Let $k$ be a number field and $F$ a $1$-variable function field over $k$ (a finitely generated extension of $k$, of transcendence degree $1$, in which $k$ is algebraically closed). If $F$ becomes the rational function field over every completion $k\_v$ of $k$, then $F$ is the rational function field over $k$. This is a restatement of the local-to-global principle for the existence of rational points on $k$-conics.


Now suppose that $F$ is a $2$-variable function field over $k$, and that $F$ becomes the rational functional field over every completion $k\_v$ of $k$. Does it follow that $F$ is the rational function field over $k$ ? (In other words, if a $k$-surface is birational of ${\mathbb P}\_2$ over every $k\_v$, is it $k$-birational to ${\mathbb P}\_2$ ?)


I don't think the answer is known, but perhaps someone can put together what is known about the group of $\bar k$-automorphisms of the rational function field ${\bar k}(x,y)$ (the Cremona group) to decide one way or the other.



A local-to-global principle for being a rational surface
It seems to me that there are irrational surfaces over $\mathbb Q$ that are $\mathbb Q\_v$-rational for all $v$. (I couldn't find them in the literature, but didn't look very hard. Almost certainly they are to be found there, in papers by either Iskovskikh or Colliot-Thelene.) 


Take the affine surface $S$ given by $y^2+byz+cz^2=f(x)$, where $f$ is an irreducible cubic and $b^2-4c$ equals the discriminant $D(f)$ of $f$, up to a square in $\mathbb Q^\*$, and $D(f)$ is not a square. According to Beauville-Colliot--Thelene--Sansuc--Swinnerton-Dyer $S$ is not $\mathbb Q$-rational, but is stably rational. (Irrationality is Iskovskikh I think, in fact.) Via projection to the $x$-line a projective model $V$ of $S$ is a conic bundle over $\mathbb P^1$ with $4$ singular fibers (one is at infinity). There is an embedding of $V$ into a weighted projective space $\mathbb P(2,2,1,1)$; the defining equation is $Y^2+bYZ+cZ^2=F(X,T)T$, where $F$ is the homogeneous version of $f$. By construction the Galois action on the $8$ lines that comprise the singular fibers is via the symmetric group $S\_3$: the two lines in the fiber at infinity are conjugated, and the other six are permuted transitively. 


Claim: Assume that $D(f)$ is square-free and prime to $6$. Then $S$ is $\mathbb Q\_v$-rational for all $v$. 


Proof: Suppose that the decomposition group $G\_v$ at $v$ is cyclic. Whatever its order ($1,2$ or $3$) there are at least $2$ disjoint lines among the $8$ that are $G\_v$-conjugate, so they can be blown down to give a conic bundle over $\mathbb P^1$ with at most $2$ singular fibres and a $\mathbb Q\_v$-point; it is well known that such a surface is $\mathbb Q\_v$-rational.


Now suppose that $G\_v= S\_3$. Then $v$ is non-archimedean and $V$ has bad reduction there. In fact, exactly two of the singular fibers are equal modulo $v$; it follows that $G\_v=S\_3$ is impossible, and we are done.


E.g., $f=x^3+x+1$, of discriminant $-31$, $c=8$, $b=1$.


(This doesn't use stable rationality, but rather the fact that these surfaces, although irrational, are very close to being rational, in the sense that the action of $Gal\_{\mathbb Q}$ on the lines is as small as possible subject to the surface being irrational, and the action of the decomposition groups is even smaller.)


