Is any quadric birational to a product of Brauer-Severi varieties?

Question

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

Is $X$ birational to a product of Brauer-Severi varieties over $k$?

Recall that a Brauer-Severi variety over $k$ is a non-singular projective variety over $k$ which becomes isomorphic to some projective space over $\bar k$.

Answer 1

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 $V \sim \prod_i W_i$.

Answer 2

Of course that is true for any field $k$ where all such quadrics have $k$-points, e.g., finite fields. However, I expect that it is generally false.

Assume that $X$ is birational to a product $P$ of Severi-Brauer varieties (including, possibly $\mathbb{P}^1$ factors). If $k$ is infinite, then for a general $2$-plane section $C$ of $X$, $C$ would map into each Severi-Brauer factor of $P$. Using the Esnault-Levine-Wittenberg indices as in Kollár's recent article, this forces each Severi-Brauer factor to be a curve. So $P$ is a product of copies of $C$ and $\mathbb{P}^1$. In particular, this means that the splitting fields of $X$ are precisely the splitting fields of $C$. But I expect this is "typically" false, i.e., there should be many more splitting fields for $X$ than for $C$.

For instance, if $X$ is given by the diagonal quadratic form, $$ Q(x_0,x_1,\dots,x_n) = x_0^2 + t_1 x_1^2 + \dots + t_n x_n^2,$$ over $\mathbb{C}(t_1,\dots,t_n)$, and if the $2$-plane is the locus where $x_i=a_ix_2+b_ix_1$ for $i\geq 3$ and given $a_i,b_j\in \mathbb{C}$, then the splitting fields of $$ G(x_0,x_1,x_2) = x_0^2 + (t_1+b_3^2t_3+ \dots + b_n^2t_n)x_1^2 + (t_2 + a_3^2t_3 + \dots + a_n^2t_n)x_2^2$$ should be far fewer than splitting fields of $Q$. For instance, for every $i=1,\dots,n$, the field extension $k(\sqrt{-t_i})$ is a splitting field of $Q$, but I do not see why that should be a splitting field for $C$.

Answer 3

If you project from a general line in $\mathbb{P}^n$ you get a map from the blowup $\tilde{Q}$ of $Q$ to $\mathbb{P}^{n-2}$ which is a conic bundle. Over an open subset of $\mathbb{P}^{n-2}$ (which is the complement of a quadric) this conic bundle is nondegenerate, and so it is a Severi-Brauer variety. So, in the end you get a birational isomorphism of $Q$ with a severi Brauer variety over an open subset of $\mathbb{P}^{n-2}$. Of course, this is not completely what you asked for.

Answer 4

Of course this can happen, but generally this is not the case. As Jason suggested, the issue comes down to splitting fields. First, note that if $X$ and $Y$ are smooth projective varieties over a field $k$, and $X \dashrightarrow Y$ is a rational map, then for a field extension $L/F$, $X(L) \neq \emptyset$ implies that $Y(L) \neq \emptyset$. This can be proved by induction on the dimension of $X$, and can be reduced to the case that $L = k$ (by extending scalars). The induction case follows by blowing up a rational point on $X$, and noting that one has a rational map from the exceptional divisor, which has smaller dimension.

In particular, it follows that if $X$ and $Y$ are birational, then they have points over the same set of fields. Now, if we have a birational isomorphism $Q \sim X_1 \times \cdots \times X_r$, for Brauer-Severi varieties $X_i$ and a quadric $Q$, it follows that since $Q$ has points in quadratic extensions, each $X_i$ does as well. Of course, at least one of the $X_i$ must have no rational points, since if they all did, so would $Q$. Therefore we have a map $Q \to X$ for some nontrivial Brauer-Severi $X$. Writing $X = BS(A)$ for a central simple algebra $A$, we find that since $X$ has a point after a quadratic extension, $A$ must be index $2$. One can then define a rational projection $X \dashrightarrow C$ where $C$ is the associated Brauer-Severi conic curve for the underlying quaternion division algebra for $A$.

The existence of the rational map $Q \dashrightarrow C$ obtained by composition will give the contradiction in general.

Consider, for example an anisotropic Pfister quadric associated to a quadratic form of dimension at least 8. For example, a dimension 8 example would look like:

$q = x_0^2 + a x_1^2 + b x_2^2 + c x_3^2 + ab x_4^2 + ac x_5^2 + bc x_6^2 + abc x_7^2$.

Such that $q$ has no nontrivial zeros over the ground field $k$. This could be arranged, for example, by making the variables $a, b, c$ indeterminates.

The associated quadric (vanishing set) $Q = X(q)$, has the property that it is "strongly 2-incompressible," (for the definition of this see http://www.math.jussieu.fr/~karpenko/publ/icm-r.pdf or http://www.math.ucla.edu/~merkurev/papers/survey-update3.pdf). This implies that if one has a rational map $Q \dashrightarrow Y$ with $\dim Y < \dim Q$, then $ind_2(Y) < ind_2(Q)$. Here, $ind_2$ is the largest power of $2$ dividing the degree of any closed point. But since $Q$ has degree $2$ closed points (by intersection with lines), it follows that $Y$ would have to have a point over some odd degree extension. Considering $Y = C$ above gives the contradiction.