9
$\begingroup$

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 $\bar k$.

$\endgroup$
8
$\begingroup$

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 $V \sim \prod_i W_i$.

$\endgroup$
5
$\begingroup$

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 $C$.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ In fact, not at all. Daniel is asking about Severi-Brauer varieties over $k$. $\endgroup$ – abx Apr 16 '14 at 12:17
  • 2
    $\begingroup$ This is what I meant in the last sentence. $\endgroup$ – Sasha Apr 16 '14 at 15:53
  • $\begingroup$ Even though it doesn't answer the question, it is still very interesting to know, thanks! I wonder if this is "the" relationship between Brauer-Severi and quadrics, rather than the relationship of the kind I asked for in my question. $\endgroup$ – Daniel Loughran Apr 16 '14 at 16:23
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for?Browse other questions tagged or ask your own question.