine hyperplane $H\subset \mathbb A^n_k$ not going through $p$ to get your birational morphism.
This means that to $x\in X$ you associate $f(x)\in H$, the point intersection of the line $\overline {px}$ with $H$.
This $f$ is the required birational map.
This map $p$ is not a morphism because it is not defined at $p$ nor, more generally, at the points of intersection of $X$ with the affine hyperplane through $p$ parallel to $H$.
Also, $f$ is not injective along the lines lying completely on $X$ and passing through $p$.
But the charm of birational geometry is that these little defects do not prevent $f$ from being a birational map from $X$ to $H$ !

I forgot to say that of course $H$ is isomorphic to $\mathbb A^{n-1}_k$, so that $X$ is birational to $\mathbb A^{n-1}_k$, and the $m$ in your question is necessarily equal to $n-1$.

Second edit
Here is a sample computation.
Let $X$ be the quadric $z=xy$ in $\mathbb A^3_k$. Take the origin for the projection center and project onto the affine (hyper)plane $H$ of equation $z=1$.
The birational map $f$ is then given by
$$f:X --\to H\cong \mathbb A^2_k: (a,b,c)\mapsto (\frac {a}{c},\frac {b}{c},1)\cong (\frac {a}{c},\frac {b}{c})$$ It is defined on the points of $X$ with $c\neq 0$, which means outside the union of the two lines $a=c=0$ and $b=c=0 $ lying on $X$.
The inverse birational map is $$f^{-1}:H\cong \mathbb A^2_k --\to X: (a,b,1)\cong (a,b)\mapsto (\frac {1}{b},\frac {1}{a},\frac {1}{ab})$$ It is defined outside of the union of the lines $a=0$ and $b=0$ in $\mathbb A^2_k$

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Thanks, but I have a question , where we are using the fact that the point is smooth? – Daniel Jan 5 '13 at 2:13
To prevent for example that the point be the vertex of a cone, in which case $f$ would squash $Q$ onto a lower dimensional quadric. – Georges Elencwajg Jan 5 '13 at 2:19
Yes , but that is intuitive, because my definition of non singular point , is just a point such that all the derivates does not take the zero value on it , how can I prove that in a completely formal way? – Daniel Jan 5 '13 at 2:24
Write your quadric $Q$ as $q(t_1,...,t_n)=0$ with $q$ homogeneous of degree $2$ and you will see that the origin is both a singular point and a vertex for $Q$. – Georges Elencwajg Jan 5 '13 at 2:29
And what is a vertex? – Daniel Jan 5 '13 at 7:31

Let $k$ be an algebraically closed field. Take the projective closure $\overline{X}\subset\mathbb{P}^n$ of $X$. Let $p\in\overline{X}$ be a smooth point and let $H\subset\mathbb{P}^n$ be an hyperplane not passing through $p$. Consider the projection $$\pi_p:\overline{X}\setminus\{p\}---> H,\: (x)\mapsto \left\langle x,p\right\rangle \cap H.$$ Now, $deg(\overline{X}) = 2$. Therefore the general line $\left\langle x,p\right\rangle$ intersects $\overline{X}\setminus\{p\}$ exactly in $x$. Therefore $\pi_p$ is birational. Finally, $\overline{X}$ is rational and therefore $X$ is rational as well.

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