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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!

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2 Answers 2

up vote 2 down vote accepted

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.

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Let $q(x_1,\cdots,x_n)=a_1x_1^2+\cdots+a_nx_n^2$ be a diagonalization of $q$. We have $F(q)\simeq F[x_1,\cdots,x_n]/(q(x_1,\cdots,x_n))$. The elements $\bar{x_1},\cdots,\bar{x_{n-1}}\in F(q)$ are algebraically independent over $F$ and $F(q)=K(\sqrt{\alpha})$ where $K=F(\bar{x_1},\cdots,\bar{x_{n-1}})$ and $\alpha=-({a_1\bar{x_1}^2+\cdots+a_{n-1}\bar{x_{n-1}}^2})/{a_{n}}$.

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