Assume $F$ is a field of characteristic $\neq 2$. Let $(V,q)$ be a quadratic space such that $\rm dim~ q\geq 3$. When $q$ is irreducible it is known that


there exist a purely transcendental field extension $K/F$ such that $[F(q):K]=2$. Here $F(q)$ is a function field of a scheme $X\_q:=\rm Proj (S(V^\*)/q)$ associated to $q$, where $S(V)$ is a symmetric algebra.


I want to show above fact, but I am not able to do it. I was trying to find some isotropic sub-form of codimension 2 of $q$, because in that case this sub-form will have function field which will be purely transcendental over $F$. Also note that the form $q$ will be irreducible since $\rm dim ~q\geq 3$. 


Can somebody suggest proof or a reference to the proof.


Thanks!



Quadratic subextension of the function field of quadric.
Let $v \in V$ be a non-isotropic vector, then the natural projection $\operatorname{Proj}(S (V^\*)/q)\to \operatorname{Proj}(S (v^\perp)) $ is a finite surjective map of algebraic varieties of degree $2$, so it corresponds to a degree $2$ extension of function fields. The second algebraic variety is just $\mathbb P^{d-2}$, so its function field is purely transcendental. 


