# In what sense can we "describe" the rational points on a unirational surface?

This is not a completely precise question, but I hope someone can offer an interesting perspective on my problem. In the field of Diophantine geometry, an important question is deciding whether a geometrically rational surface $X$ defined over a number field $k$ admits a dominant rational map $\mathbb{P}^2_k \dashrightarrow X$; if it does, we say $X$ is unirational (or unirational over $k$, but the ground field is often understood to be part of the data of $X$ so really we don't need to mention it explicitly).

So for example, if $X$ is a del Pezzo surface of degree $\geq 3$, and $X(k) \neq \emptyset$, we know $X$ is unirational. On the other hand, the question is open in general for del Pezzo surfaces of degrees $1$ and $2$. However, I would like to ask:

What good does it do us to know that $X$ is unirational?

To be a little more specific, I am wondering how much help it is to know that $X$ is unirational, if what we're really interested in is as good a description of the set $X(k)$ of rational points on $X$ as we can get.

Of course, if $X$ is actually rational, i.e. there exists a birational map $\mathbb{P}^2 \dashrightarrow X$, we have as good a description of $X(k)$ as we could wish. But there are plenty of examples of geometrically rational $X$ that admit a unirational parametrization, but not a birational one.

An example. Let $X/\mathbb{Q}$ be degree $4$ del Pezzo surface given by $$xy + x + y - 6 = u^2, ~~~ xy - x - y + 6 = v^2.$$ Projection to the $y$-coordinate gives $X$ a conic bundle structure $\pi : X \rightarrow \mathbb{P}^1$; however, $\pi$ does not have a section (as can be verified by checking that $\pi^{-1}(2)$ does not have any $2$-adic points). But if $y$ has the form $(t^2+1)/(t^2-1)$, then $X_y := \pi^{-1}(y)$ takes the form $t^2u^2-v^2 = P_6(t)$, for some degree $6$ polynomial $P_6$ with rational coefficients, and this clearly admits a parametrization. This shows that if we pull back $\pi:X \rightarrow\mathbb{P}^1$ along the map $t \mapsto (t^2 +1)/(t^2-1)$, and we denote the result by $\pi':X'\rightarrow \mathbb{P}^1$, then $\pi'$ has a section, which shows that $X'$ is birational to $\mathbb{P}^2$. It follows that there exists a degree $2$ dominant rational map $$\phi:\mathbb{P}^2 \stackrel{2:1}{\dashrightarrow} X.$$ We note that, since $t^2=(y+1)/(y-1)$, the corresponding extension of function fields is obtained by adjoining a square root of $(y+1)/(y-1)$. On the other hand, since the Brauer group of $X$ is $\mathbb{Z}/2\mathbb{Z}$ (if I have made no mistakes), there does not exist a birational map $\mathbb{P}^2 \dashrightarrow X$.

The problem is of course that $\phi$ does not give all rational points on $X$. Indeed, it only gives those rational points $(x_0,y_0,u_0,v_0)$ where $y_0^2-1 = t_0^2$ is the square of a rational number. So for example, the fiber $X_3$ has the point $(x,u,v)=(3,3,3)$. So while $\phi$ does give infinitely many rational points on $X$, and even a Zariski dense set of them, it certainly does not give a complete description of $X(\mathbb{Q})$.

[One half-baked idea that did occur to me at this point, is that one could consider "twists" of $\phi$. Indeed, since $\phi$ is $2$-to-$1$, we can define quadratic twists of it in the following way: for each $c \in \mathbb{Q}^{\times}$, consider the field extension $K_c:=\mathbb{Q}(X)[t]/((y-1)t^2-c(y+1))$ of the function field $\mathbb{Q}(X)$ of $X$. This extension corresponds to a dominant rational map $\phi_c:X'_c \dashrightarrow X$, where $X'_c$ is some geometrically rational surface with function field $K_c$, which may be chosen to be smooth and projective, defined over $\mathbb{Q}$. So we get a family of rational maps $\{ \phi_c : X'_c \dashrightarrow X \}$, which together give all rational points on $X$. Unfortunately, I have no indication that the arithmetic of the $X'_c$ should be any easier to analyze than that of $X$ itself. Moreover, there doesn't even seem to be a good reason why the set of all $c$ such that $X'_c(\mathbb{Q})\neq \emptyset$ (independent of the choice of $X'_c$ by Lang-Nishmura) should admit of an easy description.]

Is there any way to get around this? It feels disconcerting to me that, even in the case of varieties that are so agreeable as to be unirational, it seems a hard problem to actually describe the set of rational points in any non-trivial manner. But of course, this might just be one of those cases where life doesn't turn out to be as pleasant as one might have hoped...

• Hi René. In fact one can even show something stronger for your surface: there is no finite collection of rational maps $\phi_i: \mathbb{P}^2 \dashrightarrow X$ such that $X(\mathbb{Q})$ lies in the image of the union of the $\phi_i(\mathbb{P}^2(\mathbb{Q}))$. The union of the $\phi_i(\mathbb{P}^2(\mathbb{Q}))$ is a thin set, by definition, but the rational points on $X$ are not thin as $X$ satisfies weak weak approximation, cf. en.wikipedia.org/wiki/Thin_set_(Serre). Commented Oct 28, 2016 at 9:30
• I won't deny, I do find your question a little vague. What kind of thing are you after? I mean if $X$ is a rationally connected variety then, conjecturally, the Brauer-Manin obstruction is the only one to weak approximation. This, combined with the fact that $\mathrm{Br}(X)/ \mathrm{Br}(\mathbb{Q})$ is finite implies that $X$ satisfies weak weak approximation. So you can "write down" rational points on $X$ by choosing $p$-adic points, away from some bad set of primes $p$, and knowing that there are rational points which are arbitrarily close to the given set of $p$-adic points. Commented Oct 28, 2016 at 9:35
• Cont: Of course this argument does not use the explicit choice of a unirational map anywhere, and I'm not sure how "useful" such an explicit map is, since in general it will give you so few rational points, as you say. Commented Oct 28, 2016 at 9:37
• @DanielLoughran: You are right, my question is vague, as I indicated. But thank you for answering anyway -- I like your first observation a lot! This definitely goes some way towards an answer. (In fact, if you'd want to post it as an answer, I will accept it.) I had been wondering myself whether one could use the Hilbert irreducibility theorem to show that the set of values of $y_0\in\mathbb{P}^1$ for which $X_{y_0}(\mathbb{Q}) \neq \emptyset$ does not equal the union of the images of a finite set of maps $\mathbb{P}^1 \to \mathbb{P}^1$. I have a feeling that this too should work somehow.
– R.P.
Commented Nov 2, 2016 at 9:41

In fact one can even show something stronger for your surface: there is no finite collection of rational maps $\phi_i:\mathbb{P}^2 \to X$ such that $X(ℚ)$ lies in the image of the union of the $\phi_i(\mathbb{ℙ}^2(ℚ))$.

To see this, we note that $X$ satisfies so-called weak weak approximation. This follows from the fact that the Brauer group of $X$ modulo constants is finite, and that the Brauer-Manin obstruction is only obstruction to weak approximation for $X$ (this being a theorem of Skorobogatov and Salberger)

The union of the $\phi_i(\mathbb{ℙ}^2(ℚ))$ is a thin set, by definition, but the rational points on X are not thin as X satisfies weak weak approximation by a theorem of Ekedahl, cf. the discussion in https://en.wikipedia.org/wiki/Thin_set_(Serre)

This property should hold more generally for any (geometrically) rationally connected non-rational smooth projective variety $X$ over a number field (the same argument applies in this case, assuming Colliot-Thélène's conjecture that the Brauer-Manin obstruction is the only one to weak approximation).

For such systems of equations.

\left\{\begin{aligned}&xy+x+y-a=u^2\\&xy-x-y+a=v^2\end{aligned}\right.

It is better to use an algebraic approach. He gives at once rational decisions.

$$x=\frac{k^2+2kt+2t^2}{at^2}$$

$$y=\frac{k^2+2(1-a)kt+(a^2-2a+2)t^2}{at^2}$$

$$u=\frac{k^2+(2-a)kt+2t^2}{at^2}$$

$$v=\frac{2(a-1)t^2+(a-2)kt-k^2}{at^2}$$