2 archimedean corrected to non-archimedean

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 archimedean 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.)

1

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 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.)