An irreducible quadric hypersurface is rational? Here quadric hypersurface just means it is generated by a polynomial with degree 2.
I can guess the idea is to project the hypersurface from a fixed point P, to some plane by drawing a line through P. And because the degree is 2, we can get one point on the plane? And the expression should be rational function.
But how to give a formal proof? That is to prove there exists a open set of the hypersurface that is isomorphic to a open set of some $A^m$.
 A: I will assume that the field is algebraically closed, since otherwise what you say is not true. Also, characteristic not 2, since quadrics are not so well behaved then. Given this, by completing square trick, you can reduce any quadric to the form $x_1^2+x_2^2+\cdots+x_k^2$, for some $k$. If $k=1$, the quadric is non-reduced and for $k=2$, the quadric is reducible. So, assume that $k\geq 3$. Then we can write the above as $(x_1+\sqrt{-1}x_2)(x_1-\sqrt{-1}x_2)+x_3^2+\cdots+x_k^2$. Changing variables, this is same as $x_1x_2+x_3^2+\cdots$. Now, the open set $x_1\neq 0$ is given by $x_2=-\sum_{i\geq 3} x_i^2/x_1$ which is clearly the same as the open set $x_1\neq 0$ in the affine space with co-ordinates $x_1,x_3,\ldots$.
A: Pick a point on the hypersurface $X$, say $P$, the projection from Fixed point $P$ is a rational map $\phi$ from $x\in X\setminus P$ to a hyperplane $P^{n-1}$of ambient space $P^n$. you know $\phi$ corresponds to the linear system $\{$ hyperplane  sections  of  $X$ passing through point P $\}$ with base point $P$. Then By Bezout theorem, away from $P$, $\phi$ is one to one. this map is finite morphism away from $P$. so it is homeomorphism. also check that it induces local isomorphism on sheaf structure i.e it is isomorphism of schemes. For this, you should work with specific equation of the hypersurface. it is a simple exercise.
