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Show that every (irreducible) quadric in $\mathbb{P}^n$ is birational to $\mathbb{P}^{n-1}?$

It is easy to work on examples, like $xt-yz=0$ in $\mathbb{P}^3$ where we first project it to $\mathbb{P}^2$ from $[0:0:0:1]$ i.e. $[x:y:z:t] \mapsto [x:y:z]$ and the inverse (dominant) rational map is given by $[x:y:z] \mapsto [x^2:xy:xz:yz],$ but I do not know how to construct the inverse map for a general quadric!

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Over a field of characteristic not two, you can make a linear change of variables to turn every irreducible quadric into one whose equation is a sum of squares. –  Mariano Suárez-Alvarez Nov 20 '12 at 6:20
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up vote 3 down vote accepted

Over a field $k$ of characteristic $\neq 2$ every irreducible quadric $Q\subset \mathbb P^n_k$ has equation $q(x)=x_0x_1+x_2^2+...+x_n^2=0$ in suitable coordinates .
Projecting from $p=(1:0:0:\cdots:0)\in Q$ to the hyperplane $H\subset \mathbb P^n_k$ of equation $x_0=0$ will give the required birational isomorphism.
Explicitly, the projection is the birational map $$\pi: Q--\to H:(a_0:a_1:\cdots:a_n)\mapsto (0:a_1:\cdots:a_n)$$ You can compute the inverse rational map and find

$$\pi^{-1}:H--\to Q:(0:a_1:\cdots:a_n) \mapsto (-(a_2^2+\cdots +a_n^2):a_1\cdot a_1:\cdots:a_1\cdot a_n)$$

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Dear @Georges, thank you, but the correct inverse map has to be $(0:a_1:\cdots:a_n) \mapsto (-(a_2^2+\cdots +a_n^2):a_1\cdot a_1:\cdots:a_1\cdot a_n).$ –  Ehsan M. Kermani Nov 20 '12 at 8:59
    
And what if $char k=2?$ just curious. –  Ehsan M. Kermani Nov 20 '12 at 9:01
    
Dear ehsanno, you are absolutely right: I have corrected that typo. As for characteristic $2$, I think the calculation above remains correct but I don't know which quadratic forms can be reduced to $q$ by a change of coordinates. –  Georges Elencwajg Nov 20 '12 at 9:19
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