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 $k=\mathbb{Q}$?



Quotients of rational surfaces
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. 


