ON INTEGRAL POINTS ON DEGREE FOUR DEL PEZZO SURFACES

JÖRG JAHNEL^{‡} AND DAMARIS SCHINDLER^{∗,‡}

Abstract. We report on our investigations concerning algebraic and transcendental Brauer-Manin obstructions to integral points on complements of a hyperplane section in degree four del Pezzo surfaces. We discuss moreover two concepts of an obstruction at an archimedean place. Concrete examples are given of pairs of non-homogeneous quadratic polynomials in four variables representing (0, 0) over ℚ and over ℤ_{p} for all primes p, but not over ℤ. By blow-up, these yield cubic polynomials in three variables all integral solutions of which satisfy a gcd condition.

1. Introduction

One says that a variety over a number field k satisfies the Hasse principle if the existence of a rational point over every completion k_{ν} of k is enough to imply the existence of a k-rational point. This principle takes its name from the classical Hasse-Minkowski theorem, which states that the Hasse principle holds for the class of quadric hypersurfaces. Already for cubic curves and cubic surfaces however, the Hasse principle can fail. There has been much work on constructing and controlling such failures, particularly in recent time.

Many but not all (cf. [Sk99, Theorem 2]) examples of failures of the Hasse principle that are known today can be explained by one and the same argument, which was discovered by Yu. I. Manin [Ma] around 1970. In fact, a global Brauer class α ∈ Br(X) is usually responsible for the non-existence of k-rational points on the underlying proper variety X over k. This phenomenon is called the Brauer-Manin obstruction to rational points. Its mechanism works as follows.

Let 𝔭 be an arbitrary prime ideal of k. The Grothendieck-Brauer group is a contravariant functor from the category of schemes to the category of abelian groups. In particular, given a scheme X and a k_{ν_{𝔭}}-rational point x: Spec k_{ν_{𝔭}} → X, there is a restriction homomorphism x^{∗}: Br(X) → Br(k_{ν_{𝔭}}) ≅ ℚ/ℤ. For a Brauer class α ∈ Br(X), one calls

∗The second author was supported by the NSF under agreement No. DMS-1128155.

‡All computations are with magma [BCP].

Date: August 3, 2017.

2010 Mathematics Subject Classification. Primary 14G05; Secondary 11D85, 11E12, 14P25, 14F22, 14G25, 14J20.

Key words and phrases. Integral point, Integral Hasse principle, Obstruction at an archimedean place, Brauer-Manin obstruction, transcendental Brauer class, degree four del Pezzo surface, log K3 surface.

ev_{α,ν_{𝔭}}: X(k_{ν_{𝔭}}) ⟶ ℚ/ℤ , x ↦ x^{∗}(α) ,

the local evaluation map, associated with α. Analogously, when ν is a real place, there is the local evaluation map ev_{α,ν}: X(k_{ν}) → \frac{1}{2}ℤ/ℤ while, for a complex place, one has the zero map.

If X is a proper k-scheme then the local evaluation maps are continuous with respect to the ν-adic topologies on X(k_{ν}). Moreover, ev_{α,ν} is constant for all but finitely many places [Ja, Chapter IV, Proposition 2.3.a.ii)]. Thus, only adelic points x = (x_{ν})_{ν∈Ω} ∈ X(𝔸_{k}) satisfying ∑_{ν∈Ω} ev_{α,ν}(x_{ν}) = 0 ∈ ℚ/ℤ may possibly be approximated by k-rational points.

Brauer-Manin obstruction to integral points. The case of a non-proper variety U has been studied only much more recently. Here, for α ∈ Br(U), there is usually no finite set S of places such that ev_{α,ν} would be constant for all ν ∉ S. Thus, the local evaluation maps seem to be useless.

The picture changes however when integral points are considered instead. So let U be quasi-projective over k and 𝒰 be a model of U that is defined over Spec 𝒪_{k}, for 𝒪_{k} the ring of integers in k. Then, similarly to the above, there exists a finite set S = S_{𝒰 ,α} [Sk01, §5.2] of places, including all archimedean ones, such that the restriction

ev_{α,ν} |_{𝒰 (𝒪_{k,ν})} : 𝒰 (𝒪_{k,ν}) ⟶ ℚ/ℤ

is the zero map for every place ν ∉ S. Consequently, for the set of all S-integral points, one has the inclusions (cf. [CX, §1])

𝒰 (𝒪_{k,S}) ⊂ ( ∏_{ν∈S} U(k_{ν}) × ∏_{ν∉S} 𝒰 (𝒪_{k,ν}))^{Br(U)} ⊂ ∏_{ν∈S} U(k_{ν}) × ∏_{ν∉S} 𝒰 (𝒪_{k,ν})

for

( ∏_{ν∈S} U(k_{ν}) × ∏_{ν∉S} 𝒰 (𝒪_{k,ν}))^{Br(U)} :=

{(x_{ν})_{ν} ∈ ∏_{ν∈S} U(k_{ν}) × ∏_{ν∉S} 𝒰 (𝒪_{k,ν}) | ∀α ∈ Br(U): ∑_{ν} ev_{α,ν}(x_{ν}) = 0} .

In very much the same way as Yu. I. Manin did in his book [Ma] for counterexamples to the Hasse principle for rational points, J.-L. Colliot-Thélène and F. Xu [CX] explained many classical counterexamples to the integral Hasse principle or strong approximation (cf. [DW, Definition 2.1] or [PR, §7.1]) off certain primes by the Brauer-Manin obstruction to integral points. In [Xu], F. Xu even succeded in showing that the Brauer-Manin obstruction is the only obstruction to strong approximation off infinity for certain quadric fibrations, given that the set of real points is non-compact.

Colliot-Thélène and Xu already considered not only algebraic, but also transcendental Brauer classes in their work (see Remark 2.11 and Remark 3.8 in [CX]), and more examples have been constructed recently by A. Kresch and Yu. Tschinkel [KT].

Obstruction at infinity.

When asking for solutions in integers, there is another point to be considered, besides local solubility and the Brauer-Manin obstruction. Consider, for example, the equation X^{2} + 23Y^{2} = 2, which is insoluble in ℤ^{2}, although there exist rational solutions, as well as p-adic solutions for every prime p. The most natural argument to show insolubility over ℤ is certainly to notice that the equation defines a compact submanifold of ℝ^{2}, which, as integral points are sought for, leaves only finitely many cases. The same effect occurs for every system of equations in an arbitrary number of variables that includes Q(X_{1}, …, X_{m}) = C, for Q a definite quadratic form. Cf. [CW, Exemple 5.9], which is exactly of this type. Such examples are obstructed at the real place. And they are in fact obstructed in the crudest possible sense.

On the other hand, consider the equation X^{2} − Y^{2} = 2. Here, the submanifold defined by the equation is non-compact, but still there is an obstruction at the real place, in the sense below. The equation implies that |X − Y | ≤ \sqrt{2} or |X + Y | ≤ \sqrt{2}. Thus, considering only integral points, one is reduced to finitely many algebraic subvarieties of lower dimension. The same effect occurs, of course, whenever an equation of the form F_{1}(X_{1}, …, X_{m})F_{2}(X_{1}, …, X_{m}) = C is considered, for F_{1} and F_{2} coprime, non-constant forms in an arbitrary number of variables.

When working over a number field, the latter type of obstruction may even occur at a complex place. For example, every solution in ℤ[i]^{2} of the equation X^{2}+Y^{2} = C fulfils either |X − iY | ≤ \sqrt{|C|} or |X + iY | ≤ \sqrt{|C|}.

Concerning the concept of unobstructedness at infinity, it is our impression that a definite form for it has yet to be found. We think, however, that the examples just considered, despite being so simple, are typical in a certain sense. Thus, in section 2, we suggest two notions, called weak and strong unobstructedness, that reflect the effects just indicated, and compare them to results existing in the literature. We also show that the assumption of strong unobstructedness is in some sense necessary, when one asks for strong approximation or Zariski-density of integral points.

Our results.

The goal of this article is to present some theory and more examples, in which strong approximation off {∞}, strong approximation off a larger set of primes, or even the integral Hasse principle are violated. We consider open subschemes of del Pezzo surfaces of degree four that are complements of an effective divisor from the anticanonical class. These are, in particular, affine log K3 surfaces of Picard rank 5. Contrary to the examples previously studied, in this case both algebraic and transcendental Brauer classes may well occur.

The article is organised as follows. Below this introduction, we first deal with obstructions at infinity (Section 2), followed by generalities about degree four del Pezzo surfaces and complements of hyperplane sections through them (Section 3), as well as the Brauer-Manin obstruction to integral points (Section 4). We compute the list of all groups that may occur as the algebraic part of the Brauer groups of the surfaces considered in Theorem 4.8. This turns out to comprise 0, ℤ/2ℤ, (ℤ/2ℤ)^{2}, (ℤ/2ℤ)^{3}, (ℤ/2ℤ)^{4}, ℤ/4ℤ, ℤ/2ℤ×ℤ/4ℤ, and (ℤ/2ℤ)^{2}×ℤ/4ℤ, which is much richer than the corresponding list in the case of a proper degree four del Pezzo surface. For our results on the transcendental part of the Brauer group, see Theorem 4.9 and Corollary 4.10.

Section 5 then deals with the question how to evaluate a Brauer class practically. We present model cases, in which this may be done in satisfactory manner. Finally, Sections 6, 7, and 8 are devoted to explicit examples.

In a classical language, these may be formulated in the way below. A pair of (non-homogeneous) quadratic polynomials q_{1}, q_{2} ∈ ℤ[X_{0}, …, X_{3}] in four variables does not represent (0, 0) in integers, although the system of equations

q_{1}(X_{0}, …, X_{3}) = 0
q_{2}(X_{0}, …, X_{3}) = 0

is soluble over ℚ, as well as over ℤ_{p} for every prime number p.

For instance, in Section 6, we consider the following system of equations,

X_{0}X_{1} + X_{2}^{2} = X_{3} ,
X_{3}(2X_{1} + X_{2} + X_{3}) + X_{0}^{2} = X_{1} .

We show that every ℤ[\frac{1}{17}]-valued solution of this system fulfils the relation (x_{1}, x_{3})_{2} = 1, for (. , .)_{2} the 2-adic Hilbert symbol, although the analogous relation is not true for all ℤ_{2}-valued solutions. In particular, strong approximation off {17, ∞} is violated.

From this system, it is easy to deduce another one that has no integral solution at all, thereby violating the integral Hasse principle. Moreover, blowing up the point (0:1:0:0:0), we are led to the affine cubic surface

128Y_{0}^{3} + 144Y_{0}^{2} + 32Y_{0}Y_{1}Y_{2} + 8Y_{0}Y_{1} + 128Y_{0}Y_{2}^{2} + 80Y_{0}Y_{2} + 66Y_{0}

− 16Y_{1}^{2}Y_{2} − 3Y_{1}^{2} − 4Y_{1}Y_{2} + 80Y_{2}^{2} +40Y_{2} + 12 = 0

with the amusing property that each of its integral points must satisfy the condition that gcd(8y_{0} + 3, 16y_{2} + 3) > 1.

Neither of these examples can be explained by an obstruction at the real place, even in a strong sense. Observe, for instance, that the homogeneous parts \underline{q}_{1}, \underline{q}_{2} of degree 2 always generate a pencil consisting entirely of indefinite quadratic forms. This implies that the real manifold defined by (q_{1}, q_{2}) is non-compact [Fi, Satz 1] (see also [TWMW, Tables 1–3]) and, hence, has limit points at the boundary. In addition to that, both our surfaces in Sections 6 and 7 are strongly unobstructed at ∞.

The arithmetic behaviour we describe is entirely explained by algebraic and transcendental Brauer classes. In this respect, our examples are similar to the ones given in [CX], but different from those recently presented by Y. Harpaz [Harp, §4.2] and U. Derenthal and D. Wei [DW, Example 6.2].

The surface we examine in Section 8 is somewhat different. Its arithmetic is explained by Brauer classes together with effects caused by the underlying real manifold U(ℝ) being disconnected into non-compact components and a compact one. So, similarly to [DW, Example 6.2], an interaction of effects caused by Brauer classes occurs with an obstruction at infinity. This example shows that any reasonable definition of unobstructedness at infinity must include requirements on all connected components. In particular, it is insufficient just to require the existence of a single well-behaved one.

Acknowledgements: We thank Prof. J.-L. Colliot-Thélène for comments on an earlier version of this paper and Dr. M. Bright for pointing out an oversight in our original formulation of Corollary 4.10. The second author is supported by a NWO grant 016.Veni.173.016.

2. Obstruction at infinity

Example 2.1. Consider the subscheme 𝒳 ⊂ P_{ℤ}^{4} defined by the system of equations

2X_{0}^{2} + X_{1}^{2} + X_{2}^{2} = 26X_{4}^{2} ,
3X_{1}^{2} + X_{2}^{2} + X_{3}^{2} = 13X_{4}^{2}

and put 𝒰 := 𝒳 \ℋ , for ℋ := V (X_{4}) ⊂ P_{ℤ}^{4}. Then the generic fibre X of 𝒳 is a degree four del Pezzo surface. Moreover, there exist ℚ-rational points on the generic fibre U of 𝒰 , as well as ℤ_{p}-valued points on 𝒰 for every prime number p. Examples are (\frac{18}{7} : \frac{1}{7} : \frac{25}{7} : \frac{3}{7} :1) and (\frac{54}{19} : \frac{23}{19} : \frac{55}{19} : \frac{9}{19} :1). Yet, there is no integral point.

This is, however, not at all an interesting example. The reason is simply that there is an obstruction at infinity. Indeed, the two equations imply |x_{0}|, …, |x_{3}| ≤ 3 for every ℚ-rational point (x_{0} : …: x_{3} : 1) on 𝒰 . For integral points, this leaves us with only finitely many cases and it turns out that actually none of them corresponds to an integral point. This motivates the first part of the following definition. The second part is motivated by examples of the type X^{2} − Y^{2} = 2, as discussed in the introduction.

Definition 2.2. Let k be a number field and ν an archimedean place of k. Furthermore, suppose that X ⊂ P_{k}^{n} is an irreducible, closed subscheme, l ∈ Γ(P_{k}^{n}, 𝒪(1)) a linear form, and H := V (l) ⊂ P_{k}^{n} the corresponding k-rational hyperplane. Put U := X \H.

i) We say that U is weakly unobstructed at ν if every connected component of U(k_{ν}) has a limit point on H. ii) We call U strongly unobstructed at ν if every connected component U′ of U(k_{ν}) fulfils the following condition.

Given an integer d > 0, finitely many k-rational forms s_{1}, …, s_{r} ∈ Γ(X, 𝒪(d)), neither being a scalar multiple of l^{d}, and a constant c ∈ ℝ, there always exists a point x ∈ U′ such that ∥s_{i}(x)/l^{d}(x)∥_{ν} ≥ c for all i ∈ {1, …, r}. iii) If U is not weakly unobstructed (strongly unobstructed) at ν then we say that U is strongly obstructed (weakly obstructed) at ν.

Remarks 2.3. i) As P^{n}(k_{ν}) is a compact topological space, the property of being weakly unobstructed simply means that U(k_{ν}) does not have a bounded connected component. We note that if ν is a complex place and dim U ≥ 1 then U is weakly unobstructed at ν. ii) On the other hand, strong unobstructedness states that no connected component of U(k_{ν}) is contained in a finite union of tubular neighbourhoods of hypersurfaces of the form ∥s_{i}(x)/l^{d}(x)∥_{ν} ≤ c. iii) Example 2.1 is strongly obstructed. In fact, the topological boundary of 𝒰 (ℝ) is given by 2X_{0}^{2} + X_{1}^{2} + X_{2}^{2} = 3X_{1}^{2} + X_{2}^{2} + X_{3}^{2} = 0, which defines the empty set in P^{3}(ℝ). iv) As already remarked in the introduction, it seems to us that a definite form for the concept of unobstructedness at infinity has yet to be found. For instance, in [DW, Theorems 1.1 and 1.2], U. Derenthal and D. Wei included some unboundedness conditions at infinity and showed that they are, in general, necessary. In the case of a fibration over the affine line A^{1}, our notion of being strongly unobstructed implies unboundedness in the sense of Derenthal and Wei. Our definition is partly motivated by their examples, which however come with a natural projection to some affine space. Our definition aims to suggest a general notion of unobstructedness at infinity, which does not require such a fibration. It is inspired, too, by the results of J.-L. Colliot-Thélène and F. Xu [CX, Theorems 3.7, 4.5, and 8.3], as well as by the recent work of Y. Harpaz [Harp]. In fact, in the case that U(k_{ν}) is connected, Harpaz’ concept of being split [Harp, Def. 1.0.8] is slightly stronger than weak unobstructedness. It seems, however, that split varieties do not behave well, in general, when U(k_{ν}) is disconnected, cf. Section 8. Y. Harpaz’ example of a surface that is split, has no Brauer-Manin obstruction, yet no integral point, is weakly obstructed at infinity, according to our definition, as we show below. It is exactly this property of being obstructed that is used in his work to show that the surface has no integral points.

Example 2.4 (cf. [Harp, §4.2]). Let X ⊂ P_{ℚ}^{3} be defined by the equation

((11X_{0} + 5X_{3})X_{1} + 3X_{3}^{2})X_{2} = (3X_{0} + X_{3})X_{3}^{2}

and put U := X \H, for H := V (X_{3}) ⊂ P_{ℚ}^{4} . Then U is weakly unobstructed at ∞, but not strongly unobstructed. Proof. It is not hard to see that U(ℝ) is connected. Hence, weak unobstructedness follows from U being split in the terminology of [Harp, Def. 1.0.8].

In order to disprove strong unobstructedness, observe that every real point x = (x_{0} : x_{1} : x_{2} : 1) ∈ U(ℝ) must fulfil the equation ((11x_{0} + 5)x_{1} + 3)x_{2} = 3x_{0} + 1, i.e. x_{0}(11 − \frac{3}{x_{1}x_{2}} ) = \frac{1}{x_{1}x_{2}} − \frac{3}{x_{1}} − 5. This immediately shows that |x_{1}|, |x_{2}| ≥ 1 implies |x_{0}| ≤ \frac{9}{8}. □

Example 2.5. Let X ⊂ P_{ℚ}^{6} be defined by the equation

(X_{0}^{2} + X_{1}^{2} − X_{2}^{2})(X_{3}^{2} + X_{4}^{2} − X_{5}^{2}) = (X_{0}^{2} + X_{1}^{2} − X_{2}^{2} + X_{3}^{2} + X_{4}^{2} − X_{5}^{2})X_{6}^{2}

and put U := X \H, for H := V (X_{6}) ⊂ P_{ℚ}^{4} . Then U is weakly unobstructed at ∞, but not strongly unobstructed.

Proof. It is not hard to see that U(ℝ) has exactly two connected components and that (1:0:0:0:0:0:0) is a limit point for either of them.

On the other hand, every real point x = (x_{0} : x_{1} : x_{2} : x_{3} : x_{4} : x_{5} : 1) ∈ U(ℝ) must fulfil the equation Q(x_{0}, x_{1}, x_{2})Q(x_{3}, x_{4}, x_{5}) = Q(x_{0}, x_{1}, x_{2}) + Q(x_{3}, x_{4}, x_{5}), and hence |Q(x_{0}, x_{1}, x_{2})| ≤ 2 or |Q(x_{3}, x_{4}, x_{5})| ≤ 2 for Q the ternary quadratic form Q(X_{0}, X_{1}, X_{2}) := X_{0}^{2} + X_{1}^{2} − X_{2}^{2}. □

The next theorem shows that the notion of being strongly unobstructed is essentially necessary. In fact otherwise, under the assumption that U(ℝ) is connected, the integral points on 𝒰 are contained in a union of finitely many lower-dimensional subschemes. In particular, they are not Zariski dense and strong approximation off {∞} cannot be fulfilled.

Theorem 2.6. Let 𝒳 ⊂ P_{ℤ}^{n} be an irreducible, closed subscheme, l ∈ Γ(P_{ℤ}^{n}, 𝒪(1)) a linear form, ℋ := V (l) ⊂ P_{ℤ}^{n}, and 𝒰 := 𝒳 \ ℋ . Denote the generic fibres of 𝒳 and 𝒰 by X and U, respectively.

Assume that U(ℝ) is connected and that U is weakly obstructed at infinity. Then the set 𝒰 (ℤ) of integral points is not Zariski-dense in 𝒰 .

Proof. Since U is weakly obstructed at infinity, there exist an integer d > 0 and forms s_{1}, …, s_{r} ∈ Γ(X, 𝒪_{X}(d)), none of them a multiple of l^{d}, as well as a positive constant c > 0, such that the following is true. For every x ∈ U(ℚ), there is an 1 ≤ i ≤ r such that

|\frac{s_{i}(x)}{l^{d}(x)}| ≤ c .

Taking global sections commutes with arbitrary flat base extensions. In particular, Γ(X, 𝒪_{X}(d)) = Γ(𝒳, 𝒪_{𝒳} (d)) ⊗_{ℤ} ℚ. As a consequence of this, we see that there exists a non-zero integer M ∈ ℤ such that Ms_{1}, …, Ms_{r} ∈ Γ(𝒳, 𝒪_{𝒳} (d)).

Consider now an integral point x ∈ 𝒰 (ℤ). Its restriction to the generic fibre has the property that

|\frac{Ms_{i}(x)}{l^{d}(x)}| ≤ Mc ,

for some 1 ≤ i ≤ r. Thus, for every such integral point, there exists an integer B such that |B| ≤ Mc and

\frac{Ms_{i}(x)}{l^{d}(x)} = B .

In particular, this system defines a Zariski-closed subset Z ⊂ U. Moreover, if we had Z = U then s_{i} would be equal to B on U, a contradiction to none of the s_{i} being a scalar multiple of the form l^{d}. Hence, Z ⫋ U.

We conclude that there is a finite union of such proper Zariski-closed subsets of U that contains all restrictions of integral points P ∈ 𝒰 (ℤ) to the generic fibre. Hence, the integral points on 𝒰 cannot be Zariski-dense. □

We note that Theorem 2.6 may be seen as a generalisation of Theorem 4.1.1. of Harpaz’s work [Harp]. Finally, we close this section in giving a useful criterion to detect varieties that are strongly unobstructed.

Theorem 2.7. Let k be a number field, ν an archimedean place of k, X ⊂ P_{k}^{n} be a normal, projective variety, l ∈ Γ(P_{k}^{n}, 𝒪(1)) a linear form, H := V (l) ⊂ P_{k}^{n} the corresponding k-rational hyperplane, and put U := X \H. Suppose that

• the scheme (H ∩ X)_{k_{ν}} is reduced and irreducible and that

• every connected component of U(k_{ν}) has a limit point x ∈ (H ∩ X)(k_{ν}) that is non-singular as a point on H ∩ X. Then U is strongly unobstructed at ν.

Note that Theorem 2.7 does not apply to example 2.4, despite X being a normal cubic surface, since H ∩ X is a union of three lines.

Proof of Theorem 2.7. If ν is a complex place then U(k_{ν}) is clearly connected, but the same is not necessarily true for a real place. Thus, let U′ be a connected component of U(k_{ν}). There is a limit point x′ ∈ (H ∩ X)(k_{ν}) that is nonsingular on the limit set. Therefore, there exists a neighbourhood V ∋ x′ such that [U′ ∪ (H ∩ X)(k_{ν})] ∩ V is diffeomorphic to some 𝔻^{m}, ℝ^{m} or half space ℝ^{m−1}×ℝ_{≥0}.

Now let degree d forms s_{1}, …, s_{r} not being multiples of l^{d} and a constant c be given. We choose a linear form, which we denote by l′, in such a way that l′(x′) ≠ 0. Shrinking V, if necessary, we may assume that the same holds on the whole of V.

By our assumptions, (H ∩ X)_{k_{ν}} defines a prime Weil divisor on X_{k_{ν}}. Put d_{i} := ν_{(H∩X)_{kν}} (s_{i}) for i = 1, …, r. Since s_{1}, …, s_{r} are not multiples of l^{d}, we have 0 ≤ d_{i} < d for each i. The rational functions

f_{i} := \frac{s_{i}}{l^{d_{i}}l^{′d−d_{i}}}

have neither zeroes nor poles along the divisor (H ∩ X)_{k_{ν}}.

In particular, there exists a point x′′ ∈ (H ∩ X)(k_{ν}) ∩ V such that f_{i}(x′′) ≠ 0 for i = 1, …, r. Say, we have |s_{i}(x′′)/l^{d_{i}}(x′′)l^{′d−d_{i}}(x′′)| ≥ ε for every i = 1, …, r and some ε > 0, which we may assume to fulfil ε < 2c. Moreover, l(x′′) = 0, since x′′ ∈ (H ∩ X).

Next, we approximate x′′ by a point x′′′ ∈ U′ ∩ V outside the limit set. We may enforce that |s_{i}(x′′′)/l^{d_{i}}(x′′′)l^{′d−d_{i}}(x′′′)| ≥ ε/2 for every i = 1, …, r and that 0 < |l(x′′′)/l′(x′′′)| ≤ \frac{ε}{2c}. This yields

|\frac{s_{i}(x′′′)}{l^{d}(x′′′)}| = |\frac{s_{i}(x′′′)}{l^{d_{i}}(x′′′)l^{′d−d_{i}(x′′′)}} |·|\frac{l′(x′′′)}{l(x′′′)}|^{d−d_{i}} ≥ \frac{ε}{2} · (\frac{2c}{ε})^{d−d_{i}} ≥ \frac{ε}{2} · \frac{2c}{ε} = c

for i = 1, …, r, as required. □

3. Del Pezzo surfaces of degree four and complements of hyperplane sections

We are interested in quasi-projective schemes U ⊂ P_{k}^{4}, given by systems of equations of the type

Q_{1}(X_{0}, …, X_{4}) = 0 ,
Q_{2}(X_{0}, …, X_{4}) = 0 ,
X_{4} ≠ 0 ,

for homogeneous quadrics Q_{1}, Q_{2} ∈ 𝒪_{k}[X_{0}, …, X_{4}]. Then a rational point on U is the same as a solution (x_{0}, …, x_{3}, 1) for x_{0}, …, x_{3} ∈ k, while an integral point may be thought of as a solution (x_{0}, …, x_{3}, 1) for x_{0}, …, x_{3} ∈ 𝒪_{k}. The scheme U itself is an open subscheme of a del Pezzo surface of degree four.

By definition, a del Pezzo surface is a non-singular, proper algebraic surface X over a field k with an ample anti-canonical sheaf 𝒦 ^{−1}. Over an algebraically closed field, every del Pezzo surface of degree d ≤ 7 is isomorphic to P^{2}, blown up in (9−d) points in general position [Ma, Theorem 24.4.iii)].

According to the adjunction formula, a non-singular complete intersection of two quadrics in P^{4} is del Pezzo. The converse is true, as well. For every del Pezzo surface of degree four, its anticanonical image is the complete intersection of two quadrics in P^{4} [Do, Theorem 8.6.2].

Thus, associated with a degree four del Pezzo surface X, there is a pencil (μQ_{1} + νQ_{2})_{(μ:ν)∈P^{1}} of quadrics in P^{4}, the base locus of which is X. This pencil is uniquely determined up to an isomorphism of P^{4}. It contains exactly five degenerate fibres, each of which is of rank precisely 4 [Wi, Proposition 3.26.iv)].

Over an algebraically closed field, a degree four del Pezzo surface contains exactly 16 lines [Ma, Theorem 26.2.iii)] and exactly ten one-dimensional linear systems of conics [VAV, §2.3]. The latter may be obtained as follows. Take the five degenerate quadrics in the pencil. As they are of rank 4, each of them contains two onedimensional linear systems of planes. Intersecting with a second quadric from the pencil, one finds one-dimensional linear systems of conics.

Facts 3.1 (The geometric Picard group). Let k be an algebraically closed field and X a del Pezzo surface of degree four over k. Then i) Pic(X) ≅ ℤ^{6}. The classes of the 16 lines form a generating system for Pic(X). Moreover, the canonical class 𝒦 ∈ Pic(X) is not divisible by any integer ≠ ±1. ii) The orthogonal complement Pic^{0}(X) := ⟨𝒦 ⟩^{⊥} ⊂ Pic(X) is, up to sign, isometric to the lattice D_{5}. iii) The group W of all automorphisms of Pic(X) respecting 𝒦 and the intersection pairing is isomorphic to the Weyl group W (D_{5}). iv) The operation of W permutes the classes of the 16 lines in Pic(X). This operation is faithful. The image of the corresponding group homomorphism W → S_{16} is the largest subgroup of S_{16} that respects the intersection matrix.

v) The operation of W permutes the classes in Pic(X) of the 10 linear systems of conics. This operation is faithful, too. The image of the corresponding group homomorphism W → S_{10} is isomorphic to (ℤ/2ℤ)^{4} ⋊S_{5}. Here, (ℤ/2ℤ)^{4} ⊂ (ℤ/2ℤ)^{5} means the subgroup

{(x_{0}, …, x_{4}) ∈ (ℤ/2ℤ)^{5} | x_{0} + … + x_{4} = 0}

being acted upon by S_{5} in the natural way.

Proof. i) follows immediately from the description of Pic(X) given in [VAV, Proposition 2.2], together with [Ma, Theorem 26.2.i)], while ii), iii) and iv) are parts of [Ma, Theorem 23.9]. Finally, v) is known for a long time, as well. It follows, for example, from the discussion in [KST, pp. 8–10]. □

Remark 3.2. The existence of an isomorphism W (D_{5}) ≅ (ℤ/2ℤ)^{4}⋊S_{5} may, of course, be seen directly from the construction of the root system [Hu, §12]. Moreover, Aut(D_{5}) ≅ (ℤ/2ℤ)^{5} ⋊ S_{5}, in which W (D_{5}) is a subgroup of index 2.

Lemma 3.3. Let k be an algebraically closed field, X a del Pezzo surface of degree four over k, and D ⊂ X an irreducible divisor such that 𝒪(−D) = 𝒦 ∈ Pic(X) is the canonical class. Put U := X \D. Then

i) Pic(U) ≅ Pic(X)/⟨𝒦 ⟩ . ii) The orthogonal projection π: Pic(X) → ⟨𝒦 ⟩^{⊥} ⊗_{ℤ} ℝ induces an isomorphism

Pic(U) ⟶^{≅} D_{5}^{∗} ,

for D_{5}^{∗} the lattice dual to D_{5}.

Proof. i) This is [Hart, Proposition II.6.5.c)]. ii) By Fact 3.1, the kernel of π is exactly ⟨𝒦 ⟩. Moreover, for every ℒ ∈ Pic(X), one has ⟨π(ℒ), 𝒞 ⟩ = ⟨ℒ , 𝒞 ⟩ ∈ ℤ for all 𝒞 ∈ ⟨𝒦 ⟩^{⊥} = D_{5}. Hence π(ℒ) ∈ D_{5}^{∗} and, consequently, im π ⊆ D_{5}^{∗}.

In order to see that in fact equality is true, we first recall that D_{5} is of discriminant 4 [Hu, §11, Exc. 2]. Thus, ⟨𝒦 ⟩^{⊥} is of discriminant (−4). As ⟨𝒦 , 𝒦 ⟩ = 4, this yields that ⟨𝒦 ⟩ ⊕ ⟨𝒦 ⟩^{⊥} is of discriminant (−16). Moreover, Pic(X) is unimodular.

Therefore, ⟨𝒦 ⟩ ⊕ ⟨𝒦 ⟩^{⊥} must be of index 4 in Pic(X). Hence # im π/D_{5} = 4. But #D_{5}^{∗}/D_{5} = 4, again by [Hu, §11, Exc. 2]. □

Corollary 3.4. Let X and U be as above and G be a group operating on the 16 lines of X such that the intersection matrix is respected.

i) Then the induced operation of G on Pic(X) respects the intersection pairing as well as the canonical class. ii) The induced operation of G on Pic(U) ≅ D_{5}^{∗} takes place via isometries. The associated homomorphism G → Aut(D_{5}^{∗}) factors via W (D_{5}).

Proof. i) The operation of G on the free ℤ-module Div(X) over the 16 lines respects the intersection pairing and, by Fact 3.1.i), Pic(X) is the quotient of Div(X) modulo its radical. Finally, the canonical class is automatically respected, as the divisor being the sum over all 16 lines defines the invertible sheaf 𝒦 ^{⊗(−4)}. ii) The first assertion is a direct consequence of i). The second follows from Fact 3.1.iii). □

Under some minor assumptions, U is an example of a log K3 surface. Let us shortly recall the definition.

Definition 3.5 (cf. [Harp, Def. 2.0.15]). A log K3 surface is a scheme U that may be written in the form U = X \D, for X a non-singular, connected scheme that is proper of dimension two over a field k and D ⊂ X an effective divisor such that i) 𝒪(−D) ∈ Pic(X) is the canonical class and ii) π_{1}^{ét}(U\overline{k}, . ) = 0.

Theorem 3.6. Let X ⊂ P^{4} a non-singular complete intersection of two quadrics over a field k of characteristic 0 and U := X \H be the complement of a non-sin-gular hyperplane section. Then U is a log K3 surface.

Proof. We have π_{1}^{ét}(X\overline{k}, . ) = 0, since X\overline{k} is isomorphic to P^{2}, blown up in five points. Moreover, [Harp, Lemma 3.3.6] yields that π_{1}^{ét}(U\overline{k}, . ) = 0, too. Indeed, each of the 16 lines on X\overline{k} intersects the irreducible divisor H ∩ X transversely and in exactly one point. Finally, the adjunction formula shows that 𝒦_{X} ≅ 𝒪_{P}^{4}(−5H + 2H + 2H)|_{X} ≅ 𝒪_{X}(−H), which implies the claim. □

4. The Grothendieck-Brauer group

Generalities.

By definition [GrBrII, Remarque 2.7], the cohomological Grothendieck-Brauer group Br(U) of a scheme U is the étale cohomology group H_{ét}^{2}(U, 𝔾_{m}). If U is defined over a field k then the Hochschild-Serre spectral sequence [SGA4, Exp. VIII, Proposition 8.4]

H^{p}(Gal(\overline{k}/k), H_{ét}^{q}(U\overline{k}, 𝔾_{m})) ⟹ H_{ét}^{p+q}(U, 𝔾_{m})

carries essential information about Br(U). It yields a three-step filtration

0 ⊆ Br_{0}(U) ⊆ Br_{1}(U) ⊆ Br(U) ,

the subquotients of which we are going to describe below. In order to avoid unnecessary complications, let us assume that

Γ_{ét}(U\overline{k}, 𝔾_{m}) = \overline{k}^{∗} .  (1)

Lemma 4.1. Let k be a field, X a non-singular, proper scheme over k, and U = X \ D be the complement of a divisor D. Assume that D splits geometrically into irreducible components D_{1}, …, D_{r}, the intersection matrix of which has rank r. Then (1) is fulfilled.

Proof. The group Γ_{ét}(U\overline{k}, 𝔾_{m}) consists of all rational functions s on X\overline{k} such that div s = k_{1}[D_{1}] + … + k_{r}[D_{r}] for some integers k_{1}, …, k_{r}. Since div s is a principal divisor on X\overline{k}, its intersection numbers with D_{1}, …, D_{r} must all vanish. By our assumption, this is possible only for k_{1} = … = k_{r} = 0. I.e., s must be constant and one sees that Γ_{ét}(U\overline{k}, 𝔾_{m}) = \overline{k}^{∗} , as required. □

Remark 4.2. The assumptions on D are satisfied, in particular, if D is a geometrically irreducible divisor such that D^{2} ≠ 0.

4.3. i) In general, Br_{0}(U) is the image of a natural homomorphism

H^{2}(Gal(\overline{k}/k), Γ_{ét}(U\overline{k}, 𝔾_{m})) ⟶ Br(U) .  (2)

Under the assumption of (1), this shows that Br_{0}(U) is the image of the natural homomorphism Br(k) → Br(U). The homomorphism (2) is an injection, thereby ensuring that Br_{0}(U) ≅ Br(k), under some minor assumption. A sufficient condition is that U has a k-rational point or, if k is a number field, that U has an adelic point (cf. [Co, Proposition 1.3.4.1]). The part Br_{0}(U) ⊆ Br(U) does not contribute to the Brauer-Manin obstruction. ii) The subquotient Br_{1}(U)/ Br_{0}(U) is, in general, isomorphic to

ker d_{2}^{1,1} : H^{1}(Gal(\overline{k}/k), Pic(U\overline{k})) ⟶ H^{3}(Gal(\overline{k}/k), Γ_{ét}(U\overline{k}, 𝔾_{m})) ,

for d_{2}^{1,1} the differential in the Hochschild-Serre spectral sequence. Under the assumption of (1), the right hand side simplifies to H^{3}(Gal(\overline{k}/k), \overline{k}^{∗}). Moreover, if k is a number field then, as a by-product of class field theory [Ta, section 11.4], it is known that H^{3}(Gal(\overline{k}/k), \overline{k}^{∗}) = 0. Consequently, one has in fact

Br_{1}(U)/ Br_{0}(U) ≅ H^{1}(Gal(\overline{k}/k), Pic(U\overline{k})) .

The subquotient Br_{1}(U)/ Br_{0}(U) is called the algebraic part of the Brauer group. The effects of algebraic Brauer classes towards the Brauer-Manin obstruction are called the algebraic Brauer-Manin obstruction. iii) The subgroup Br_{1}(U) is nothing but the kernel of the natural homomorphism Br(U) → Br(U\overline{k}). Thus, there is a natural injection

Br(U)/ Br_{1}(U) ↪ Br(U\overline{k})^{Gal(\overline{k}/k)} .

It seems to be hard, even in concrete cases, to decide which Galois invariant Brauer classes on U\overline{k} actually descend to U. For interesting particular results, the reader is advised to study the paper [CS] of J.-L. Colliot-Thélène and A. N. Skorobogatov. The quotient Br(U)/ Br_{1}(U) is called the transcendental part of the Brauer group. The effects of transcendental Brauer classes towards the Brauer-Manin obstruction are called the transcendental Brauer-Manin obstruction.

The algebraic part of the Brauer group.

4.4. Let k be a number field, X ⊂ P_{k}^{4} a degree four del Pezzo surface, H ⊂ P_{k}^{4} a k-rational hyperplane such that H ∩ X is geometrically irreducible, and U := X \H. Then, (H ∩X)^{2} = 4. Hence, by Lemma 4.1 and 4.3.ii),

Br_{1}(U)/ Br_{0}(U) ≅ H^{1}(Gal(\overline{k}/k), Pic(U\overline{k})) .

Moreover, by Lemma 3.3.ii), Pic(U\overline{k}) ≅ D_{5}^{∗} and, under this isomorphism, the operation of Gal(\overline{k}/k) goes over into a homomorphism Gal(\overline{k}/k) ⟶ Aut(D_{5}^{∗}) factoring via W (D_{5}).

4.5. W (D_{5}) has 197 conjugacy classes of subgroups. It is known that each of them occurs already for a degree four del Pezzo surface defined over ℚ. Indeed, this was shown by B. È. Kunyavskij, A. N. Skorobogatov, and M. A. Tsfasman [KST]. The reader might want to compare [EJ], which reports on a more recent investigation.

We calculated H^{1}(H, D_{5}^{∗}) for each conjugacy class of subgroups in W (D_{5}), using magma’s functionality for the computation of the cohomology of finite groups. We partially reproduce the resulting list below in Figure 1.

Remarks 4.6 (Some details on the implementation).

i) It should not be difficult to create the ℤW(D_{5})-module D_{5}^{∗} within a computer algebra system. To be concrete, in magma, the command

w d5 := TransitiveGroup(16,1328);

creates the image of W (D_{5}) in S_{16}. Then the command

subs2 := Subsets(Set(GSet(w d5)), 2);

shows the decomposition of {1, …, 16} × {1, …, 16} into exactly three orbits under the W (D_{5})-operation, from which the intersection matrix is easily constructed, cf. [JL, §3.1]. On the other hand, the command

Div := PermutationModule(wd5, Integers());

creates a ℤW(D_{5})-module Div that, as a ℤ-module, is free over {1, …, 16} and carries the operation of W (D_{5}) that is induced by the one on that set. Thus, D_{5}^{∗} ≅ Div / Div_{0}, for Div_{0} ⊂ Div the submodule of rank 11 given by the condition that ⟨x, e_{1}⟩ = … = ⟨x, e_{16}⟩. Here, ⟨., .⟩ denotes the pairing encoded by the intersection matrix. ii) As our version of magma is unwilling to compute the quotient of two ℤW(D_{5})-modules and allows the user to do this only for kW(D_{5})-modules, for k

14 JÖRG JAHNEL AND DAMARIS SCHINDLER

a field, we worked modulo a large prime and recovered the structure of the ℤW(D_{5})-module D_{5}^{∗} from this. It turned out that the image of W (D_{5}) in GL_{5}(ℤ) is generated by the two matrices

( \begin{matrix}1 & 0 & −1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ −1 & −1 & 0 & −1 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ −2 & 0 & 1 & −1 & −1 \end{matrix} ) and ( \begin{matrix}2 & 1 & 0 & 1 & 1 \\ −1 & −1 & 0 & −1 & 0 \\ 0 & 1 & 1 & 0 & −1 \\ −2 & 0 & 1 & −1 & −1 \\ −1 & −1 & 0 & 0 & −1 \end{matrix} ) .

Figure 1. List of algebraic Brauer groups

1, 1, [ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 ], [],
 []
2, 2, [ 2, 2, 2, 2, 2, 2, 2, 2 ], [ 2 ],
 [ 2, 2, 2 ]
3, 2, [ 2, 2, 2, 2, 2, 2, 2, 2 ], [ 2 ],
 [ 2 ]
4, 2, [ 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2 ], [ 2 ],
 []
5, 2, [ 1, 1, 1, 1, 2, 2, 2, 2, 2, 2 ], [ 2 ],
 []
6, 2, [ 2, 2, 2, 2, 2, 2, 2, 2 ], [ 2 ],
 [ 2 ]
7, 3, [ 1, 1, 1, 1, 3, 3, 3, 3 ], [ 3 ],
 []
8, 5, [ 1, 5, 5, 5 ], [ 5 ],
 []
9, 4, [ 4, 4, 4, 4 ], [ 2, 2 ],
 [ 2, 2, 2, 2 ]
10, 4, [ 2, 2, 2, 2, 2, 2, 2, 2 ], [ 2, 2 ],
 []
11, 4, [ 4, 4, 4, 4 ], [ 4 ],
 [ 2, 2, 4 ]

⋮

181, 96, [ 16 ], [ 2, 2, 2 ],
 [ 2, 2 ]

⋮

189, 128, [ 16 ], [ 2, 2, 2 ],
 [ 2 ]
190, 192, [ 16 ], [ 6 ],
 [ 2 ]
191, 192, [ 16 ], [ 2 ],
 [ 2 ]
192, 192, [ 8, 8 ], [ 2 ],
 []
193, 192, [ 16 ], [ 2, 2, 2 ],
 [ 2 ]
194, 320, [ 16 ], [ 4 ],
 []
195, 384, [ 16 ], [ 2, 2 ],
 [ 2 ]
196, 960, [ 16 ], [],
 []
197, 1920, [ 16 ], [ 2 ],
 []

ON INTEGRAL POINTS ON DEGREE FOUR DEL PEZZO SURFACES 15

The list shows in every second line, in this order, the number of the conjugacy class in the numbering chosen by magma, the order of the corresponding subgroups, the orbit type on the 16 lines the subgroups yield, and the type of the abelian quotient. The resulting group cohomology group is indicated on the following line. Remarks 4.7. i) To summarise, we find that H^{1}(H, D_{5}^{∗}) is

• 0 in 59 cases,

• ℤ/2ℤ in 62 cases,

• (ℤ/2ℤ)^{2} in 44 cases,

• (ℤ/2ℤ)^{3} in 16 cases,

• (ℤ/2ℤ)^{4} in three cases,

• ℤ/4ℤ in nine cases,

• ℤ/2ℤ × ℤ/4ℤ in three cases, and

• (ℤ/2ℤ)^{2} × ℤ/4ℤ in one case. In comparison with the Brauer groups of proper degree four del Pezzo surfaces, which may be only 0, ℤ/2ℤ, or (ℤ/2ℤ)^{2} [Ma, Section 31, Table 3] (see also [SD]), these figures are unexpectedly rich. ii) The running time was a little less than one second on one core of a AMD Phenom II X4 955 processor, using magma, version 2.20.8. iii) Note that Manin’s formula [Ma, Proposition 31.3] (see also [Ja, Chapter III, Proposition 8.18]) does not apply to the computation of H^{1}(H, D_{5}^{∗}), as D_{5}^{∗} is not a unimodular lattice.

In the theorem below, we list some cases of particular interest. Theorem 4.8. Let k be a number field, X ⊂ P_{k}^{4} a degree four del Pezzo surface, H ⊂ P_{k}^{4} be a k-rational hyperplane such that H ∩ X is geometrically irreducible, and put U := X \H.

a) Then Br_{1}(U)/ Br_{0}(U) is isomorphic to 0, ℤ/2ℤ, (ℤ/2ℤ)^{2}, (ℤ/2ℤ)^{3}, (ℤ/2ℤ)^{4}, ℤ/4ℤ, ℤ/2ℤ×ℤ/4ℤ, or (ℤ/2ℤ)^{2}×ℤ/4ℤ.

b) Suppose that

i) the Galois group faithfully operating on the 16 lines on X is isomorphic to the full W (D_{5}) or its index two subgroup. Then Br_{1}(U)/ Br_{0}(U) = 0. ii) the Galois group faithfully operating on the 16 lines on X is isomorphic to the index five subgroup in W (D_{5}). Then Br_{1}(U)/ Br_{0}(U) = ℤ/2ℤ. iii) two of the five degenerate quadrics in the pencil associated with X are defined over k and the Galois group faithfully operating on the 16 lines on X is of index 20 in W (D_{5}). Then Br_{1}(U)/ Br_{0}(U) = (ℤ/2ℤ)^{2}. Proof. a), b.i) and b.ii) are immediate consequences from the list above. b.iii) According to Facts 3.1.iii and v), there is a canonical isomorphism

W (D_{5}) ⟶ (ℤ/2ℤ)^{4} ⋊ S_{5} , where the factor S_{5} is the group permuting the five degenerate quadrics in the pencil associated with X. Thus, by our first assumption, the Galois group H ↪ W (D_{5}) faithfully operating on the 16 lines is mapped isomorphically onto a subgroup of (ℤ/2ℤ)^{4} ⋊ S_{3} ⊂ (ℤ/2ℤ)^{4} ⋊ S_{5}, which is of order 96. On the other hand, order 96 is assumed, so that H goes over exactly into that semidirect product. It is not hard to check that 181 is the corresponding number in the list. □

The transcendental part of the Brauer group.

Theorem 4.9. Let k be an algebraically closed field of characteristic 0 and U := X \H, for X ⊂ P_{k}^{4} a degree four del Pezzo surface and H ⊂ P_{k}^{4} a hyperplane such that D := H ∩ X is non-singular. Then there is a canonical isomorphism

Br(U) ≅ H_{ét}^{1}(D, ℚ/ℤ) .

Proof. Let n be any positive integer. Then, the Gysin sequence [SGA5, Exp. VII, Corollaire 1.5] (or [Mi, Chapter VI, Remark 5.4.b)]) reads

H_{ét}^{0}(D, ℤ/nℤ) ⟶ H_{ét}^{2}(X, μ_{n}) ⟶ H_{ét}^{2}(U, μ_{n}) ⟶ H_{ét}^{1}(D, ℤ/nℤ) ⟶ H_{ét}^{3}(X, μ_{n}) .

Here, as X is a del Pezzo surface, one has H_{ét}^{2}(X, μ_{n}) ≅ Pic(X)/n Pic(X) and H_{ét}^{3}(X, μ_{n}) = 0. Moreover, H_{ét}^{0}(D, ℤ/nℤ) = ℤ/nℤ and the generator is mapped to the anticanonical class in Pic(X)/n Pic(X). Therefore, in view of [Hart, Proposition II.6.5.c)], the sequence simplifies to

0 ⟶ Pic(U)/n Pic(U) ⟶ H_{ét}^{2}(U, μ_{n}) ⟶ H_{ét}^{1}(D, ℤ/nℤ) ⟶ 0 ,

which shows that Br(U)_{n} ≅ H_{ét}^{1}(D, ℤ/nℤ). As n was arbitrary, this implies that Br(U)_{tors} ≅ H_{ét}^{1}(D, ℚ/ℤ).

However, for non-singular schemes in general, it is known that the cohomological Brauer group is torsion [GrBrII, Proposition 1.4]. □

Corollary 4.10. Let k be a number field and U := X \H, for X ⊂ P_{k}^{4} a degree four del Pezzo surface and H ⊂ P_{k}^{4} a k-rational hyperplane such that D := H ∩ X is non-singular. Then, for every n ≥ 2, there is a canonical monomorphism

[Br(U)/ Br_{1}(U)]_{n} ↪ (J(D)(k̄)_{n} ⊗_{ℤ/nℤ} μ_{n}^{∨})^{Gal(\overline{k}/k)} ,  (3)

for J(D) the Jacobian variety of D.

Proof. By 4.3.iii), there is a canonical injection Br(U)/ Br_{1}(U) ↪ Br(U\overline{k})^{Gal(\overline{k}/k)}. Moreover, Theorem 4.9 yields a canonical isomorphism

Br(U\overline{k})_{n} ≅ H_{ét}^{1}(D\overline{k}, ℤ/nℤ) ≅ H_{ét}^{1}(D\overline{k}, μ_{n}) ⊗_{ℤ/nℤ} μ_{n}^{∨} .

Here, H_{ét}^{1}(D\overline{k}, μ_{n}) ≅ Pic(D\overline{k})_{n} by virtue of [Mi, Chapter III, Proposition 4.11]. Moreover, as D is a non-singular curve, one has Pic(D\overline{k})_{n} ≅ J(D)(\overline{k})_{n}, according to the Picard interpretation of the Jacobian. □

Remarks 4.11. i) For the rest of the paper, we only consider 2-torsion classes. In this case, the inclusion (3) simplifies to

[Br(U)/ Br_{1}(U)]_{2} ↪ J(D)(k)_{2} .

Moreover, if D has a k-rational point O then we may choose this point to convert D into an elliptic curve and get an injection into the k-rational 2-torsion points of the elliptic curve D. ii) As indicated for a more general situation in paragraph 4.3.iii), it may well be a hard problem to decide which elements on the right hand side of (3) belong to the image of this monomorphism. In some particular cases, however, we are able to write down transcendental Brauer classes, explicitly. Cf. Theorem 5.5 below. iii) By Poincaré duality [Mi, Chapter VI, Corollary 11.2], we have a perfect, Galoisinvariant paring

H_{ét}^{1}(Dk̄, μ_{n}) × H_{ét}^{1}(Dk̄, ℤ/nℤ) → H_{ét}^{2}(Dk̄, μ_{n}) ≅ ℤ/nℤ ,

which induces an isomorphism

H_{ét}^{1}(Dk̄, \frac{1}{n}ℤ/ℤ) ≅ Hom(H_{ét}^{1}(Dk̄, μ_{n}), \frac{1}{n}ℤ/ℤ) ≅ Hom(J(D)(k̄)_{n}, ℚ/ℤ) .

Hence one obtains, similarly as in Corollary 4.10, a monomorphism

[Br(U)/ Br_{1}(U)]_{n} ↪ Hom_{Gal(\overline{k}/k)}(J(D)(k̄)_{n}, ℚ/ℤ) .

In particular, a proper n-torsion class in Br(U)/ Br_{1}(U) causes a k-rational n-isogeny J(D) ↠ D′ to an elliptic curve D′ with a k-rational proper n-torsion point.

5. Explicit Brauer classes

The tame symbol.

Let X be an irreducible, non-singular scheme of dimension ≤ 2 over a field k of characteristic 0. Then there is a canonical monomorphism Br(X) ↪ Br(k(X)). Moreover, within Br(k(X)), one has [GrBrII, Proposition 2.3]

Br(X) ≅ ⋂_{x∈X^{(1)}}Br(Spec 𝒪_{X,x}) ,

where X^{(1)} denotes the set of all codimension-one points on the scheme X.

In other words, a Brauer class of the function field k(X) extends to the whole of X if and only if it is unobstructed at any prime divisor. And if this is the case then the extension is unique.

The obstruction at a prime divisor x is tested by the so-called ramification homomorphism. This is the following composition [GrBrIII, Proposition 2.1],

ram_{x}: Br(k(X)) ⟵^{≅} H_{ét}^{2}(X, 𝔾_{m,k(X)}) ⟶ H_{ét}^{2}(X, i_{∗}ℤ_{x}) ⟵^{≅} H_{ét}^{1}(Spec k(x), ℚ/ℤ)

= H^{1}(Gal(\overline{k(x)}/k(x)), ℚ/ℤ) .

Here, the first map is an isomorphism due to [Ja, Chapter III, Corollary 5.6], the second is the homomorphism induced by the valuation map

val_{x}: 𝔾_{m,k(X)} → i_{∗}ℤ_{x} ,

and the third is the connecting homomorphism induced by the short exact sequence 0 → ℤ → ℚ → ℚ/ℤ → 0.

A way to explicitly write down a Brauer class of the function field is provided by the cyclic algebras. For these, let n ∈ ℕ and suppose that k(X) contains a primitive n-th root of unity ζ_{n}. Then, for f, g ∈ k(X)^{∗},

(f, g; ζ_{n}) := ⨁_{0≤a,b<n}u^{a}v^{b}k(X) for u^{n} = f, v^{n} = g, vu = uvζ_{n} ,

is an Azumaya algebra [Pi, §15.4, Proposition] and therefore defines a class in Br(k(X)). The corresponding cohomology class in H^{2}(Gal(\overline{k(X)}/k(X)), \overline{k(X)}^{∗}) may be described by the cocycle Φ such that [Pi, §15.1, Proposition a]

Φ(σ, τ) = \{ \begin{matrix}1 & if i + j < n , \\ f & if i + j ≥ n , \end{matrix}

for 0 ≤ i, j < n chosen by the condition that σ(\sqrt[n]{g}) = ζ_{n}^{i} \sqrt[n]{g} and τ(\sqrt[n]{g}) = ζ_{n}^{j} \sqrt[n]{g}.

Finally, recall that there is the tame symbol

δ_{x}: k(X)^{∗} × k(X)^{∗} ⟶ k(x)^{∗}, (f, g) ↦ (−1)^{ν_{x}(f)ν_{x}(g)}\overline{(\frac{f^{νx(g)}}{g^{νx(f)}} )} ,

where the bar denotes the natural residue map 𝒪_{X,x} → k(x).

Proposition 5.1. Let n ∈ ℕ and fix a primitive n-th root of unity ζ_{n} ∈ k(X)^{∗}. Then the following diagram commutes,

\xymatrix{k(X)^{∗}× k(X)^{∗} \ar[r]^{δ_{x}} \ar[d]_{R_{n,ζn}} & k(x)^{∗} \ar[d]^{π} \\ Br(k(X)) \ar[r]^{_{ram_{x}}} & H^{1}(Gal(\overline{k(x)}/k(x)), ℚ/ℤ) .}

Here, the downward arrow R_{n,ζ_{n}} to the left sends (f, g) to the Brauer class of (f, g; ζ_{n}), while the one to the right is the canonical map

k(x)^{∗} ↠ k(x)^{∗}/(k(x)^{∗})^{n} ⟶^{≅} H^{1}(Gal(\overline{k(x)}/k(x)), μ_{n}) ,

followed by the isomorphism induced by the identification μ_{n} → \frac{1}{n}ℤ/ℤ, ζ_{n} ↦ −\frac{1}{n}.

Proof. If ν_{x}(f) = ν_{x}(g) = 0 then δ_{x}(f, g) = 1 such that (f, g) is mapped to 0 via the upper right corner.

On the other hand, the cohomology class of R_{n,ζ_{n}}(f, g) = (f, g; ζ_{n}) comes via inflation from H^{2}(Gal(k(X)(\sqrt[n]{g})/k(X)), \overline{k(X)}^{∗}) and is represented by a cocycle whose image is \{1, f\}. Since the field extension k(X)(\sqrt[n]{g})/k(X) is unramified at x and one has ν_{x}(1) = ν_{x}(f) = 0, it follows that ram_{x}(f, g; ζ_{n}) is represented by the cocycle that is constantly zero. Hence, ram_{x}(f, g; ζ_{n}) = 0. Thus, via the lower left corner, (f, g) is mapped to 0, too.

Next, we observe that the maps δ_{x} and R_{n,ζ_{n}} are bimultiplicative and antisymmetric. Thus, in order to complete the proof, it suffices to consider the case that ν_{x}(f) = 1 and ν_{x}(g) = 0. Upper right path: We have δ_{x}(f, g) = \overline{(1/g)}, which is mapped under π to the class of the homomorphism

σ ↦ \frac{σ(\sqrt[n]{\overline{(1/g)}})}{\sqrt[n]{\overline{(1/g)})}} = \frac{\sqrt[n]{\overline{g}}}{σ(\sqrt[n]{\overline{g}})}

in H^{1}(Gal(k(x)(\sqrt[n]{\overline{g}})/k(x)), μ_{n}) ↪ H^{1}(Gal(\overline{k(x)}/k(x)), μ_{n}). In other words, π(δ_{x}(f, g)) maps the element σ ∈ Gal(k(x)(\sqrt[n]{\overline{g}})/k(x)) such that \sqrt[n]{\overline{g}} ↦ ζ_{n}^{i} \sqrt[n]{\overline{g}} to ζ_{n}^{−i}, which finally gets identified with \frac{i}{n}. Lower left path: Here, (f, g) is sent first by R_{n,ζ_{n}} to (f, g; ζ_{n}), which is described by the cocycle Φ such that

Φ(σ, τ) = \{ \begin{matrix}1 & if i + j < n , \\ f & if i + j ≥ n , \end{matrix}

for 0 ≤ i, j < n chosen in the way that σ(\sqrt[n]{g}) = ζ_{n}^{i} \sqrt[n]{g} and τ(\sqrt[n]{g}) = ζ_{n}^{j} \sqrt[n]{g}. Applying the valuation ν_{x}, one finds

(ν_{x}∘Φ)(σ, τ) = \{ \begin{matrix}0 & if i + j < n , \\ 1 & if i + j ≥ n . \end{matrix}

This, however, is exactly the image of

Ψ: σ ↦ \frac{i}{n} , for σ(\sqrt[n]{g}) = ζ_{n}^{i} \sqrt[n]{g} ,

understood as an element in H^{1}(Gal(k(X)(\sqrt[n]{g})/k(X)), ℚ/ℤ) , under the connecting homomorphism associated with 0 → ℤ → ℚ → ℚ/ℤ → 0. Thus, ram_{x}(R_{n,ζ_{n}}(f, g)) maps the element σ ∈ Gal(k(x)(\sqrt[n]{\overline{g}})/k(x)) sending \sqrt[n]{\overline{g}} ↦ ζ_{n}^{i} \sqrt[n]{\overline{g}} to \frac{i}{n}. This completes the proof. □

Corollary 5.2. Let X be an irreducible, non-singular scheme of dimension ≤ 2 over a field k of characteristic 0, n ∈ ℕ, ζ_{n} ∈ k(X) a primitive n-th root of unity, f, g ∈ k(X)^{∗}, and x ∈ X^{(1)} a prime divisor. Then the Brauer class of (f, g; ζ_{n}) is unobstructed at x if and only if δ_{x}(f, g) is an n-th power in k(x). □

Remarks 5.3. i) In Theorem 4.9, the Gysin homomorphism appeared instead of the tame symbol. It is, in fact, true that these two maps are compatible. Cf., for example, [BMT, Lemma 2.5]. ii) In [AM, §3], M. Artin and D. Mumford provide a more complete picture of the ramification homomorphism. For example, the following is shown. Let X be a proper surface over an algebraically closed field and U := X\D for D an irreducible non-singular curve. Suppose that (f, g; ζ_{n}) defines a Brauer class on U.

Then the divisor div t of the rational function t := δ_{x}(f, g) ∈ k(D), for x the generic point of D, is divisible by n. Of course, on a curve of genus ≥ 1, this is not enough to imply that t has to be an n-th power.

Two model cases.

Theorem 5.4 (An algebraic Brauer class). Let k be any field of characteristic 0, X ⊂ P_{k}^{4} a del Pezzo surface of degree four over k, and U := X \ H for H := V (X_{4}) ⊂ P_{k}^{4}. Suppose that the pencil of quadrics associated with X contains a k-rational quadric of rank 4 that may be written in the form

l_{1}l_{2} − l_{3}^{2} + dl_{4}^{2} ,

for linear forms l_{1}, …, l_{4} and d ∈ k^{∗} a non-square.

i) Then

(\frac{l_{1}}{X_{4}} , d; −1)

defines an algebraic Brauer class α ∈ Br_{1}(U)_{2}. ii) If V (X_{4}) ∩ X is reduced and has a k-rational point then α does not extend to X and is, in particular, nontrivial.

Proof. i) Let D ⊂ U be any prime divisor and x ∈ U^{(1)} its generic point. Then ν_{x}(d) = 0 and ν_{x}(\frac{l_{1}}{X_{4}} ) = 0, unless D is a component of V (l_{1}) ∩ X. Thus, if D ⊄ V (l_{1}) ∩ X then δ_{x}(\frac{l_{1}}{X_{4}} , d) = 1.

On the other hand, if D ⊂ V (l_{1}) ∩ X then δ_{x}(\frac{l_{1}}{X_{4}} , d) = \frac{1}{d^{νx(l_{1})}} , which is a square on D, since the equation −l_{3}^{2} + dl_{4}^{2} = 0 implies

d = \frac{l_{3}^{2}}{l_{4}^{2}} .

The Brauer class α is clearly algebraic, as it gets annihilated under base extension to k(\sqrt{d}). ii) Take now D to be a component of V (X_{4}) ∩ X. Then δ_{x}(\frac{l_{1}}{X_{4}} , d) = d for x ∈ U^{(1)} the generic point of D, which cannot be a square when D has a k-rational point. □

Theorem 5.5 (A transcendental Brauer class). Let k be any field of characteristic 0 and X ⊂ P_{k}^{4} a del Pezzo surface of degree four over k that is given by a system of equations of the type

l_{1}l_{2} + au^{2} = X_{4}l_{3} ,
l_{3}l_{4} + bv^{2} = X_{4}l_{1} ,

for linear forms l_{1}, …, l_{4}, u, v, and a, b ∈ k^{∗}. Assume that the forms l_{1}, l_{3}, u, and v are linearly independent. Put U := X \H for H := V (X_{4}) ⊂ P_{k}^{4}.

i) Then

(\frac{bl_{1}}{X_{4}} , \frac{al_{3}}{X_{4}} ; −1)

defines a Brauer class τ ∈ Br(U)_{2}. ii) If D := V (X_{4})∩X is geometrically integral and \frac{l_{1}}{l_{3}} is not the square of a rational function on D\overline{k} then τ_{U\overline{k}} does not extend to X\overline{k}. In particular, τ is transcendental.

Proof. i) Let D ⊂ U be any prime divisor and x ∈ U^{(1)} its generic point. Then ν_{x}(\frac{bl_{1}}{X_{4}} ) = 0 and ν_{x}(\frac{al_{3}}{X_{4}} ) = 0, unless D ⊂ V (l_{1}) ∩ X or D ⊂ V (l_{3}) ∩ X. Thus, if D is not a component of either of these two subschemes then δ_{x}(\frac{bl_{1}}{X_{4}} , \frac{al_{3}}{X_{4}} ) = 1.

On the other hand, if D ⊂ V (l_{1}) ∩ X, but D ⊄ V (l_{3}) ∩ X, then one finds that δ_{x}(\frac{bl_{1}}{X_{4}} , \frac{al_{3}}{X_{4}} ) = (\frac{X_{4}}{al_{3}} )^{ν_{x}(l_{1})}, which is a square on D, since the equation au^{2} = X_{4}l_{3} implies

\frac{X_{4}}{al_{3}} = \frac{u^{2}}{l_{3}^{2}} .

Similarly, if D ⊂ V (l_{3}) ∩ X, but D ⊄ V (l_{1}) ∩ X, then δ_{x}(\frac{bl_{1}}{X_{4}} , \frac{al_{3}}{X_{4}} ) = (\frac{bl_{1}}{X_{4}} )^{ν_{x}(l_{3})}, which is a square on D, since bv^{2} = X_{4}l_{1} yields

\frac{bl_{1}}{X_{4}} = \frac{l_{1}^{2}}{v^{2}} .

Finally, there cannot be a divisor that is contained in both, V (l_{1})∩X and V (l_{3})∩X, as this would imply l_{1} = l_{3} = u = v = 0, which, according to our assumptions, is fulfilled only by a single point. ii) Let now x be the generic point of V (X_{4}) ∩ X. Then δ_{x}(\frac{bl_{1}}{X_{4}} , \frac{al_{3}}{X_{4}} ) = \frac{al_{3}}{bl_{1}} . Since \frac{a}{b} is a square in \overline{k}, this immediately implies the claim. □

Constancy of the local evaluation maps.

Remark 5.6. Let now k be a number field and U be a k-scheme, on which a Brauer class α ∈ Br(U) is given by (f, g; −1), for rational functions f, g ∈ k(U). Moreover, let ν be any place of k and x ∈ U(ℚ_{ν}), x ∉ supp div f ∪ supp div g, a point. Then

ev_{α,ν}(x) = \{ \begin{matrix}0 & if (f(x), g(x))_{ν} = 1 , \\ \frac{1}{2} & if (f(x), g(x))_{ν} = −1 , \end{matrix}

where (., .)_{ν}: k_{ν}^{∗} × k_{ν}^{∗} → { 1, −1 } denotes the usual ν-adic Hilbert symbol.

Indeed, according to the definition, ev_{α,ν}(x) = x^{∗}α and our assumptions guarantee that the pull-back may be obtained in a naive way as the cyclic (quaternion) algebra (f(x), g(x); −1) over k_{ν}. Finally, the invariant of such an algebra is directly related, by the above rule, to the Hilbert symbol [Pi, §18.4, Exercise 4].

Lemma 5.7. Let k be a number field, 𝒳 ⊂ P_{𝒪_{k}}^{4} the closed subscheme defined by two quadratic forms with coefficients in 𝒪_{k}, and 𝒰 := 𝒳 \ℋ for ℋ := V (X_{4}) ⊂ P_{𝒪_{k}}^{4} . Suppose that the generic fibre X of 𝒳 is a non-singular surface and let U and α ∈ Br(U)_{2} be as in Theorem 5.4. Assume that one of the quadratic forms defining 𝒳 may be written in the form

l_{1}l_{2} − l_{3}^{2} + dl_{4}^{2} ,

for l_{1}, …, l_{4} linear forms having coefficients in 𝒪_{k} and d ∈ 𝒪_{k} a non-square. Furthermore, let 𝔭 be a prime ideal in 𝒪_{k} such that 𝔭 ∤ 2d,

• the linear forms (l_{1} mod 𝔭), …, (l_{4} mod 𝔭) are 𝒪_{k}/𝔭-linearly independent, and • the reduction modulo 𝔭 of the cusp given by l_{1} = … = l_{4} = 0 does not lie on the reduction 𝒰_{𝔭}. Then the local evaluation map

ev_{α,ν_{𝔭}}: 𝒰 (𝒪_{k,𝔭}) ⟶ ℚ/ℤ

is constantly zero.

Proof. Let us denote integral points x ∈ 𝒰 (𝒪_{k,𝔭}) in the form x = (x_{0} : …: x_{3} : 1), for x_{0}, …, x_{3} ∈ 𝒪_{k,𝔭}. The local evaluation map ev_{α,ν_{𝔭}} is known to be locally constant with respect to the 𝔭-adic topology on U(k_{ν_{𝔭}}) [Ja, Chapter IV, Proposition 2.3.a.ii)]. Consequently, it suffices to prove the assertion for points x ∈ 𝒰 (𝒪_{k,𝔭}) such that l_{1}(x) ≠ 0 and l_{2}(x) ≠ 0. Thus, what we have to show is

(l_{1}(x_{0}, …, x_{3}, 1), d)_{ν_{𝔭}} = 1 ,

for every point x = (x_{0} : …: x_{3} : 1) ∈ U(𝒪_{k,𝔭}) satisfying the inequalities above.

For this, we first observe that the equation given yields l_{1}(x)l_{2}(x) = l_{3}^{2}(x)−dl_{4}^{2}(x), which is a norm from k_{ν_{𝔭}}(\sqrt{d}). Therefore, in the Hilbert symbol, we may replace l_{1}(x) by l_{2}(x).

If (d mod 𝔭) ∈ 𝒪_{k}/𝔭 is a square then assertion is certainly true. Thus assume that (d mod 𝔭) ∈ 𝒪_{k}/𝔭 is a non-square. Then (l_{1}(x), d)_{ν_{𝔭}} = −1 would mean that ν_{𝔭}(l_{1}(x)) is odd. In which case, ν_{𝔭}(l_{2}(x)) is also odd. In particular, 𝔭|l_{1}(x) and 𝔭|l_{2}(x), which directly implies that 𝔭|l_{3}(x) and 𝔭|l_{4}(x), too.

However, according to our assumptions, such an 𝒪_{k,𝔭}-valued point on 𝒰 does not exist. Indeed, its reduction modulo 𝔭 would coincide with that of the cusp, which is supposed not to lie on 𝒰_{𝔭}. □

Remark 5.8. In particular, we see that if 𝔭 ∤ 2 is a prime ideal of good reduction of 𝒳 then ev_{α,𝔭} is constantly zero.

Indeed, then 𝒳_{𝔭} is a degree four del Pezzo surface over 𝔽_{𝔭}. Hence, the rank of a degenerate quadric in the pencil cannot drop below 4 [Wi, Proposition 3.26.iv)]. This implies immediately that 𝔭 ∤ d, as well as linear independence of (l_{1} mod 𝔭), …, (l_{4} mod 𝔭). Moreover, the reduction of the cusp would be a singular point on 𝒰_{𝔭}, which, by assumption, does not exist.

Lemma 5.9. Let k be a number field and 𝒳 ⊂ P_{𝒪_{k}}^{4} be the closed subscheme defined by a system of equations of the type

l_{1}l_{2} + au^{2} = X_{4}l_{3} ,
l_{3}l_{4} + bv^{2} = X_{4}l_{1} ,

for linear forms l_{1}, …, l_{4}, u, v having coefficients in 𝒪_{k} and a, b ∈ 𝒪_{k}^{∗}. Assume that the forms l_{1}, l_{3}, u, and v are k-linearly independent and that the generic fibre X of 𝒳 is a non-singular surface. Put 𝒰 := 𝒳 \ℋ for ℋ := V (X_{4}) ⊂ P_{𝒪_{k}}^{4} and let U and τ ∈ Br(U)_{2} be as in Theorem 5.5. Then, for every prime ideal 𝔭 in 𝒪_{k} such that 𝔭 ∤ 2ab, the local evaluation map

ev_{τ,ν_{𝔭}}: 𝒰 (𝒪_{k,𝔭}) ⟶ ℚ/ℤ

is constantly zero.

Proof. Let us denote integral points x ∈ 𝒰 (𝒪_{k,𝔭}) in the form x = (x_{0} : …: x_{3} : 1) for x_{0}, …, x_{3} ∈ 𝒪_{k,𝔭}. As the local evaluation map ev_{τ,ν_{𝔭}} is locally constant with respect to the 𝔭-adic topology on U(k_{ν_{𝔭}}), it suffices to prove the assertion for points x ∈ 𝒰 (𝒪_{k,𝔭}) such that l_{i}(x) ≠ 0, for i = 1, …, 4. Thus, what we have to show is

(bl_{1}(x), al_{3}(x))_{ν_{𝔭}} = 1 ,

for every point x = (x_{0} : …: x_{3} : 1) ∈ 𝒰 (𝒪_{k,𝔭}) satisfying the inequalities above.

First of all, observe that a and b are 𝔭-adic units. Furthermore, the first equation yields (−a)l_{1}(x)l_{2}(x) = a^{2}u^{2}(x) − al_{3}(x), which is a norm from k_{ν_{𝔭}}(\sqrt{al_{3}(x)}). Therefore, in the Hilbert symbol, we may replace bl_{1}(x) by (−ab)l_{2}(x).

Analogously, we have (−b)l_{3}(x)l_{4}(x) = b^{2}v^{2}(x) − bl_{1}(x) being a norm from k_{ν_{𝔭}}(\sqrt{bl_{1}(x)}). Thus, we may as well replace al_{3}(x) by (−ab)l_{4}(x).

This shows, if one of the numbers l_{1}(x), l_{2}(x) is of even 𝔭-adic valuation, as well as one of the numbers l_{3}(x), l_{4}(x), then assertion is certainly true. Thus, suppose without restriction that ν_{𝔭}(l_{1}(x)) and ν_{𝔭}(l_{2}(x)) are both odd. We have to show in this situation that al_{3}(x) ∈ k_{ν_{𝔭}}^{∗} is a square.

For this, as ν_{𝔭}(l_{1}(x)) is odd and ν_{𝔭}(bv^{2}(x)) is even, the second equation implies

ν_{𝔭}(l_{3}(x)l_{4}(x)) = min(ν_{𝔭}(bv^{2}(x)), ν_{𝔭}(l_{1}(x)))
≤ ν_{𝔭}(l_{1}(x)) ,

hence

ν_{𝔭}(l_{3}(x)) ≤ ν_{𝔭}(l_{3}(x)l_{4}(x)) ≤ ν_{𝔭}(l_{1}(x)) < ν_{𝔭}(l_{1}(x)l_{2}(x)) .

Note here that ν_{𝔭}(l_{2}(x)) ≥ 1, the valuation of l_{2}(x) being odd. Consequently, the first equation implies that al_{3}(x) is a square in k_{ν_{𝔭}}^{∗}. This completes the proof. □

6. Examples

Recall [DW, Definition 2.1] (cf. [PR, §7.1]) that strong approximation off a certain set S of places is said to hold for an algebraic variety U defined over a number field k if the image of the set U(k) of k-rational points on U is dense in the space U(𝔸_{k}^{S}) of adelic points on U outside S.

Del Pezzo surfaces of degree four. In our first example, only a transcendental Brauer class occurs.

Example 6.1. Let X ⊂ P_{ℚ}^{4} be given by the system of equations

X_{0}X_{1} + X_{2}^{2} = X_{4}X_{3} ,
X_{3}(2X_{1} + X_{2} + X_{3}) + X_{0}^{2} = X_{4}X_{1}

and put U := X \H, for the hyperplane H := V (X_{4}) ⊂ P_{ℚ}^{4} .

i) Then X is a del Pezzo surface of degree four. ii) D := H ∩ X is a non-singular curve of genus one. Furthermore,

D(ℚ) = {(0:1:0:0:0), (0:−1:0:2:0)} .

iii) The manifold X(ℝ) is connected, its submanifold U(ℝ) is connected, too, and U is strongly unobstructed at ∞. iv) One has Br(U)/ Br_{0}(U) ≅ ℤ/2ℤ, a representative of the nontrivial element being

τ: (\frac{X_{1}}{X_{4}} , \frac{X_{3}}{X_{4}} ; −1) .

Moreover, Br_{1}(U)/ Br_{0}(U) = 0. I.e., the Brauer class τ is transcendental.

Proof. i) and ii) magma reports that both X and D are non-singular. More precisely, D is isomorphic to the elliptic curve E: y^{2}z + xyz = x^{3} + 4xz^{2}, the Mordell-Weil group of which is isomorphic to ℤ/2ℤ. iii) The pencil of quadrics in P^{4} corresponding to X contains only one degenerate quadric that is defined over ℝ. According to [VAV, Theorem 3.4 and Lemma 5.1], this implies Br(X_{ℝ} )/ Br(ℝ) = 0, which is enough to show that X(ℝ) is connected [Si, Théorème 1.4].

As X has no real point such that x_{1} = x_{4} = 0, the rational map π: X− [figure] P^{1}, given by x ↦ x_{1}/x_{4}, is well-defined on real points. Furthermore, U(ℝ) = π^{−1}(A^{1}(ℝ)). A calculation using Gröbner bases shows that π has exactly 10 singular fibres, among which only two, those over 0 and ∞, are real. Finally, experiments with t = 1 and t = −1 indicate that the fibres π^{−1}(t), for t > 0 as well as for t < 0, are genus-one curves with a connected set of real points. As connectedness of π^{−1}(0) is directly checked, this is enough to guarantee that U(ℝ) is connected.

Since U(ℝ) is connected and D is a non-singular, geometrically irreducible scheme having real points, Theorem 2.7 shows that U is strongly unobstructed at the real place. iv) A calculation using Gröbner bases shows that the Galois group operating on the 16 lines on X is the full W (D_{5}) of order 1920. Therefore, Theorem 4.8.b.i) immediately implies that Br_{1}(U)/ Br_{0}(U) = 0.

On the other hand, we have #D(ℚ) = 2. Moreover, E has two ℚ-rational 4-isogenies, but no nontrivial n-isogenies for any other n ≥ 3. As the isogenous curves have no proper ℚ-rational 4-torsion points, Corollary 4.10 shows that there is an injection Br(U)/ Br_{1}(U) ↪ ℤ/2ℤ. Consequently, Br(U)/ Br_{0}(U) is of order at most 2. As far as a concrete description is asked for, Theorem 5.5 applies. It shows that (\frac{X_{1}}{X_{4}} , \frac{X_{3}}{X_{4}} ; −1) defines a Brauer class on U.

In order to prove that this is indeed a nontrivial class, the simplest argument is probably that its local evaluation map ev_{τ,ν_{2}} is non-constant, which we show in Example 6.2.i) below. □

Example 6.2 (continued). Let X, U, and τ be as above. Moreover, let 𝒳 ⊂ P_{ℤ}^{4} be defined by the same system of equations as X and put 𝒰 := 𝒳 \ ℋ , for ℋ := V (X_{4}) ⊂ P_{ℤ}^{4}. Then i) For all primes p ≠ 2 including the archimedean one, the local evaluation map ev_{τ,ν_{p}}: 𝒰 (ℤ_{p}) → ℚ/ℤ is constantly zero. On the other hand, ev_{τ,ν_{2}} : 𝒰 (ℤ_{2}) → ℚ/ℤ is non-constant.

ii) For p = 17, even the local evaluation map ev_{τ,ν_{17}} : U(ℚ_{17}) → ℚ/ℤ is constantly zero.

Proof. i) For all primes with the exception of the infinite one, this follows directly from Lemma 5.9. To prove constancy for ev_{τ,ν_{∞}}, we have to show that, for a real point x = (x_{0} :…:x_{3} :1) ∈ U(ℝ), the coordinates x_{1} and x_{3} cannot both be negative.

Assume the contrary. In that case, the first equation shows x_{0} > 0. Moreover, x_{2}^{2} + |x_{3}| = |x_{0}x_{1}| and hence x_{2}^{2} ≤ |x_{0}x_{1}|, implying x_{2} ≤ x_{0} or x_{2} ≤ −x_{1}. In the first case,

x_{3}(2x_{1} + x_{2} + x_{3}) + x_{0}^{2} = 2x_{1}x_{3} + x_{2}x_{3} + x_{3}^{2} + x_{0}^{2}
≥ 2x_{1}x_{3} + x_{0}x_{3} + x_{3}^{2} + x_{0}^{2} > 2x_{1}x_{3} + (x_{0} + x_{3})^{2} > 0 ,

in contradiction to the second equation. And in the second case, the same is true, since x_{2}x_{3} ≥ −x_{1}x_{3}.

Finally, non-constancy of ev_{τ,ν_{2}} may easily be seen directly. Indeed, for the integral point x = (−1 : −1 : −1 : 2 : 1) ∈ 𝒰 (ℤ_{2}), one has (x_{1}, x_{3})_{2} = (−1, 2)_{2} = 1, hence ev_{τ,ν_{2}}(x) = 0. On the other hand, the ℤ_{2}-valued point x′ = (\frac{2}{5} : \frac{2}{5} : \frac{1}{5} : \frac{1}{5} :1) ∈ 𝒰 (ℤ_{2}) yields (x_{1}, x_{3})_{2} = (\frac{2}{5}, \frac{1}{5})_{2} = −1 and ev_{τ,ν_{2}}(x) = \frac{1}{2}.

ii) The local evaluation map ev_{τ,ν_{17}} is locally constant with respect to the 17-adic topology on U(ℚ_{17}). Thus, it suffices to prove the claim for points x = (x_{0} : …: x_{4}) ∈ U(ℚ_{17}) such that x_{1} ≠ 0 and x_{3} ≠ 0. Moreover, the assumption implies that x_{4} ≠ 0. Let us assume that the coordinates are normalised such that x_{0}, …, x_{4} ∈ ℤ_{17} and at least one is a unit.

If x_{4} is a unit then x defines a ℤ_{17}-valued point on 𝒰 . In this case, ev_{τ,ν_{17}}(x) = 0, as follows from i). Thus, assume that 17|x_{4}. Then x reduces modulo 17 to a point on the reduction D_{17} of the curve D. A direct inspection shows that D_{17} has exactly 17 𝔽_{17}-rational points, that ξ_{1}ξ_{3} ≠ 0 for exactly 14 of these, and that ξ_{1}/ξ_{3} ∈ 𝔽_{17}^{∗} is a square for each of them. Therefore, if 17 ∤ x_{1} and 17 ∤ x_{3} then x_{1}/x_{3} ∈ ℚ_{17}^{∗} is a square and, consequently, (\frac{x_{1}}{x_{4}} , \frac{x_{3}}{x_{4}} )_{17} = (−\frac{x_{1}}{x_{3}} , \frac{x_{3}}{x_{4}} )_{17} = 1.

Moreover, if 17|x_{4} and 17|x_{3} then the equations of the surface imply 17|x_{0} and 17|x_{2}, so that x_{1} is a unit. As the second equation shows that −x_{3}(2x_{1} +x_{2} +x_{3}) is a norm from ℚ_{17}(\sqrt{x_{1}/x_{4}}) , one has (\frac{x_{1}}{x_{4}} , \frac{x_{3}}{x_{4}} )_{17} = (\frac{x_{1}}{x_{4}} , \frac{−2x_{1}−x_{2}−x_{3}}{x_{4}} )_{17} = (\frac{x_{1}}{x_{4}} , \frac{−2x_{1}}{x_{4}} )_{17} = 1.

Finally, assume that 17|x_{4} and 17|x_{1}. Then the equations of the surface imply 17|x_{2} and that x_{0} and x_{3} are units such that x_{0}^{2} + x_{3}^{2} ≡ 0 (mod 17). This is enough to ensure that x_{0}/x_{3} ∈ ℚ_{17}^{∗} is a square. Since the first equation shows that −x_{0}x_{1} is a norm from ℚ_{17}(\sqrt{x_{3}/x_{4}}) , one has (\frac{x_{1}}{x_{4}} , \frac{x_{3}}{x_{4}} )_{17} = (−\frac{x_{0}}{x_{4}} , \frac{x_{3}}{x_{4}} )_{17} = (−\frac{x_{3}}{x_{4}} , \frac{x_{3}}{x_{4}} )_{17} = 1, which completes the proof. □

Remarks 6.3. i) Thus, the Brauer class τ causes a violation of strong approximation on X off \{17, ∞\}. In fact, a ℤ[\frac{1}{17}]-valued point such that x_{1} ≠ 0 and x_{3} ≠ 0 must necessarily fulfil (\frac{x_{1}}{x_{4}} , \frac{x_{3}}{x_{4}} )_{2} = 1, although not all ℤ_{2}-valued points satisfy this relation. ii) This restriction applies, in particular, to integral points. iii) There are infinitely many integral points on 𝒰 . Indeed, the curve defined by V (X_{3}) yields the family

(−n^{2} :n^{4} :±n^{3} :0:1) .

On the other hand, the curve defined by V (2X_{1} + X_{2} + X_{3}) is elliptic. It carries the six integral points (0:0:0:0:1), (0:0:−1:1:1), (−2:4:−1:−7:1), (−2:4:0:−8:1), (−14:196:−49:−343:1), (−14:196:48:−440:1), and no others. iv) A rather naively implemented search for integral points on 𝒰 delivered 28 of height <50 000 that are not of the forms mentioned above. These are the following,

(0 : −2 : 1 : 1 : 1), (−1 : −1 : −1 : 2 : 1), (−1 : −3 : 1 : 4 : 1), (4 : −8 : 6 : 4 : 1), (−6 : 4 : 4 : −8 : 1), (−8 : 18 : −11 : −23 : 1), (14 : −28 : 20 : 8 : 1), (16 : −56 : 30 : 4 : 1), (76 : −696 : −230 : 4 : 1), (−97 : 521 : −223 : −808 : 1), (−105 : 1413 : 381 : −3204 : 1), (263 : −829 : 467 : 62 : 1), (−556 : 912 : 712 : −128 : 1), (−708 : 1278 : −951 : −423 : 1), (839 : −1595 : 1157 : 444 : 1), (1004 : −1648 : −1288 : 4352 : 1), (−2073 : 3573 : −2721 : −2988 : 1), (2238 : −6876 : 3924 : 9288 : 1), (−2264 : 3840 : −2948 : −3056 : 1), (−2916 : 5832 : 4122 : −15228 : 1), (3324 : −15678 : −7219 : 289 : 1), (3879 : −6183 : −4899 : 16344 : 1), (−5450 : 14688 : 8947 : −791 : 1), (−5809 : 11231 : −8077 : −2950 : 1), (8908 : −16476 : 12115 : 5017 : 1), (−10194 : 6948 : 8415 : −15687 : 1), (22238 : −38044 : 29087 : 31097 : 1), and (−26396:44152:−34138:−33148:1).

v) The exceptional behaviour at the prime 17 is possible only because D has bad reduction at the prime 17. The curve D_{17} has exactly one singular point, at (1:1:13:1:0). Using Gröbner bases, one may easily verify that, on D_{17}, there is the relation

\frac{X_{1}}{X_{3}} = 8(\frac{X_{0}X_{2}X_{3} − 2X_{2}^{3} + X_{0}X_{2}^{2} + X_{0}^{2}X_{2}}{X_{0}^{2}(4X_{0} + X_{2})} )^{2} .

Example 6.4. Let 𝒳 ′ ⊂ P_{ℤ}^{4} be given by the system of equations

X_{0}(8X_{1} + 3X_{4}) + X_{2}^{2} = X_{4}(8X_{3} + 2X_{4}) ,
(8X_{3} + 2X_{4})(16X_{1} + X_{2} + 8X_{3} + 8X_{4}) + X_{0}^{2} = X_{4}(8X_{1} + 3X_{4}) .

Put 𝒰′ := 𝒳 ′\ℋ for the hyperplane ℋ := V (X_{4}) ⊂ P_{ℤ}^{4} and denote the generic fibre of 𝒰′ by U′. Then 𝒰′(ℤ_{p}) ≠ ∅ for every prime number p and U′(ℚ) ≠ ∅, but there are no ℤ[\frac{1}{17}]-valued points, 𝒰′(ℤ[\frac{1}{17}]) = ∅ . I.e, the Hasse principle for ℤ[\frac{1}{17}]-valued points is violated. In particular, a failure of the integral Hasse principle occurs. The violations can be explained by a transcendental Brauer class.

Proof. There is a morphism π: 𝒰′ → 𝒰 to the scheme from Example 6.2 that is given by

π: (x_{0} :x_{1} :x_{2} :x_{3} :1) ↦ (x_{0} :(8x_{1} + 3):x_{2} :(8x_{3} + 2):1) .

This morphism yields bijections between the sets of ℚ-rational points, as well as between the sets of ℤ_{p}-valued points for p ≠ 2. In particular, U′(ℚ) ≠ ∅ and 𝒰′(ℤ_{p}) ≠ ∅, except possibly for p ≠ 2. Moreover, there is the ℤ_{2}-valued point (3:−\frac{4}{3} :5:0:1) ∈ 𝒰′(ℤ_{2}).

There is, however, no ℤ[\frac{1}{17}]-valued point on 𝒰′. In fact, the image of such a point would fulfil (8x_{1} + 3, 8x_{3} + 2)_{2} = (3, 2)_{2} = −1, in contradiction to what was proven about 𝒰, cf. Remark 6.3.i). □

Remarks 6.5. i) Observe that the non-existence of integral points on 𝒰′ may not be explained by a lack of real points. In fact, the ℚ-scheme U′ is strongly unobstructed at ∞, simply because U is. ii) The result may be formulated in a more classical manner as follows. The pair (q_{1}, q_{2}) of quadratic polynomials in four variables, consisting of

q_{1} :=  X_{0}(8X_{1} + 3) + X_{2}^{2} − (8X_{3} + 2) ,
q_{2} := (8X_{3} + 2)(16X_{1} + X_{2} + 8X_{3} + 8) + X_{0}^{2} − (8X_{1} + 3) ,

does not represent (0, 0) integrally, although it does so over the rationals and in p-adic integers for every prime number p.

Blowing up a point. A cubic surface.

Example 6.6. Let 𝒮 ⊂ P_{ℤ}^{3} be given by the equation

Y_{0}^{3} + Y_{0}Y_{1}Y_{2} + Y_{0}Y_{2}^{2} − 2Y_{1}^{2}Y_{2} + Y_{1}^{2}Y_{3} + 2Y_{2}^{2}Y_{3} − Y_{2}Y_{3}^{2} = 0

and put 𝒱 := 𝒮 \ℰ, for the hyperplane ℰ := V (Y_{3}) ⊂ P_{ℤ}^{3}. Then every integral point (y_{0} : y_{1} : y_{2} : 1) ∈ 𝒱 (ℤ) such that y_{0}y_{2} ≠ 0 satisfies the condition that (y_{0}(y_{2} − y_{1}^{2}), y_{2})_{2} = 1 or gcd(y_{0}, 2y_{2} − 1) > 1.

Proof. The scheme 𝒮 is obtained from the scheme 𝒳 of Example 6.1 by blowing up the point (0 : 1 : 0 : 0 : 0). From the computational viewpoint, this means to eliminate X_{1} from the equations defining 𝒳 . Moreover, we replaced the coordinate sections X_{0}, X_{2}, X_{3}, X_{4}, in this order, by Y_{0}, Y_{1}, Y_{2}, and Y_{3}. An integral point on 𝒱 thus corresponds to a ℚ-rational point (x_{0} : x_{1} : x_{2} : x_{3} : 1) ∈ U(ℚ) such that x_{0}, x_{2}, and x_{3} are integers, but x_{1} not necessarily.

If, however, gcd(x_{0}, 2x_{3} − 1) = 1 then the equations x_{0} · x_{1} = x_{3} − x_{2}^{2} and (2x_{3} − 1) · x_{1} = −(x_{0}^{2} + x_{2}x_{3} + x_{3}^{2}) together imply that x_{1} has to be an integer, as well. Then, by Example 6.2.i), (x_{1}, x_{3})_{2} = 1. The assertion follows, as 𝒮 does not have integral points such that y_{2} − y_{1}^{2} = 0, but y_{0}y_{2} ≠ 0. □

Remarks 6.7. i) Write S ⊂ P_{ℚ}^{3} and E ⊂ P_{ℚ}^{3} for the generic fibres of 𝒮 and ℰ, respectively, and put V := S\E. Then, under the blow-down morphism π: S → X, one has π^{−1}(H) = E ∪ L and, hence, π^{−1}(U) = V \L, for L ⊂ S the line given by Y_{0} = 2Y_{2} − Y_{3} = 0. Note that a point on the hyperplane H has been blown up. The effect deduced above by an elementary argument is, of course, that of the Brauer class π^{∗}τ ∈ Br(V \L). The points such that gcd(y_{0}, 2y_{2} − 1) = 1 are exactly those that are integral with respect to the obvious integral model of V \L. ii) Blowing up a point outside H leads to π^{−1}(H) being a non-plane genus-one curve on S. In this case, the corresponding notion of integrality for points is further away from the classical meaning. iii) There exist integral points on 𝒱 of all three kinds that are allowed by the statements above. For instance, for (−1 : −1 : 2 : 1), the gcd is 1 and the Hilbert symbol is 1. On the other hand, for (−17 : 15 : −8 : 1), the gcd is 17 > 1 and the Hilbert symbol is 1, while for (3:5:2:1), the gcd is 3 > 1 and the Hilbert symbol is equal to (−1).

Example 6.8. Let 𝒮′ ⊂ P_{ℤ}^{3} be given by the equation

128Y_{0}^{3} + 144Y_{0}^{2}Y_{3} + 32Y_{0}Y_{1}Y_{2} + 8Y_{0}Y_{1}Y_{3} + 128Y_{0}Y_{2}^{2} + 80Y_{0}Y_{2}Y_{3} + 66Y_{0}Y_{3}^{2}

− 16Y_{1}^{2}Y_{2} − 3Y_{1}^{2}Y_{3} − 4Y_{1}Y_{2}Y_{3} + 80Y_{2}^{2}Y_{3} + 40Y_{2}Y_{3}^{2} + 12Y_{3}^{3} = 0

and put 𝒱 ′ := 𝒮′\ℰ, for the hyperplane ℰ := V (Y_{3}) ⊂ P_{ℤ}^{3}. Then every integral point (y_{0} :y_{1} :y_{2} :1) ∈ 𝒱 ′(ℤ) satisfies gcd(8y_{0} + 3, 16y_{2} + 3) > 1.

Proof. There is a morphism π: 𝒱 ′ → 𝒱 to the scheme from Example 6.6 that is given by

π: (y_{0} :y_{1} :y_{2} :1) ↦ ((8y_{0} + 3):(2y_{1} + 1):(8y_{2} + 2):1) .

As (8y_{0} +3)[(8y_{2} +2)−(2y_{1} +1)^{2}] ≡ 3 (mod 8) and 8y_{2} +2 ≡ 2 (mod 8), the Hilbert symbol is always equal to (−1), as required. □

Remark 6.9. The surface 𝒱 ′ contains infinitely many integral points. Indeed, define the two sequences c and c′ in ℤ^{3} recursively by

c_{1} := [0, −2, 0], c_{2} := [−48, 170, −24], c_{i+2} := −110c_{i+1} − c_{i} − [48, 48, 24] ,

c′_{1} := [0, 2, 0], c′_{2} := [−48, −266, −24], c′_{i+2} := −110c′_{i+1} − c′_{i} − [48, 48, 24] . (4)

Then, for each i ∈ ℕ, (c_{i1} : c_{i2} : c_{i3} : 1) ∈ 𝒱 ′(ℤ) and (c′_{i1} : c′_{i2} : c′_{i3} : 1) ∈ 𝒱 ′(ℤ). Observe that the intersection of 𝒮′ with the plane given by Y_{0} = 2Y_{2} contains the exceptional line and, therefore, splits off a conic.

There are further integral points on 𝒱 ′ not being of this particular form, for instance (−1536:5414:−803:1) and (20 706:−344 632:534:1). Moreover, both are the smallest members of infinite sequences of integral points of the same kind as (4). However, the second member of the sequence starting at (−1536 : 5414 : −803 : 1) involves 1340-digit integers, already.

7. More examples

A del Pezzo surface of degree four. In the example below, an algebraic Brauer class interacts with transcendental ones.

Example 7.1. Let X ⊂ P_{ℚ}^{4} be given by the system of equations

X_{0}X_{1} + X_{2}^{2} = X_{4}X_{3} ,
X_{3}(X_{1} + X_{3}) + X_{0}^{2} = X_{4}X_{1}

and put U := X \H, for the hyperplane H := V (X_{4}) ⊂ P_{ℚ}^{4} .

i) Then X is a del Pezzo surface of degree four. ii) D := H ∩ X is a non-singular curve of genus one. Furthermore,

D(ℚ) = {(0:1:0:0:0), (0:1:0:−1:0)} .

iii) The manifold X(ℝ) is connected, its submanifold U(ℝ) consists of two connected components, and U is strongly unobstructed at ∞. iv) One has Br(U)/ Br_{0}(U) ≅ (ℤ/2ℤ)^{2}, the three nontrivial elements being represented by

α: (\frac{X_{1}}{X_{4}} , −1; −1) ,
τ: (\frac{X_{1}}{X_{4}} , \frac{X_{3}}{X_{4}} ; −1) ,  and
α + τ: (\frac{X_{1}}{X_{4}} , −\frac{X_{3}}{X_{4}} ; −1) .

Moreover, Br_{1}(U)/ Br_{0}(U) = ⟨α⟩ ≅ ℤ/2ℤ.

Proof. i) and ii) magma reports that both X and D are non-singular. More precisely, D is isomorphic to the elliptic curve E: y^{2}z = x^{3} + xz^{2} of j-invariant 1728, an isomorphism ι: D → E being given by

(X_{0} :X_{1} :X_{2} :X_{3} :0) ↦ (X_{2}^{2} :−X_{1}X_{2} :−X_{1}X_{3}) .

This explains, in particular, why D has no ℚ-rational points other than the two obvious ones. iii) The pencil of quadrics in P^{4} corresponding to X contains exactly three degenerate quadrics that are defined over ℝ. Among the corresponding rank-four discriminants, two are negative but the third one is positive. According to [VAV, Theorem 3.4 and Lemma 5.1], this already implies that Br(X_{ℝ} )/ Br(ℝ) = 0. Consequently, X(ℝ) is connected [Si, Théorème 1.4].

As the cubic polynomial x^{3} + x has only one real root, we see that D(ℝ) is connected, too. Such a curve may cut X(ℝ) into not more than two components. On the other hand, U(ℝ) is clearly disconnected. Indeed, for a real point (x_{0} :x_{1} :x_{2} :x_{3} :1) ∈ U(ℝ), the second equation implies (x_{3} + \frac{x_{1}}{2} )^{2} + x_{0}^{2} = \frac{1}{4}[x_{1}^{2} + 4x_{1}], hence x_{1} ≥ 0 or x_{1} ≤ −4. And there exist points of both kinds, for example (−1:1:0:−1:1) and (0:−4:\sqrt{2}:2:1).

Since D is a non-singular, geometrically irreducible curve having real points and both components of U(ℝ) have limit points in D(ℝ), Theorem 2.7 yields that U is strongly unobstructed at the real place. iv) A calculation using Gröbner bases shows that the Galois group operating on the 16 lines on X is of order exactly 384. Thus, Theorem 4.8.b.ii) immediately implies that Br_{1}(U)/ Br_{0}(U) ≅ ℤ/2ℤ. Moreover, the second equation is only of rank 4 and may be written in the form X_{1}(X_{4} − X_{3}) − X_{3}^{2} − X_{0}^{2} = 0. Therefore, Theorem 5.4 shows that (\frac{X_{1}}{X_{4}} , −1; −1) defines a nontrivial algebraic Brauer class on U.

On the other hand, we have #D(ℚ) = 2. Again, E has two nontrivial 4-isogenies, but no nontrivial n-isogenies for any other n ≥ 3. Only one of the isogenous curves has a ℚ-rational 4-torsion point, which, however, is not in the image of the 4-torsion of E. Hence, Corollary 4.10 provides us with an injection Br(U)/ Br_{1}(U) ↪ ℤ/2ℤ. Consequently, Br(U)/ Br_{0}(U) is of order at most 4. As far as a concrete description is asked for, Theorem 5.5 applies. It shows that (\frac{X_{1}}{X_{4}} , \frac{X_{3}}{X_{4}} ; −1) defines a Brauer class on U.

In order to prove that this is indeed a transcendental class, we need to show that \frac{X_{1}}{X_{3}} is not the square of a rational function on D\overline{ℚ} . For this, we may argue as follows. One has

−\frac{X_{1}}{X_{3}} = −\frac{X_{1}X_{3}}{X_{3}^{2}}
= \frac{X_{0}^{2}+X_{3}^{2}}{X_{3}^{2}}
= \frac{X_{0}^{2}X_{1}^{2}+X_{1}^{2}X_{3}^{2}}{X_{1}^{2}X_{3}^{2}}
= \frac{X_{2}^{4}+X_{1}^{2}X_{3}^{2}}{X_{1}^{2}X_{3}^{2}}

and this is the product of F := \frac{X_{2}^{2}+iX_{1}X_{3}}{X_{1}X_{3}} together with its complex conjugate. One directly finds that

div F = 2[(1:0:0:i:1)] − 2[(0:1:0:0:1)] ,

hence

div \overline{F} = 2[(1:0:0:−i:1)] − 2[(0:1:0:0:1)] ,

and

div(\frac{X_{1}}{X_{3}} ) = 2[(1:0:0:i:1)] + 2[(1:0:0:−i:1)] − 4[(0:1:0:0:1)] .

In particular, (1 : 0 : 0 : i: 1) − (0 : 1 : 0 : 0 : 1) ∈ J(D)(\overline{ℚ} ) is a proper 2-torsion point, and (1:0:0:−i:1) − (0:1:0:0:1) ∈ J(D)(\overline{ℚ} ) is another.

Therefore, their sum (1:0:0:i:1) + (1:0:0:−i:1) − 2(0:1:0:0:1) ∈ J(D)(\overline{ℚ} ) is the third 2-torsion point, in particular it is non-zero. As a consequence of this, we see that \frac{X_{1}}{X_{3}} can not be the square of a rational function on D\overline{ℚ} , as required. □

Example 7.2 (continued). Let X, U, α, and τ be as above. Moreover, let 𝒳 ⊂ P_{ℤ}^{4} be defined by the same system of equations as X and put 𝒰 := 𝒳 \ ℋ , for ℋ := V (X_{4}) ⊂ P_{ℤ}^{4}. Then

i) for all primes p ≠ 2, ∞, the local evaluation maps ev_{α,ν_{p}}: 𝒰 (ℤ_{p}) → ℚ/ℤ, ev_{τ,ν_{p}}: 𝒰 (ℤ_{p}) → ℚ/ℤ, and ev_{α+τ,ν_{p}}: 𝒰 (ℤ_{p}) → ℚ/ℤ are constantly zero. ii) The local evaluation map ev_{τ,ν_{∞}}: U(ℝ) → \frac{1}{2}ℤ/ℤ is constantly zero, but ev_{τ,ν_{2}} : 𝒰 (ℤ_{2}) → ℚ/ℤ is non-constant. iii) The local evaluation map ev_{α+τ,ν_{∞}}: U(ℝ) → \frac{1}{2}ℤ/ℤ is non-constant, but ev_{α+τ,ν_{2}} : 𝒰 (ℤ_{2}) → ℚ/ℤ is constantly zero. iv) The maps ev_{α,ν_{2}} : 𝒰 (ℤ_{2}) → ℚ/ℤ and ev_{α,ν_{∞}}: U(ℝ) → \frac{1}{2}ℤ/ℤ are both nonconstant. v) For p ≡ 1 (mod 4), even the local evaluation map ev_{α,ν_{p}}: U(ℚ_{p}) → ℚ/ℤ is constantly zero.

Proof. i) For ev_{τ,ν_{p}}, this is directly Lemma 5.9. And for ev_{α,ν_{p}}, Lemma 5.7 applies, since d = −1, the linear forms X_{1}, X_{4} − X_{3}, X_{3}, and X_{0} are linearly independent modulo every prime p, and the cusp (0 : 0 : 1 : 0 : 0) does not satisfy the second equation X_{0}X_{1} + X_{2}^{2} = X_{4}X_{3} modulo any prime.

ii) The second equation excludes the possibility that X_{1} and X_{3} might both be negative. Hence, ev_{τ,ν_{∞}} is constantly zero.

Furthermore, for the integral point x = (−1 : 1 : 0 : −1 : 1) ∈ 𝒰 (ℤ_{2}), one has (x_{1}, x_{3})_{2} = (1, −1)_{2} = 1 and thus ev_{τ,ν_{2}}(x) = 0. On the other hand, the ℤ_{2}-valued point x′ = (−\frac{1}{3} : \frac{1}{3} :−\frac{2}{3} : \frac{1}{3} :1) ∈ 𝒰 (ℤ_{2}) yields (x_{1}, x_{3})_{2} = (\frac{1}{3}, \frac{1}{3})_{2} = −1, hence ev_{τ,ν_{2}}(x) = \frac{1}{2}.

iii) First of all, the integral point x = (−1 : 1 : 0 : −1 : 1) ∈ U(ℝ) yields (x_{1}, −x_{3})_{∞} = (1, 1)_{∞} = 1, therefore ev_{α+τ,ν_{∞}}(x) = 0. On the other hand, there is the real point x′ = (0:−4:\sqrt{2}:2:1) such that (x′_{1}, −x′_{3})_{∞} = (−4, −2)_{∞} = −1, hence ev_{α+τ,ν_{∞}}(x) = \frac{1}{2}.

Finally, constancy of ev_{α+τ,ν_{2}} is elementary, too, but quite involved. The idea of the proof is roughly as follows. It suffices to show (x_{1}, −x_{3})_{2} = 1 for every ℤ_{2}-valued point x = (x_{0} :x_{1} :x_{2} :x_{3} :x_{4}) ∈ 𝒰 (ℤ_{2}) such that x_{1} ≠ 0 and x_{3} ≠ 0.

In order to do this, several cases have to be distinguished. It turns out that the assumption 8 ∤ x_{3} leads to only finite many cases and that (x_{1}, −x_{3})_{2} = 1 is fulfilled in each of them. On the other hand, if 8|x_{3} then the second equation x_{0}^{2}+x_{3}^{2} = (1−x_{3})x_{1} shows that (x_{1}, −1)_{2} = 1. Moreover, x_{1} is automatically a square, except for the case that |ν_{2}(x_{0}) − ν_{2}(x_{3})| ≤ 1. In this case, however, ν_{2}(x_{0}x_{1}) ≥ ν_{2}(x_{3}) + 3 such that the first equation implies that x_{3} is a square. The assertion follows.

iv) is an immediate consequence of ii) and iii), while v) follows from the fact that (−1) is a square in ℚ_{p} for p ≡ 1 (mod 4). □

Remarks 7.3. i) The algebraic Brauer class α causes a violation of strong approximation off S_{1} := \{p prime | p ≡ 1 (mod 4)\}. In fact, a ℤ[\frac{1}{S_{1}} ]-valued point such that x_{1} ≠ 0 must necessarily fulfil (\frac{x_{1}}{x_{4}} , −1)_{2} + (\frac{x_{1}}{x_{4}} , −1)_{∞} = 0, although not all adelic points outside S_{1} satisfy this relation.

ii) Furthermore, strong approximation off {∞} is violated. In fact, the Brauer class τ yields a restriction for integral points. Namely, an integral point such that x_{1} ≠ 0 and x_{3} ≠ 0 must necessarily fulfil (x_{1}, x_{3})_{2} = 1, although not all ℤ_{2}-valued points satisfy this relation.

iii) The Brauer class α + τ yields another restriction for integral points. Indeed, an integral point such that x_{1} ≠ 0 and x_{3} ≠ 0 must necessarily fulfil (x_{1}, −x_{3})_{∞} = 1, although there exist real points violating this. iv) There are infinitely many integral points on 𝒰 . In fact, the curves defined by V (X_{3}) and V (X_{1} + X_{3}) yield the families

(−n^{2} :n^{4} :±n^{3} :0:1) and (−(n^{2} + 1):(n^{2} + 1)^{2} :±(n^{2} + 1)n:−(n^{2} + 1)^{2} :1) .

v) A search for integral points on 𝒰 delivered exactly 16 of height <50 000 that are of neither of the two types above. These are

(−5 : 13 : ±8 : −1 : 1), (−58 : 676 : ±198 : −4 : 1), (−268 : 4240 : ±1064 : −4224 : 1), (−1297 : 11437 : ±3850 : −11289 : 1), (−2416 : 6736 : ±4034 : −1020 : 1), (−4513 : 9685 : ±6611 : −3084 : 1), (−6668:13456:±9472:−5824:1), and (−11681:27061:±17779:−6700:1).

Example 7.4. Let 𝒳 ′ ⊂ P_{ℤ}^{4} be given by the system of equations

X_{0}(4X_{1} + 3X_{4}) + X_{2}^{2} = X_{4}(4X_{3} + 3X_{4}) ,
X_{3}(4X_{1} + 4X_{3} + 6X_{4}) + X_{0}^{2} = X_{4}(4X_{1} + 3X_{4})

and put 𝒰′ := 𝒳 ′\ℋ , for the hyperplane ℋ := V (X_{4}) ⊂ P_{ℤ}^{4}. Denote the generic fibre of 𝒰′ by U′. Then 𝒰′(ℤ_{p}) ≠ ∅ for every prime number p and U′(ℚ) ≠ ∅, but there are no integral points, 𝒰′(ℤ) = ∅. In other words, the integral Hasse principle is violated. The violation can be explained by a Brauer-Manin obstruction.

Proof. There is a morphism π: 𝒰′ → 𝒰 to the scheme from the example before that is given by

π: (x_{0} :x_{1} :x_{2} :x_{3} :1) ↦ (x_{0} :(4x_{1} + 3):x_{2} :(4x_{3} + 3):1) .

This morphism yields bijections between the sets of ℚ-rational points, as well as between the sets of ℤ_{p}-valued points for p ≠ 2. In particular, U′(ℚ) ≠ ∅ and 𝒰′(ℤ_{p}) ≠ ∅, except possibly for p ≠ 2. Moreover, there is a ℤ_{2}-valued point such that x_{1} = 0 and x_{3} = 2.

There is, however, no integral point on 𝒰′. In fact, the image of such a point would fulfil (4x_{1} + 3, 4x_{3} + 3)_{2} = (3, 3)_{2} = −1, in contradiction to what was proven, cf. Remark 7.3.i). □

Remark 7.5. Again, the ℚ-scheme U′ is strongly unobstructed at ∞. Concerning this, the situation is completely analogous to the one described in Remark 6.5.i).

Blowing up a point. A cubic surface.

Example 7.6. Let 𝒮 ⊂ P_{ℤ}^{3} be given by the equation

Y_{0}^{3} + Y_{0}Y_{2}^{2} − Y_{1}^{2}Y_{2} + Y_{1}^{2}Y_{3} + Y_{2}^{2}Y_{3} − Y_{2}Y_{3}^{2} = 0 .

Put 𝒱 := 𝒮 \ℰ for the hyperplane ℰ := V (Y_{3}) ⊂ P_{ℤ}^{3}. Then every integral point (y_{0} : y_{1} : y_{2} : 1) ∈ 𝒱 (ℤ) such that y_{0}y_{2} ≠ 0 satisfies that (y_{0}(y_{2} − y_{1}^{2}), y_{2})_{2} = 1. In particular, strong approximation off {∞} is violated.

Proof. The scheme 𝒮 is obtained from the scheme 𝒳 from Example 7.1 by blowing up the point (0 : 1 : 0 : 0 : 0). We eliminated X_{1} from the equations defining 𝒳 and replaced the coordinate sections X_{0}, X_{2}, X_{3}, X_{4}, in this order, by Y_{0}, Y_{1}, Y_{2}, and Y_{3}. An integral point on 𝒱 corresponds to a ℚ-rational point (x_{0} :x_{1} :x_{2} :x_{3} :1) ∈ U(ℚ) such that x_{0}, x_{2}, and x_{3} are integers, but x_{1} not necessarily.

For the main assertion, let us first observe that there are no integral points such that y_{2} − y_{1}^{2} = 0, but y_{0}y_{2} ≠ 0. Thus, the symbol is properly defined. Moreover, Example 7.2.ii) shows that (y_{0}(y_{2} − y_{1}^{2}), y_{2})_{∞} = 1. We will complete the proof by verifying (y_{0}(y_{2} − y_{1}^{2}), y_{2})_{p} = 1 for every prime p ≠ 2.

In order to do this, assume that p ∤ (x_{3} − 1) first. Then (x_{3} − 1)·x_{1} = −(x_{0}^{2} + x_{3}^{2}) implies that x_{1} is a p-adic integer. Consequently, by Example 7.2.i), one has (y_{0}(y_{2} − y_{1}^{2}), y_{2})_{p} = (x_{1}, x_{3})_{p} = 1. On the other hand, if p | (x_{3} − 1) then x_{3} is a square in ℤ_{p}^{∗} and therefore (y_{0}(y_{2} − y_{1}^{2}), y_{2})_{p} = (x_{1}, x_{3})_{p} = 1, as well. The assertion follows.

Finally, we note that, for the point (y_{0} :y_{1} :y_{2} :1) = (−\frac{1}{3} :−\frac{2}{3} : \frac{1}{3} :1) ∈ 𝒱 (ℤ_{2}) , one has that (y_{0}(y_{2}−y_{1}^{2}), y_{2})_{2} = −1. Hence, this ℤ_{2}-valued point cannot be approximated by integral ones. □

Remarks 7.7. Write S ⊂ P_{ℚ}^{3} and E ⊂ P_{ℚ}^{3} for the generic fibres of 𝒮 and ℰ, respectively, and put V := S \E. Then, under the blow-down morphism π: S → X, one has π^{−1}(H) = E ∪ L and, hence, π^{−1}(U) = V \L, for L ⊂ S the line given by Y_{0} = Y_{2} − Y_{3} = 0.

i) The effect shown above is that of the Brauer class π^{∗}τ ∈ Br(V \L), which, as \frac{Y_{2}}{Y_{3}} = \frac{X_{3}}{X_{4}} is a square on L, turns out to be unobstructed at L and extends to the whole of V. ii) The other generator shows a behaviour similar to that indicated in Example 7.6. I.e., it yields that gcd(y_{0}, y_{2} − 1) > 1 or (y_{0}(y_{2} − y_{1}^{2}), −y_{2})_{∞} = 1 for every integral point, while neither of the two statements is always true.

Example 7.8. Let 𝒮′ ⊂ P_{ℤ}^{3} be given by the equation

16Y_{0}^{3} + 12Y_{0}^{2}Y_{3} + 16Y_{0}Y_{2}^{2} + 24Y_{0}Y_{2}Y_{3} + 12Y_{0}Y_{3}^{2} − 4Y_{1}^{2}Y_{2} − 2Y_{1}^{2}Y_{3} + 8Y_{2}^{2}Y_{3}

+ 11Y_{2}Y_{3}^{2} + 4Y_{3}^{3} = 0

and put 𝒱 ′ := 𝒮′\ℰ, for the hyperplane ℰ := V (Y_{3}) ⊂ P_{ℤ}^{3}. Then 𝒱 ′(ℤ_{p}) ≠ ∅ for every prime number p and 𝒱 ′(ℚ) ≠ ∅, but there are no integral points, 𝒱 ′(ℤ) = ∅. In other words, the integral Hasse principle is violated. The violation is explained by a Brauer-Manin obstruction.

Proof. There is a morphism π: 𝒱 ′ → 𝒱 to the scheme from Example 7.6 that is given by

π: (y_{0} :y_{1} :y_{2} :1) ↦ ((4y_{0} + 1):2y_{1} :(4y_{2} + 3):1) .

This morphism yields bijections between the sets of ℚ-rational points, as well as between the sets of ℤ_{p}-valued points for p ≠ 2. In particular, 𝒱 ′(ℚ) ≠ ∅ and 𝒱 ′(ℤ_{p}) ≠ ∅, except possibly for p ≠ 2. Moreover, there is the ℤ_{2}-valued point (−\frac{1}{3} :−\frac{1}{3} :−\frac{2}{3} :1).

There is, however, no integral point on 𝒱 ′. In fact, the image of such a point would fulfil ((4y_{0} + 1)[4y_{2} + 3 − 4y_{1}^{2}], 4y_{2} + 3)_{2} = (3, 3)_{2} = −1, in contradiction to what was proven. □

8. Yet another example

In the example below, algebraic Brauer classes interact with effects caused by U(ℝ) being disconnected into compact and non-compact components. The two non-compact components of U(ℝ) in fact fulfil the requirements of Definition 2.2.ii). Nevertheless, strong approximation off {∞} is violated, as there are Brauer classes α_{1} and α_{2} working at the primes 2, 3, and ∞.

Just requiring 2-adic and 3-adic approximation by a sequence (x_{n})_{n∈ℕ} of integral points, these classes enforce a certain behaviour at the infinite prime, as well. In the present case, they determine the connected component of U(ℝ) the points x_{n} are lying on. The violation of strong approximation off {∞} happens when trying to approximate an adelic point, for which the x_{n} ∈ 𝒰 (ℤ), for n → ∞, turn out to be bound to to the single compact component of U(ℝ). This underlines observations made by U. Derenthal and D. Wei in [DW, Example 6.2].

As a consequence, we conclude that every serious definition of unobstructedness at infinity must include requirements on all connected components of U(ℝ). In particular, it is clearly insufficient just to require the existence of a single well-behaved one. Or to require that each irreducible component of the boundary contains limit points just from U(ℝ).

Example 8.1. Let X ⊂ P_{ℚ}^{4} be given by the system of equations

X_{0}(X_{0} + X_{1}) = X_{2}^{2} + (X_{0} + X_{4})^{2} ,
(X_{0} + X_{2})(X_{0} + 2X_{2}) = 2X_{1}^{2} + 3X_{3}^{2}

and put U := X \H, for the hyperplane H := V (X_{4}) ⊂ P_{ℚ}^{4} .

i) Then X is a del Pezzo surface of degree four.

ii) The curve D := H ∩ X is non-singular of genus one. One has D(ℚ) = ∅, but D(ℝ) ≠ ∅. Furthermore, J(D)(ℚ) ≅ ℤ.

iii) The manifold X(ℝ) consists of two connected components and its submanifold U(ℝ) decomposes into three connected components. U is not unobstructed at ∞, not even in the weak sense.

iv) One has Br(U)/ Br_{0}(U) ≅ (ℤ/2ℤ)^{2}, two generators being represented by

α_{1} : (\frac{X_{0}}{X_{4}} , −1; −1) and
α_{2} : (\frac{X_{0}+X_{2}}{X_{4}} , −6; −1) .

Moreover, Br(U) = Br_{1}(U).

Proof. i) and ii) magma reports that both X and D are non-singular. Furthermore, the system of equations defining D is easily seen to be insoluble in ℚ_{3}. Therefore D(ℚ) = ∅ . On the other hand, (1:1:1: \frac{2}{3} \sqrt{3}:0) ∈ D(ℝ) . The Jacobian of D is reported to be isomorphic to the elliptic curve E: y^{2}z = x^{3} + 6x^{2}z + 72xz^{2} − 54z^{3} over ℚ of j-invariant \frac{512 000}{603} and Mordell-Weil group E(ℚ) being free of rank one. iii) The pencil of quadrics in P^{4} defining X contains five degenerate ones, all of which are real. Among the corresponding rank-four discriminants, four are negative and the fifth is positive. According to [VAV, Theorem 3.4 and Lemma 5.1], this implies Br(X_{ℝ} )/ Br(ℝ) = (ℤ/2ℤ)^{2}, which in itself shows that X(ℝ) consists of exactly two connected components [Si, Théorème 1.4]. The evaluation of the global Brauer class on X_{ℝ} , given by (\frac{X_{0}+X_{2}}{X_{0}} , −1; −1), distinguishes the two components.

As the cubic polynomial x^{3} + 6x^{2} + 72x − 54 has exactly one real root, we see that D(ℝ) is connected. Such a curve may meet only one component of X(ℝ), it obviously meets X_{+}, the one where \frac{X_{0}+X_{2}}{X_{0}} is positive, and it cannot cut that into more than two components. On the other hand, one sees that D(ℝ) indeed decomposes X_{+} into two components, these being distinguished by the sign of \frac{X_{0}}{X_{4}} . Observe that both are non-empty, as they contain the real points (17 : 3 : 4 : −13 : 1) and (−\frac{41}{8} : \frac{3}{2} : \frac{5}{4} :−\frac{11}{8} :1), respectively.

Finally, let us note once again that the component X_{−} := X(ℝ)\X_{+} of X(ℝ) is not at all met by the curve D(ℝ). Hence, X_{−} forms as well a connected component of U(ℝ). As X_{−} is compact, Definition 2.2.i) immediately shows that U cannot be weakly unobstructed. iv) A calculation using Gröbner bases indicates that the Galois group operating on the 16 lines on X is of order exactly 96. As both quadrics used to describe X are defined over ℚ and of rank 4, Theorem 4.8.b.iii) implies that Br_{1}(U)/ Br_{0}(U) ≅ (ℤ/2ℤ)^{2}.

Moreover, Theorem 5.4 shows that (\frac{X_{0}}{X_{4}} , −1; −1) and (\frac{X_{0}+X_{2}}{X_{4}} , −6; −1) define nontrivial algebraic Brauer classes on U. They do not just differ by an element from Br(ℚ), as easily follows from the results given in Example 8.3.i) and ii), below. On the other hand, J(D) do not have any nontrivial ℚ-rational isogeny. Thus, Corollary 4.10 yields Br(U) = Br_{1}(U). □ Remarks 8.2. i) Writing down the points on U in the form (x_{0} : x_{1} : x_{2} : x_{3} : 1), the three connected components of U(ℝ) are distinguished by the signs of x_{0} and x_{0} +x_{2} (respectively that of x_{0} + 2x_{2} if x_{0} + x_{2} = 0). ii) There are no real points on U such that x_{0} > 0 and x_{0} + x_{2} < 0, in accordance with our result that U(ℝ) consists of only three components. One may see this in an elementary fashion, as follows. The first equation implies x_{1} = \frac{x_{2}^{2}}{x_{0}} + \frac{1}{x_{0}} + 2, hence x_{1} > \frac{x_{2}^{2}}{x_{0}} in the case that x_{0} > 0. Moreover, (\frac{x_{0}}{x_{2}} )^{2}(\frac{x_{0}}{x_{2}} + 1)(\frac{x_{0}}{x_{2}} + 2) > 2, which implies that \frac{x_{0}}{x_{2}} < −2.293 … < −1 if x_{2} is negative, and hence x_{0} + x_{2} = x_{2}(\frac{x_{0}}{x_{2}} + 1) > 0. iii) One may as well verify in an elementary manner that the component where x_{0} < 0 and x_{0} + x_{2} > 0 is compact. All such real points fulfil x_{2} < 3.

Indeed, the equations of X imply that (\frac{x_{0}}{x_{2}} )^{2}(\frac{x_{0}}{x_{2}} + 1)(\frac{x_{0}}{x_{2}} + 2) ≥ 2[1 + \frac{1}{x_{2}^{2}} + \frac{2x_{0}}{x_{2}^{2}} ]^{2}. If one assumes x_{2} ≥ 3 and |x_{0}| < |x_{2}| then the right hand side is strictly larger than 2[1 + 0 − \frac{2}{3}]^{2} = \frac{2}{9}. However, the polynomial t^{2}(t + 1)(t + 2) adopts its maximum in the interval [−1, 0] at the point t = \frac{−9+\sqrt{17}}{8} and this maximum is 0.201 … < \frac{2}{9}.

Example 8.3 (continued). Let X, U, α_{1}, and α_{2} be as above. Moreover, let 𝒳 ⊂ P_{ℤ}^{4} be defined by the same system of equations as X and put 𝒰 := 𝒳 \ ℋ , for ℋ := V (X_{4}) ⊂ P_{ℤ}^{4}. Then

i) for all primes p ≠ 2, 3, ∞, the local evaluation maps ev_{α_{1},ν_{p}}: 𝒰 (ℤ_{p}) → ℚ/ℤ, ev_{α_{2},ν_{p}}: 𝒰 (ℤ_{p}) → ℚ/ℤ, and ev_{α_{1}+α_{2},ν_{p}}: 𝒰 (ℤ_{p}) → ℚ/ℤ are constantly zero. ii) The local evaluation map ev_{α_{1},ν_{3}} : 𝒰 (ℤ_{3}) → ℚ/ℤ is constantly zero, while ev_{α_{1},ν_{∞}}: U(ℝ) → \frac{1}{2}ℤ/ℤ and ev_{α_{1},ν_{2}} : 𝒰 (ℤ_{2}) → ℚ/ℤ are non-constant. iii) The local evaluation maps ev_{α_{2},ν_{∞}}: U(ℝ) → \frac{1}{2}ℤ/ℤ , ev_{α_{2},ν_{2}} : 𝒰 (ℤ_{2}) → ℚ/ℤ, and ev_{α_{2},ν_{3}} : 𝒰 (ℤ_{3}) → ℚ/ℤ are all non-constant. iv) The local evaluation maps ev_{α_{1}+α_{2},ν_{∞}}: U(ℝ) → \frac{1}{2}ℤ/ℤ, ev_{α_{1}+α_{2},ν_{2}}: 𝒰 (ℤ_{2}) →ℚ/ℤ, and ev_{α_{1}+α_{2},ν_{3}}: 𝒰 (ℤ_{3}) →ℚ/ℤ are all non-constant.

Proof. i) For the ev_{α_{1},ν_{p}}, Lemma 5.7 applies, since d = −1, the linear forms X_{0}, X_{0} + X_{1}, X_{2}, and X_{0} + X_{4} are linearly independent modulo an arbitrary prime p, and the cusp (0:0:0:1:0) does not satisfy the second equation X_{0}X_{1} + X_{2}^{2} = X_{4}X_{3} modulo any prime p ≠ 3.

Similarly, for the ev_{α_{2},ν_{p}}, one has d = −6, the linear forms 2(X_{0} + X_{2}), X_{0} + 2X_{2}, X_{1}, and X_{3} are linearly independent modulo every prime p and the cusp (0:0:0:0:1) does not satisfy the first equation X_{0}(X_{0}+X_{1}) = X_{2}^{2}+(X_{0}+X_{4})^{2} modulo any prime. The assertion about ev_{α_{1}+α_{2},ν_{p}} follows. ii and iii) Constancy of ev_{α_{1},ν_{3}} follows from Lemma 5.7, too, since the cusp (0:0:0:1:0) is a point not lying on 𝒰_{3}.

Furthermore, the asserted non-constancies are directly checked. In order to do this, let us consider the points x = (5: \frac{12}{5} :−1: \frac{2}{5} :1) and x′ = (−\frac{37}{13} : \frac{21}{13} : \frac{4}{13} : \frac{5}{13} :1), for which we clearly have that x, x′ ∈ 𝒰 (ℤ_{2}) ∩ 𝒰 (ℤ_{3}) ∩ U(ℝ). Explicit calculations show that ev_{α_{1},ν}(x) = ev_{α_{2},ν}(x) = ev_{α_{1}+α_{2},ν}(x) = 0 for any place ν. On the other hand, ev_{α_{1},ν}(x′) = \frac{1}{2} for ν = ν_{∞} and ν = ν_{2}, as well as ev_{α_{2},ν}(x′) = \frac{1}{2} for ν = ν_{∞}, ν = ν_{2}, and ν = ν_{3}. This establishes ii) and iii). iv) As a consequence of ii) and iii), we see that ev_{α_{1}+α_{2},ν_{3}} is non-constant. Moreover, there is the point x′′ = (−\frac{85}{91} :−\frac{15}{91} : \frac{92}{91} : \frac{9}{91} :1) ∈ 𝒰 (ℤ_{2}) ∩ U(ℝ) , for which direct calculations show that ev_{α_{1}+α_{2},ν}(x′′) = \frac{1}{2} for ν = ν_{∞} and ν = ν_{2}. This completes the proof. □

Remarks 8.4. i) Similarly to the situation in the examples of the sections above, the Brauer classes α_{1} and α_{2} cause failures of strong approximation. There is a violation of strong approximation off S_{1} := {p prime | p ≡ 1 (mod 4)}, due to α_{1}. A ℤ[\frac{1}{S_{1}} ]-valued point must necessarily fulfil (\frac{x_{0}}{x_{4}} , −1)_{2} + (\frac{x_{0}}{x_{4}} , −1)_{∞} = 0, although not all adelic points outside S_{1} satisfy this relation.

Similarly, α_{2} causes a violation of strong approximation off

S_{2} := \{p prime | (\frac{−6}{p}) = 1\} = \{p prime | p ≡ 1, 5, 7, 11 (mod 24)\} .

A ℤ[\frac{1}{S_{2}} ]-valued point such that x_{0} + x_{2} ≠ 0 must necessarily fulfil the nontrivial relation (\frac{x_{0}+x_{2}}{x_{4}} , −6)_{2} + (\frac{x_{0}+x_{2}}{x_{4}} , −6)_{3} + (\frac{x_{0}+x_{2}}{x_{4}} , −6)_{∞} = 0. However, as U(ℝ) is disconnected and one of its connected components is compact, further effects are occurring. Cf. Example 8.5 below. ii) There is exactly one integral point on 𝒰 lying on the compact component of U(ℝ), namely (−1:0:1:0:1). A search for integral points on 𝒰 delivered the following twelve others.

(17 : 3 : 4 : ±13 : 1), (1409 : 147 : −452 : ±383 : 1), (6305 : 12972 : 9043 : ±3550 : 1), (17741:12759:15044:±20351:1), (−23293:−2328:−7367:±19622:1), and (60569:2052:11143:±44472:1).

Example 8.5 (2nd continuation). Let U be as above. Then strong approximation off {∞} is violated for U.

Proof. Consider the adelic point 𝕩 outside ∞ that is equal to (−1 : 0 : 1 : 0 : 1) at every prime p ≠ 5, ∞ and equal to (17 : 3 : 4 : 13 : 1) at p = 5. Working with the model 𝒰 as above, strong approximation off {∞} would imply that there exists a sequence (x_{n})_{n∈ℕ} of integral points x_{n} ∈ 𝒰 (ℤ) being convergent to 𝕩 simultaneously with respect to the 2-, 3-, and 5-adic topologies. The latter ensures that x_{n} ≠ (−1:0:1:0:1) for n ≫ 0.

Moreover, 2-adic convergence implies that ev_{α_{1},ν_{2}}(x_{n}) = ev_{α_{1},ν_{2}}(−1:0:1:0:1) = \frac{1}{2} and ev_{α_{2},ν_{2}}(x_{n}) = ev_{α_{2},ν_{2}}(−1 : 0 : 1 : 0 : 1) = 0 for n ≫ 0, while 3-adic convergence enforces ev_{α_{2},ν_{3}}(x_{n}) = ev_{α_{2},ν_{3}}(−1 : 0 : 1 : 0 : 1) = 0. The relations discussed in Remark 8.4.i) now imply that, necessarily, ev_{α_{1},ν_{∞}}(x_{n}) = \frac{1}{2} and ev_{α_{2},ν_{∞}}(x_{n}) = 0.

In other words, x_{n} must be contained in the same connected component of U(ℝ) as the point (−1 : 0 : 1 : 0 : 1). This component is, however, compact and does not contain any other integral point. □

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Département Mathematik, Univ.Siegen,Walter-Flex-Str.3, D-57068Siegen,Germany

E-mail address: ~jjahnel@mathematik.uni-siegen.de
URL: http://wwwuser.gwdg.de/~jjahnel

Mathematisch Instituut, Universiteit Utrecht, Budapestlaan 6, NL-3584 CD Utrecht, The Netherlands

E-mail address: d.schindler@uu.nl
URL: http://www.uu.nl/staff/DSchindler