7
$\begingroup$

Let $X$ be a del Pezzo surface over a number field $k$. (A del Pezzo surface over $k$ is a smooth, projective, geometrically connected surface whose anti-canonical class $K_X$ is ample.) Let $d := K_X^2$ be the degree of $X$. It is well-known that $d$ satisfies $1 \leq d \leq 9$.

If $d \geq 5$, and $X(k) \neq \emptyset$, then $X$ is birationally equivalent to $\mathbb{P}^2_k$. Moreover, if $d=5$ or $d=7$, the condition on $X(k)$ is automatically satisfied. (This is Theorem 9.4.7 in Bjorn Poonen's Rational Points on Varieties.) In other words, for del Pezzo surfaces of degree $5$ and higher, the only obstruction to $k$-rationality is the possible lack of $k$-rational points.

My question concerns the "first" (i.e. highest-degree) non-trivial case, as far as rationality is concerned, namely the case where $d=4$. In this case, there are examples of $X$ with $X(k)\neq \emptyset$ but where $X$ is non-rational. An obstruction to rationality is given by the (non-trivial part of the) Brauer group $\operatorname{Br}(X)$ of $X$: that is, if $\operatorname{Br}(X)/\operatorname{Br}(k) \neq 0$, then $X$ is not $k$-rational. (To be sure, such $X$ exist, even among those $X$ that have rational points. This is where the $d=4$ case differs from the higher degree cases.) Indeed, this simply follows from the fact that $\operatorname{Br}$ is a birational invariant of smooth, projective, geometrically connected varieties, and from the fact that $\operatorname{Br}(\mathbb{P}^2_k) = \operatorname{Br}(k)$. My question is:

Is the converse true? That is, if $X$ is a del Pezzo surface of degree $4$ over a number field $k$ with $X(k)\neq\emptyset$, and $\operatorname{Br}(X) = \operatorname{Br}(k)$, does it follow that $X$ is a $k$-rational surface?
Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ In the rather special but related case of diagonal cubic surfaces, what you ask follows from the calculations of Colliot-Thélène, Kanevsky and Sansuc: see Proposition 1 and the following Lemme 1 in their article. $\endgroup$
    – Martin Bright
    Commented May 11, 2016 at 7:38
  • 1
    $\begingroup$ You can assume that your $X$ is minimal, since otherwise blowing down reduces to the case of higher degree. For dp4s, minimal is equivalent to Picard rank 1. So you asking whether it's possible to have a surface with $\mathrm{H}^1(k,\mathrm{Pic}\,\bar{X})=0$ and $\mathrm{H}^0(k,\mathrm{Pic}\,\bar{X})=\mathbb{Z}$, and with a rational point. $\endgroup$
    – Martin Bright
    Commented May 11, 2016 at 8:25
  • $\begingroup$ Daniel Loughran already pointed out that the answer is no. But if you want at least a partial result in this direction, see theorem 3.36 in Wittenberg's book Intersections de deux quadriques et pinceaux de courbes de genre 1. $\endgroup$
    – Gro-Tsen
    Commented May 11, 2016 at 8:43
  • 1
    $\begingroup$ Hi Martin. It is not quite true that for a DP4, minimal is equivalent to Picard rank $1$. You can have minimal DP4s with a conic bundle structure (the example I construct in my answer is of this type). For cubic surfaces, however, of course minimal is equivalent to Picard rank $1$. $\endgroup$
    – Daniel Loughran
    Commented May 11, 2016 at 8:44
  • $\begingroup$ @Gro-Tsen: The result you mention in Wittenberg's book concerns the Hasse principle. How does this give applications to rationality? $\endgroup$
    – Daniel Loughran
    Commented May 11, 2016 at 8:50

2 Answers 2

Reset to default
7
$\begingroup$

Interesting question.

But, alas, the answer is no.

The issue is that you have missed an extra non-rationality criterion. Namely, it is possible that such a surface $X$ has $\mathrm{Br}(X) = \mathrm{Br}(k)$, yet we have $\mathrm{Br}(X_K) \neq \mathrm{Br}(K)$ for some finite extension $K/k$.

Problems of this type for cubic surfaces over finite fields are considered in the PhD thesis of Shuijing Li: http://arxiv.org/abs/0904.3555

Though she only works over finite fields, the analysis for surfaces over number fields with a cyclic splitting field is more-or-less identical.

Here is how the counter-example goes. Take your DP4 $X$ and blow it up in a rational point not on a line to obtain a cubic surface $S$. We then choose this surface so that it has conjugacy class $C_7$ in the notation of Manin's table (p176 of his book); such a surface exists over $\mathbb{Q}$ by the recent resolution of the inverse Galois problem for cubic surfaces by Elsenhans and Jahnel. Its splitting field is cyclic of order $6$.

You see from the table that the Brauer group of $S$ is constant. However this surface is non-rational. This is because the Brauer group (modulo constants) becomes $(\mathbb{Z}/2\mathbb{Z})^2$ after a cubic extension, as it becomes the class $C_3$ from Manin's table.

I don't know whether this more general "extra non-rationality criterion" I mention is sufficient over number fields. That it is sufficient over finite fields is proven in Theorem 5.3.7 in Li's thesis.

To approach this general problem over number fields, I guess one could just enumerate all conjugacy classes of subgroups of the appropriate Weyl group in magma, and study which Brauer groups arise as well as the configurations of lines, and use the non-rationality criterion given in Theorem 5.3.1 in Li.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Thanks Daniel! The "extra non-rationality criterion" that you mention looks promising, maybe I will look into it later. It should be a finite check, as you suggest. $\endgroup$
    – R.P.
    Commented May 11, 2016 at 12:23
  • $\begingroup$ @René: If you want to discuss this approach more, feel free to contact me via email. $\endgroup$
    – Daniel Loughran
    Commented May 11, 2016 at 12:32
  • $\begingroup$ In Tschinkel's talk maths.ed.ac.uk/cheltsov/edge2017/pdf/yura.pdf, on slide 15 he says that it is a conjecture of Colliot-The'le`ne that existence of a point plus vanishing of Br(X_K)/Br(K) for all field extensions K/k implies stable rationality for del Pezzo surfaces. $\endgroup$
    – Evgeny Shinder
    Commented Mar 21, 2020 at 10:16
5
$\begingroup$

I will supplement Dan's nice answer by claiming that the answer is almost always no. Specifically, almost every degree 4 del Pezzo surface over the rational numbers with a rational point is non-rational but has trivial Brauer group.

To support this, I wrote a little Magma program that picked two random quadrics in $\mathbb{P}^4$ containing the point $(1:0:0:0:0:0)$ and with integer coefficients in $[-5,5]$. Let $X$ be the intersection of these two quadrics. I computed the Fano scheme of lines on $X$. If the Fano scheme is irreducible over $\mathbb{Q}$, that is, the lines are all Galois-conjugate, then a calculation shows that the Brauer group is trivial and the surface is not rational. I ran the program 100 times and, apart from 1 time when $X$ was singular, this was always the case. I'm sure it would be straightforward to show that the Fano scheme is always irreducible outside a thin set of the space parametrising such pairs of quadrics.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Hi Martin. I like your answer, but can you provide more details how you show non-rationality of such a surface? I mean a DP5 can have irreducible Fano scheme of lines, yet is still rational. $\endgroup$
    – Daniel Loughran
    Commented May 11, 2016 at 9:29
  • $\begingroup$ OK, maybe I need a little more. I just computed the Galois group of the field of definition of the 16 lines in one random example, and it was the whole Weyl group $W(D_5)$. I hope that's enough to guarantee non-rationality. And if that happens in one example, then I think it has to be the generic behaviour: specialising can only make the Galois action smaller. $\endgroup$
    – Martin Bright
    Commented May 11, 2016 at 9:43
  • 1
    $\begingroup$ I have realised that you can prove non-rationality in this case using a result of Iskovskih on conic bundles. Your DP4 has Picard number $1$. The blow-up in a point off a line is a cubic surface of Picard number $2$ with a line. Such cubic surfaces are non-rational by Corollary 2.6 of "Iskovskih - Rational surfaces with a sheaf of rational curves and with a positive square of canonical class". $\endgroup$
    – Daniel Loughran
    Commented May 11, 2016 at 11:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .