Let $k$ be a field with algebraic closure $\bar k$. Assume that $k$ is perfect and not of characteristic $2$ for simplicity. Let 


$$X: \quad Q(x)=0, \quad \subset \mathbb{P}^n\_k,$$
be a non-singular quadric hypersurface over $k$. Then it is well-known that $X\times\_k \bar k$ is birational to some projective space, with the birational map given by stereographic projection. My question is whether some analogue of this holds over $k$. Namely



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> > Is $X$ birational to a product of Brauer-Severi varieties over $k$?
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Recall that a Brauer-Severi variety over $k$ is a non-singular projective variety over $k$ which becomes isomorphic to some projective space over $\bar k$.



Is any quadric birational to a product of Brauer-Severi varieties?
Consider the projective quadric $V$ given by
$$
2x^2+y^2+z^2+w^2=0
$$
over $\mathbb{Q}$. Inspired by Jason Starr's remark on splitting fields, I will prove that $V$ is not birational to a product of Severi-Brauer varieties over $\mathbb{Q}$. 


Clearly $V$ has no points over $\mathbb{R}$, hence no rational points. So if $V$ is birational to a product $\prod\_i W\_i$ of (positive-dimensional) Severi-Brauer varieties, then we must have $\dim W\_i = 1$ for all $i$ since the $W\_i$ have points over a quadratic extension but not all of them have points over $\mathbb{Q}$. Hence $V \sim W\_1 \times W\_2$. Also, $V$ has points over $\mathbb{Q}\_p$ for all primes $p>2$ by Chevalley-Warning and Hensel's lemma. So both $W\_i$ have points over $\mathbb{Q}\_p$ for all primes $p>2$, which implies that the $W\_i$ are isomorphic to either $\mathbb{P}^1$ or to $W':x^2+y^2+z^2=0$, since these are the only conics over $\mathbb{Q}$ up to isomorphism that have points over $\mathbb{Q}\_p$ for all $p>2$; moreover, not both $W\_i$ can be isomorphic to $\mathbb{P}^1$. This then implies that $W\_1 \times W\_2$ has no points over $\mathbb{Q}\_2$, while $V$ contains the point $(1,1,2,\sqrt{-7})\in V(\mathbb{Q}\_2)$.


**Edit:** It might be enlightening to generalize the example a little bit. Assume that $V$ is a smooth quadric over a number field $k$ that is birationally equivalent to a finite product $\prod\_i W\_i$ of Severi-Brauer varieties. If we define $S$ as the set of places $v$ of $k$ such that $V(k\_v) = \emptyset$ and for each $i$ we define $S\_i$ as the set of places such that $W\_i(k\_v) = \emptyset$, then by Lang-Nishimura (which I used above several times) we have
$$
S = \bigcup\_i S\_i.
$$
By the reciprocity law for the Brauer group, each $S\_i$ has an *even* number of elements. It follows that $S$ cannot consist of a single place. This explains the example above, where $V(\mathbb{Q}\_p) \neq \emptyset$ for all primes $p$, but $V(\mathbb{R}) = \emptyset$; hence $\# S=1$ and there cannot exist $W\_i$ such that $V \sim \prod\_i W\_i$.


