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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 $k=\mathbb{Q}$?

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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
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2 Answers

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.

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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
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@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
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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.

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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
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