MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a projective surface defined over a field $k$ of characteristic $0$, and let $G$ be a finite group acting biregularly on $X$.

Assuming that $X$ is rational over $k$, is the quotient $X/G$ always rational?

If $k=\mathbb{C}$, we can use Castelnuovo's theorem and see that $X/G$ is unirational and hence rational. If $k=\mathbb{R}$, then $X/G$ is geometrically rational and also connected for the transcendental topology, and is thus rational.

But what happens for a general $k$, in particular when $k=\mathbb{Q}$?

share|cite|improve this question
For $k=\mathbf{Q}$ work of Saltman,(retract rational) on Noether's problem could be relevant, and Swan, Lenstra for abelian groups. But all of them talk of higher dimensional varieties. Specifically cyclically permuting $n$ variables over $\mathbf{Q}$ does not give a fixed field that is purely transcendental for $n=47$ (Swan) and $n=8$ (Lenstra). – P Vanchinathan Jan 10 '13 at 15:44
up vote 6 down vote accepted

This is not always true, and cubic threefolds give a counterexample over the field $k=\mathbb{C}(t)$. Let $\mathcal{Y}$ be a smooth cubic hypersurface in $\mathbb{P}^4_{\mathbb{C}}$. Let $L\subset \mathcal{Y}$ be a line. Denote by $\mathcal{X}$ the (locally closed) subvariety of $\mathcal{Y}\times L$ parameterizing pairs $(y,p)$ such that the intersection of $\text{Span}(L,y)$ with $\mathcal{Y}$ is a plane cubic $L\cup C$, where $C$ is a plane conic intersecting $L$ transversally at $p$. This condition on the conic $C$ is valid for all $y$ in a dense open subset of $\mathcal{Y}\setminus L$. Define an involution, $$ i:\mathcal{X} \to \mathcal{X}, \ i(y,p) = (y,q), $$ where $C\cap L$ equals $\{ p,q \}$. This involution defines an action on $\mathcal{Y}$ by the cyclic group $G$ of order. The quotient is the (dense, open) image $U$ of the projection $\text{pr}_{\mathcal{Y}}:\mathcal{X}\to \mathcal{Y}$.

How does this give a counterexample for surfaces? Let $\Pi$ be a linear $2$-plane containing $L$. Let $$\pi:(\mathbb{P}^3_{\mathbb{C}} \setminus \Pi) \to \mathbb{P}^1_{\mathbb{C}}$$ be linear projection away from $\Pi$. Let $U_\Pi$ be $U\setminus \Pi$, and let $\mathcal{X}_{\Pi}$ be the inverse image of $U_\Pi$ in $\mathcal{X}$. Of course this is a $G$-invariant, dense, open subset of $\mathcal{X}$. The claim is that a general fiber of $f\circ \text{pr}_{\mathcal{Y}}:\mathcal{X}_{\Pi} \to \mathbb{P}^1$ is a rational surface. Then letting $k$ be the function field $\mathbb{P}^1_{\mathbb{C}}$, and letting $Y$ and $X$ be the generic fiber of $f$, resp. $f\circ \text{pr}_{\mathcal{X}}$, this gives a counterexample.

Consider the morphism $$(f\circ \text{pr}_{\mathcal{Y}}, \text{pr}_{L}): \mathcal{X}_{\Pi} \to \mathbb{P}^1_{\mathbb{C}} \times_{\mathbb{C}} L.$$ A general point of the target parameterizes a pair $([H],p)$, where $H$ is a hyperplane in $\mathbb{P}^3_{\mathbb{C}}$ containing $\Pi$, and where $p$ is a point of $L$. Consider the "projective linear" tangent space to $X$ at $p$, i.e., the unique hyperplane $\Sigma$ in $\mathbb{P}^3_{\mathbb{C}}$ with maximal order of contact with $X$ at $p$. Then $\Sigma$ contains $L$. The intersection of $\Sigma$ and $H$ is a linear $2$-plane $\Xi$ that contains $L$. If $p$ and $H$ are general then $\Xi$ is not equal to $\Pi$, and the intersection of $\Xi$ with $\mathcal{Y}$ is a plane cubic $L\cup C$, where $C$ is a plane conic that intersects $L$ transversally at $p$ and $i(p)$. Thus the fiber of $(f\circ \text{pr}_{\mathcal{Y}}, \text{pr}_{L})$ over $([H],p)$ is $C\setminus \{p,i(p)\}$. Therefore, at least after passing to a dense open subset of the target, the morphism $(f\circ \text{pr}_{\mathcal{Y}}, \text{pr}_{L})$ is a dense open subset of a conic bundle. Moreover, this conic bundle has a section; namely send $([H],p)$ to the point $p$ of the conic $C$. A conic bundle with a section is birational to $\mathbb{P}^1$ over the base. Thus the composite morphism $$f\circ \text{pr}_{\mathcal{Y}}:\mathcal{X}_\Pi \to \mathbb{P}^1_{\mathbb{C}} $$ is birational to $$\text{pr}_1:\mathbb{P}^1_{\mathbb{C}} \times_{\mathbb{C}} \mathbb{P}^1_{\mathbb{C}}\times_{\mathbb{C}} L \to \mathbb{P}^1_{\mathbb{C}}.$$

If memory serves, this description of $\mathcal{X}$ as a conic bundle is described in the appendix to Clemens and Griffiths where they explain Mumford's Prym construction.

Edit. Of course the point is that the Clemens-Griffiths theorem proves that $\mathcal{Y}$ is not rational over $\mathbb{C}$. If the generic fiber $Y$ of $f$ were rational over $k=\mathbb{C}(t)$, then $\mathcal{Y}$ would be rational over $\mathbb{C}$.

Edit. I decided to add the following comment to the answer. In his book "Cubic Forms", Manin seems to give examples of quartic del Pezzo surfaces $Y$ over number fields that have a rational point, that have a degree $2$ double-cover $X$ that is rational (so that $Y$ is $X/G$ for $G$ a cyclic group of order $2$), yet with $Y$ irrational. The reference is Theorem IV.29.2, Theorm IV.29.4 and Remark IV.29.4.1, pp. 157-158 with r=5, and also Section IV.31, pp. 174--182.

share|cite|improve this answer
Thanks for the nice answer. Do you think that such a phenomenon could also be found over $\mathbb{Q}$? – Jérémy Blanc Jan 11 '13 at 13:05
Certainly you can find examples as above where the field is $\mathbb{Q}(t)$, but I guess you want the field to be $\mathbb{Q}$. I believe every del Pezzo surface of degree at least $4$ with a rational point is rational. So that suggests to focus on cubic surfaces. In Manin's "Cubic Forms", I believe he states a condition regarding the Galois action on the Picard lattice that implies that the cubic surface is irrational. So that is a place to start. – Jason Starr Jan 11 '13 at 14:52
@Jérémy: I double-checked in "Cubic Forms". Actually a del Pezzo of degree $4$ with a rational point is not automatically rational, and this gives examples as you ask, cf. Theorem IV.29.2, Theorm IV.29.4 and Remark IV.29.4.1, pp. 157-158 with $r=5$, and also Section IV.31, pp. 174--182. – Jason Starr Jan 11 '13 at 16:33
Thanks for the reference. That's perfect. – Jérémy Blanc Jan 12 '13 at 17:34

Just to round out the picture: if the characteristic of $k$ is positive and $G$ is a finite, but non-reduced group scheme (for example, the infinitesimal group scheme $\mu_p$ of $p$.the roots of unity), then the quotient $X/G$ need not even have Kodaira dimension $-\infty$ after desingularization. Moreover, if $G$ is not linearly reductive (for example, the infinitesmial group scheme $\alpha_p$), then a resolution of singularities $f:Y\to X/G$ need not satisfy $R^if_\ast{\mathcal O}_Y=0$ for $i\geq1$. Both phenomenons occur already if $\dim X=2$ - for example, there are many examples of ``unirational surfaces of general type'' in positive characteristic.

share|cite|improve this answer
Thanks for the comment. I was aware that caracteristic $p$ provided many counterexamples, it is good to have detailed this. – Jérémy Blanc Jan 11 '13 at 13:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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