EDIT: The vague question Q1 below is partially answered, while the concrete question Q2 seems to be still open.

Let $V$ be a geometrically integral variety over a field $K$. I consider the following properties:

(1) There exists a dominant rational map $\mathbb{P}_K^n\dashrightarrow V$ for $n={\rm dim}(V)$.

(2) There exists a dominant rational map $\mathbb{P}_K^n\dashrightarrow V$ for some $n$. ($V$ is unirational)

(3) There exists a dominant rational map $\mathbb{P}_K^n\dashrightarrow V$ with geometrically integral generic fiber, for some $n$.

(4) There exists a dominant rational map $\mathbb{P}_K^n\dashrightarrow V$ with a right inverse $V\dashrightarrow\mathbb{P}_K^n$, for some $n$. ($V$ is retract rational)

(5) There exists a birational map $\mathbb{P}_K^n\dashrightarrow V\times\mathbb{P}_K^m$ for some $m,n$. ($V$ is stably rational)

(6) There exists a birational map $\mathbb{P}_K^n\dashrightarrow V$ for some $n$. ($V$ is rational)

We have that $(6)\Rightarrow(5)\Rightarrow(4)\Rightarrow(3)\Rightarrow(2)\Leftrightarrow(1)$. For curves all of these properties are equivalent, but they diverge in higher dimension. From browsing the literature I gather that it is known that $(2)\not\Rightarrow(4)$ and $(5)\not\Rightarrow(6)$, and it is expected that $(4)\not\Rightarrow(5)$.

However, I am interested in property (3), which I could not find anywhere in the literature.

Q1: Does property (3) occur in the literature? Does it have a name? Is it equivalent to either (2) or (4)?

EDIT: As the answer of Daniel Loughran shows, the Châtelet surfaces described below are examples for $(3)\not\Rightarrow(5)$ over $K=\mathbb{R}$.

Phrased more down to earth, here is a very concrete question I am interested in:

Q2: Is every intermediate field $F$ of $\mathbb{R}(X,Y,Z)/\mathbb{R}$ which is algebraically closed in $\mathbb{R}(X,Y,Z)$ purely transcendental over $\mathbb{R}$?

Of course this is clear for $F$ of transcendence degree $0$, $1$ or $3$ over $\mathbb{R}$, so it is really just a question about surfaces. The equivalent question over $\mathbb{C}$ has a positive answer, as every unirational complex surface is known to be rational. The closest to a counterexample I found in the literature is the surface over $\mathbb{R}$ given by $x^2+y^2=f(z)$ with $f$ of degree $3$ with three real roots, which I think satisfies (2) but not (5), but I don't know if it satisfies (3).

EDIT: As explained in the answer of Daniel Loughran, such a surface $V$ satisfies (3). But it seems not clear whether the $n$ in $\mathbb{P}_K^n\dashrightarrow V$ with geometrically integral generic fiber can be chosen to be 3, which would be needed to answer Q2 negatively.

  • 2
    $\begingroup$ Are your maps in (1), (2), etc, $K$-defined? $\endgroup$
    – YCor
    May 8, 2020 at 13:49
  • $\begingroup$ Yes, everything defined over K. I'll change the $\mathbb{P}^n$ to $\mathbb{P}^n_K$ to make this clear. $\endgroup$
    – Arno Fehm
    May 8, 2020 at 13:50
  • 2
    $\begingroup$ I do not know if (3) has been studied much, but I have seen a similar condition but for covering families of curves in Kollár's notes arxiv.org/abs/0710.5516 Section 8 - called the Lefschetz condition. You might find some more references there. $\endgroup$
    – Frank
    May 9, 2020 at 10:01

1 Answer 1


Let $k$ be a field of characteristic $0$, $a \in k$ and $f$ a separable polynomial of degree $3$.

The projective surface $X$, given as the minimal smooth compactification of the affine surface $$X: \quad x^2 - ay^2 = f(z)$$ you have written down is an example of a Châtelet surface. (Note that $X(k) \neq \emptyset$ always as there is a rational point at infinity). These have been studied in great detail by Colliot-Thélène and his collaborators. The key paper relevant to your question is:

Arnaud Beauville, Jean-Louis Colliot-Thélène, Jean-Jacques Sansuc and Peter Swinnerton-Dyer - Variétés Stablement Rationnelles Non Rationnelles, Annals of Mathematics.

Such surfaces are non-rational provided $a$ is not a square in any of the residue fields of the irreducible factors of $f$. Moreover, in the above paper, it is shown that they are stably rational provided certain assumption hold (e.g. $f$ is irreducible with Galois group $S_3$). But as remarked in the comments, there are examples which are also not stably rational.

They prove this using universal torsors $$T \to X.$$ For a good overview of the theory of universal torsors, I would recommend the book

Skorogobatov - Torsors and rational points

I will just remark that these are torsors under the Néron-Severi torus, in particular the generic fibre is geometrically integral.

A sufficient criterion for the existence of a universal torsor is $X(k) \neq \emptyset$; but as already explained we have this property so universal torsors exist. There may be many universal torsors in general; but the twists of a given torsor gives a parametristation of the rational points of $X$. So there is always some torsor with a rational point. But it turns out that such torsors $T$ are birational to a complete intersection of two quadrics in projective space, which is shown to be a rational variety (details in the above paper). So this shows that (3) holds.

Altogether, this shows that there are Châtelet surfaces which satisfy (3), (5), but not (6), and also those which satisfy (3) but not (5) nor (6). This seems to give a complete answer to your question.

Further details on these results and constructions can be found in the seminar Bourbaki report:

Laurent Moret-Bailly - Variétés stablement rationnelles non rationnelles

  • 1
    $\begingroup$ Thanks, I will have a closer look at that paper. I am a bit confused though: $X(\mathbb{R})$ is a real manifold with two connected components, hence so is $(X\times\mathbb{A}^n)(\mathbb{R})$. Shouldn't this imply that $X\times\mathbb{A}^n$ and $\mathbb{A}^m$ are not birationally equivalent (over $\mathbb{R}$)? $\endgroup$
    – Arno Fehm
    May 8, 2020 at 20:41
  • 2
    $\begingroup$ I think you are both right. The (smooth projective model of) $X$ does not satisfy (4) or (5) because it has non-connected real locus. It satisfies (3) because a universal torsor with a real point is stably rational (by Colliot-Thélène, Sansuc and Swinnerton-Dyer, Intersections of two quadrics and Châtelet surfaces, II, Theorem 8.1). $\endgroup$ May 9, 2020 at 14:43
  • $\begingroup$ @OlivierBenoist: Thanks a lot. So if I understand correctly, the example would indeed answer the question. The only thing that is not clear to me yet is why a universal torsor with a real point necessarily exists. Is that a general fact? (A reference would be appreciated - I have not yet been able to access CT-S' La descente sur les variétés rationneles where apparently they were defined.) $\endgroup$
    – Arno Fehm
    May 10, 2020 at 7:34
  • $\begingroup$ Hi. Yes you are right - to prove stable rationality one needs some genericity assumptions, e.g. the polynomial $f$ is irreducible with Galois group $S_3$. Anyway, you seem happy with my answer so I will upgrade it and add more details. $\endgroup$ May 12, 2020 at 11:07
  • $\begingroup$ As Daniel Loughran explained to me by email, the universal torsors of these Châtelet surfaces are in fact of dimension 8, so this example does show that $(3)\not\Rightarrow(5)$, but it does not answer question Q2. $\endgroup$
    – Arno Fehm
    May 22, 2020 at 6:40

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