Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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
up vote 3 down vote accepted

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

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

share|cite|improve this answer
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
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. – DCV Jan 12 '15 at 16:55
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.