Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

up vote 4 down vote accepted

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.)

share|improve this answer
Thanks for the nice example. I have just returned from the ICM in Hyderabad. Will take a closer look at it later. –  Chandan Singh Dalawat Aug 29 '10 at 3:37
Great! I don't think there were any explicit examples in the literature; I had asked a few experts, and they didn't know any. –  Chandan Singh Dalawat Nov 26 '10 at 4:04
add comment

I do not know about the local-to-global principle for testing rationality of a surface, but weakening "rational" to "unirational" it certainly fails. Indeed, del Pezzo surfaces violating the Hasse principle are examples of such a violation: I will sketch an argument below.

Observe that

  • a smooth del Pezzo surface of degree at least two over a local field containing a point is automatically unirational (with a uniform bound on the degree of unirationality): this is classical, see for instance Manin's book on Cubic forms, the only use of the field being local is to ensure the existence of a point not lying on any exceptional curve;

  • a del Pezzo surface cannot be unirational over a field where it has no points: this is obvious (Lang-Nishimura).

Therefore, any del Pezzo surface violating the Hasse principle is unirational in every completion of the base-field (since it has points everywhere locally), but cannot be unirational over the base-field (since it has no points).

There are del Pezzo surfaces of degree four violating the Hasse principle: the following is a beautiful example of Birch--Swinnerton-Dyer of such a surface S in $\mathbb{P}^4$. \[ S \colon \,\left\{ \begin{array}{rcl} uv & = & x^2 - 5 y^5 \\\ [5pt] (u+v)(u+2v) & = & x^2 - 5 z^5 . \end{array} \right. \]

Remarks. A del Pezzo surface of degree at least five cannot be a violation, since as soon as it satisfies the Hasse principle for points, it is rational. Hence, this example is "minimal" in this sense. In this case, the degree of unirationality of S over the completions of Q is at most two; I do not know if the surfaces over the various completions are in fact rational.

share|improve this answer
Very nice, damiano! –  Wanderer Apr 13 '10 at 16:16
As I recall the history, this example is not completely "fair" because some of the local surfaces are in fact cones, and the only singular point is the vertex. There is an "honest" example by V.A. Iskovskikh, and a long paper of Colliot-Thelene-Sansuc elaborating on the "correct" notion of the "birational" Hasse principle which should be used in this case. –  VA. Apr 13 '10 at 16:46
Reducing the equations modulo 2 and 5 you do indeed get singular surfaces, but over the completions of Q you get smooth surfaces. –  damiano Apr 13 '10 at 17:02
Very nice example. Many thanks. –  Chandan Singh Dalawat Apr 14 '10 at 5:55
add comment

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.