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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...

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    $\begingroup$ 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). $\endgroup$ Oct 28 '16 at 9:30
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    $\begingroup$ 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. $\endgroup$ Oct 28 '16 at 9:35
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    $\begingroup$ 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. $\endgroup$ Oct 28 '16 at 9:37
  • $\begingroup$ @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. $\endgroup$
    – RP_
    Nov 2 '16 at 9:41
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As requested, I'm upgrading my comments to an answer.

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).

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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.

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

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

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

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

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