Beyond the Manin Obstruction

Br(X)/Br0(X)×X(Ak) → Q/Z. We shall call this the Brauer–Manin pairing. Let us define X(Ak) Br as “the right kernel” of this pairing, that is, the subset of points of X(Ak) orthogonal to all elements of Br(X). Manin made an important observation that by the global reciprocity law the image ofX(k) under the diagonal embedding X(k) →֒ X(Ak) is contained in X(Ak) . A variety X such that X(Ak) 6= ∅ whereas X(k) = ∅ is a counterexample to the Hasse principle. Such a counterexample is accounted for by the Manin obstruction if alreadyX(Ak) Br is empty. For a long time most known counterexamples to the Hasse principle could be explained by means of the Manin obstruction (to the best of my knowledge the case of the Bremner–Lewis–Morton curve 3x + 4y − 19z = 0 remains undecided). Recently Sarnak and L.Wang [SW] showed that the Manin obstruction is not the only obstruction to the Hasse principle for smooth hypersurfaces of degree 1130 in PQ if one assumes Lang’s conjecture that X(Q) is finite if XC is hyperbolic. The aim of this note is to construct a smooth proper surface over k = Q of Kodaira dimension 0 which is a counterexample to the Hasse principle but for which the Manin obstruction is not sufficient to explain the absence of Q-rational points. We exploit the same kind of surfaces which has been recently used to produce counterexamples to a conjecture of Mazur [AA]. These surfaces are quotients of a product of two curves of genus one by a fixed point free involution. Although in our example the Manin obstruction fails to provide a finite decision process for determining the existence of rational points there still is such a process. We propose a refinement of the Manin obstruction and show that for surfaces of our type it is the only obstruction to the Hasse principle. To define it we use a combination of the Manin obstruction and a descent very similar to the classical descent on elliptic curves. The point is that unlike in that classical case in the case of surfaces the Brauer group can become substantially bigger after passing to a finite unramified covering, thus the Manin obstruction can become finer. The refined obstruction depends on the choice of a finitely generated submodule of Pic(X), and it can give something