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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!

Ehsan M. Kermani
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1 Answers1

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Over an algebraically closed field $k$ of characteristic $\neq 2$ every irreducible quadric $Q\subset \mathbb P^n_k$ has equation $q(x)=x_0x_1+x_2^2+...+x_m^2=0 \quad (2\leq m\leq n)$ 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

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

Georges Elencwajg
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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 08:59
  • And what if $char k=2?$ just curious. – Ehsan M. Kermani Nov 20 '12 at 09:01
  • 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 09:19
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    A small observation: the irreducible quadric may be written as $q(x) = x_0x_1 + x_2^2 + \ldots + x_m^2$, where $2\leq m\leq n$. One has $m = n$ if and only if the quadric is smooth. Anyway, the proof still works. – Davide Cesare Veniani Jan 12 '15 at 16:55
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    Can someone tell me why every irreducible quadric can be written in this form? – Leafar Apr 21 '15 at 14:51
  • Dear Georges, a recent question pointed to your nice answer here. Apart from DCV's comment, I note also that your answer applies only to certain fields $k$, say algebraically closed. (For instance, over $\mathbf R$, the forms $xy-z^2$ and $x^2+y^2+z^2$ have different indices, hence cannot be equivalent.) –  Apr 24 '15 at 11:22
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    Dear @Asal: thanks a lot for pointing out that (now corrected) lamentable omission of mine. My fondly remembered high-school teacher, Monsieur Devroegh, who first taught me the equations of hyperbolas and ellipses would be ashamed of me... – Georges Elencwajg Apr 24 '15 at 13:33
  • Dear Georges, as an algebraic geometer it is all too easy to forget that not every field is algebraically closed. My undergraduate students are often baffled when I excitedly inform them that every quadratic form in three variables defines a curve in the projective plane, and they try to apply this piece of wisdom in the case of $x^2+y^2+z^2 \ldots$ –  Apr 24 '15 at 13:42
  • @Davide Cesare Veniani. Thanks a lot Davide , you are absolutely right. I have now corrected my answer according to your remark. – Georges Elencwajg Sep 23 '20 at 00:22