ON THE FREQUENCY OF ALGEBRAIC BRAUER CLASSES FOR CERTAIN LOG K3 SURFACES

JÖRG JAHNEL AND DAMARIS SCHINDLER^{∗}

Abstract. Given systems of two (inhomogeneous) quadratic equations in four variables, it is known that the Hasse principle for integral points may fail. Sometimes this failure can be explained by some integral Brauer-Manin obstruction. We study the existence of a non-trivial algebraic part of the Brauer group for a family of such systems and show that the failure of the integral Hasse principle due to an algebraic Brauer-Manin obstruction is rare, as for a generic choice of a system the algebraic part of the Brauergroup is trivial. We use resolvent constructions to give quantitative upper bounds on the number of exceptions.

1. Introduction

We consider a system of two rational quadratic forms in 5 variables given by

Q_{1}(x) = x^{t}Ax, Q_{2}(x) = x^{t}Bx,

where A, B ∈ M_{5×5}(ℚ) are two symmetric matrices with rational entries. For a generic choice of A and B, the intersection

S_{A,B} : Q_{1}(x) = Q_{2}(x) = 0  (1.1)

defines a del Pezzo surface of degree four in ℙ_{ℚ}^{4}. It is well known that the Hasse principle and weak approximation may fail for such surfaces (see for example [BSD]). In all known examples, for del Pezzo surfaces of degree four, the failure of the Hasse principle can be explained by a Brauer-Manin obstruction. I.e., in all of these situations one has adelic points on S_{A,B} but the Brauer-Manin set S_{A,B}(𝔸_{ℚ})^{Br}, that is the subset of adelic points that are in the kernel of the Brauer-Manin pairing with the Brauer group Br(S_{A,B}), is empty.

More recently, Colliot-Thélène and Xu [CX], initiated the study of integral Brauer-Manin obstructions. In [CX], they studied integral points on homogeneous spaces and representation problems by integral quadratic forms. In another direction, the concept of Brauer-Manin obstructions for affine varieties has been pursued in [CW] for families of affine cubic surfaces, such as representation problems of an integer by sums of three cubes. Moreover, see work of Bright and Lyczak [BL] for certain log K3 surfaces and Berg [Be] on the description of the Brauer-Manin obstruction for affine Châtelet surfaces.

Date: August 30, 2019.

∗The second author was supported by the NWO Veni Grant No. 016.Veni.173.016.

†Mathematics Subject Classification 2010, Primary 11E12; Secondary 14G20, 14G25, 14J20

In the situation of a del Pezzo surface of degree four, one may choose a hyperplane H ⊂ ℙ_{ℚ}^{4} and consider the complement U := S_{A,B} \ H. Integral points on an integral model U of U then essentially correspond to integer solutions of systems of two inhomogeneous quadratic equations in four variables. When does such a system have an integer solution? A necessary condition is that it has real solutions and solutions in ℤ_{p} for every prime p. However, these conditions may not be sufficient. Again, there could be a Brauer-Manin obstruction leading to a violation of the local-global principle. We are interested in the question how often one should expect a violation of the local-global principle due to a Brauer-Manin obstruction. A Brauer-Manin obstruction can only occur if there are non-constant Brauer classes in the Brauer group of the scheme U. The Brauer classes which vanish after a base change to the algebraic closure of ℚ are most accessible to computations. In this article we study upper bounds on how often the algebraic part of the Brauer-group is non-trivial for certain families of systems of two inhomogeneous quadratic equations in four variables. It would be very interesting to find lower bounds as well (for certain families see [JS1]), or get a prediction of the density of such examples within a given family.

In [JS1], the authors computed a list of the algebraic parts of the Brauer group that can occur for such surfaces assuming that the intersection S_{A,B} ∩H is geometrically integral. Note that these results were recently extended to del Pezzo surfaces of degree at most 7 by M. Bright and J. Lyczak [BL]. We recall that the list in the case of the complement of a hyperplane section in a del Pezzo surface of degree four consists of 0, ℤ/2ℤ, (ℤ/2ℤ)^{2}, (ℤ/2ℤ)^{3}, (ℤ/2ℤ)^{4}, ℤ/4ℤ, ℤ/2ℤ×ℤ/4ℤ and (ℤ/2ℤ)^{2}×ℤ/4ℤ. Under the assumption that H ∩S_{A,B} is geometrically irreducible (for a more general criterion see [JS1, Lemma 4.1]), the algebraic part of the Brauer group is given by the first Galois cohomology group

Br_{1}(U)/ Br_{0}(U) ≅ H^{1}(Gal(ℚ̄/ℚ), Pic(Uℚ̄)).

Moreover, any hyperplane H lies in the anticanonical class and hence the algebraic part of the Brauer group is independent of which rational hyperplane we remove as long as we assume H ∩ S_{A,B} to be geometrically irreducible.

In this note, we study the question how often one typically expects to find a non-trivial algebraic part of the Brauer group Br_{1}(U)/ Br_{0}(U), where we vary over certain subfamilies of S_{A,B}. We fix a matrix A ∈ M_{5×5}(ℤ) with det(A) ≠ 0 and vary over integral matrices B. Note that if the intersection in (1.1) is smooth of codimension 2 as a projective scheme in ℙ^{4} then S_{A,B} is indeed a del Pezzo surface of degree four.

We define N_{2}(P) to be the number of symmetric matrices B = (b_{ij})_{1≤i,j≤5} ∈ M_{5×5}(ℤ) with |b_{ij}| ≤ P for all 1 ≤ i ≤ j ≤ 5, such that S_{A,B} is smooth of codimension 2 and Br_{1}(U)/ Br_{0}(U) ≠ 0 for any hyperplane H with H ∩ S_{A,B} geometrically irreducible.

Theorem 1.1. Let A ∈ M_{5×5}(ℤ) and assume that det(A) ≠ 0. Then, for every ε > 0, one has the upper bound

N_{2}(P) ≪_{ε,A} P^{14+1/5+ε}.

In total, there are about P^{15} integer matrices B that we consider, and hence the exceptional ones, counted by N_{2}(P), are sparse in this very precise sense. In particular, for any choice of B outside of the exceptional set counted by N_{2}(P), there is no algebraic Brauer-Manin obstruction to integral points for the system

Q_{1}(x) = Q_{2}(x) = 0, L(x) = ±1,

where L(x) is a sufficiently general linear form in the variables x with integral coefficients. It would be interesting to understand the transcendental part of the Brauer group of U as well, see for example [JS1] for some discussions.

The frequency of violations of the local-global principle has recently also been studied by Mitankin [Mi] for affine quadric surfaces. Even more is known for a few selected examples of families of projective varieties, see for example [JS16] and [BB2].

Our strategy of proof for Theorem 1.1 is inspired by work of Dietmann [Di]. He bounds the number of monic integer polynomials of degree n of bounded height which have a certain Galois group strictly smaller than S_{n}.

Under the assumptions of Theorem 1.1, the set of matrices B, counted by N_{2}(P), is a thin subset in 15-dimensional affine space (see Remark 4.1). Therefore, sieve methods, as used in Theorem 13.1 in [Se], would lead to the bound

N_{2}(P) ≪ P^{14+1/2} log P.  (1.2)

With our methods, we improve upon the exponent in this estimate.

Acknowledgements: We would like to thank the referees for carefully reading this paper and for their suggestions that improved the presentation of the material.

2. Preparations

With two symmetric 5×5 matrices A and B, we associate the characteristic polynomial

f(λ, μ; A, B) := det(λA + μB),

which is homogeneous of degree 5 in λ, μ. Sometimes, we write f(λ, μ) for f(λ, μ; A, B) when the matrices A and B are considered fixed. We recall that we can read off from the polynomial f(λ, μ) whether the associated variety S_{A,B} is smooth of codimension 2.

Lemma 2.1 (Proposition 3.26 in [Wi]). Let A, B ∈ M_{5×5}(ℚ) be two symmetric matrices and S_{A,B} be defined as in (1.1), where Q_{1} and Q_{2} are the two quadratic forms associated with A and B. Then S_{A,B} is smooth over ℚ and pure of codimension 2 as a variety in ℙ^{4} if and only if the characteristic polynomial f(λ, μ) is not identically zero and separable.

Let Disc(f) be the discriminant of f(λ, μ) (see for example [Sch]). If we fix A ∈ M_{5×5}(ℤ) then Disc(f) is a polynomial in the coefficients (b_{ij}) of B. Moreover, S_{A,B} is smooth of codimension 2 if and only if Disc(f) ≠ 0. Hence we can trivially handle the count of matrices B of bounded height, for which S_{A,B} is not a del Pezzo surface of degree 4. Note that Disc(f) is not identically zero if we assume that det(A) ≠ 0. In this case, we have

♯{B ∈ M_{5×5}(ℤ) : B^{T} = B, |b_{ij}| ≤ P ∀1 ≤ i, j ≤ 5, Disc(f) = 0} ≪ P^{14}. (2.1)

Note that in this estimate the implied constant does not depend on the matrix A.

The roots of the characteristic polynomial f(λ, μ) correspond to pairs of pencils of conics contained in S_{A,B}. These pencils of conics generate Pic(Uℚ̄) up to index two and the algebraic part of the Brauer group of U is determined by the operation of the Galois group on these conics. Moreover, one can already read off from the characteristic polynomial f(λ, μ) some information about 2-torsion elements contained in Br_{1}(U)/ Br_{0}(U). Note that, if H ∩ S_{A,B} is geometrically integral and S_{A,B} is smooth of codimension 2, then U = X \H is an open del Pezzo surface of degree four as in the language of [JS2, Definition 2.5]. As we work over the base field ℚ, we have an isomorphism

Br_{1}(U)/ Br_{0}(U) ≅ H^{1}(Gal(ℚ̄/ℚ), Pic(Uℚ̄)).

The latter group is analysed in detail in [JS2] for an open degree four del Pezzo surface, and we now recall a result from that paper which forms one of the key inputs for our estimates for N_{2}(P).

Theorem 2.2 (Corollary 3.11 in [JS2]). Let S_{A,B} be a smooth del Pezzo surface of degree 4, as defined in (1.1), and H ⊂ ℙ^{4} a hyperplane such that H ∩ S_{A,B} is geometrically integral. If Br_{1}(U)/ Br_{0}(U) ≠ 0 then the characteristic polynomial f(λ, μ) has a rational root or splits off a factor of degree 2 over the rationals.

For the first part of our estimates to follow, we use the following result due to Bombieri and Pila [BP].

Theorem 2.3 (Theorem 4 in [BP]). Let G(x, y) ∈ ℤ[x, y] be an absolutely irreducible polynomial of absolute degree d. Let N ≥ exp(d^{6}) be an integer. Then one has the bound

♯\{x, y ∈ ([0, N] ∩ ℤ)^{2} : G(x, y) = 0\} ≤ N^{1/d} exp (12\sqrt{d log N log log N}) .

Note that this result is more precise than what we need, as in our application the degree d is fixed.

3. Irreducibility results and a resolvent construction

In our proof of Theorem 1.1, we need the following result on geometric irreducibility of fibers. It can be found in online lecture notes by B. Osserman [Os, Proposition 2.3]. For convenience of the reader, we give our own proof here.

Proposition 3.1. Let A be a Noetherian ring, set X := Spec A and let g ∈ A[x_{1}, …, x_{n}] be a polynomial of total degree d. Then there is a Zariski closed subset Z ⊂ X with the following property.

A prime ideal 𝔭 ∈ X is contained in Z if and only if there exists an algebraically closed field extension k of the residue field κ(𝔭) such that g is reducible or has degree strictly less than d when considered as a polynomial over k.

Remark 3.2. Note that the last statement in Proposition 3.1 is equivalent to saying that for every algebraically closed extension field k of κ(𝔭) the polynomial g is reducible or has degree less than d when considered as a polynomial over k. (See for example [Ha, Exercise II.3.15].)

Proof. Step 1: First, we consider the homogenization G(x_{0}, …, x_{n}) of the polynomial g(x_{1}, …, x_{n}), defined by

G(x_{0}, …, x_{n}) := x_{0}^{d}g (\frac{x_{1}}{x_{0}} , …, \frac{x_{n}}{x_{0}} ) .

Then G defines a hypersurface V ⊂ ℙ_{A}^{n}. Let 𝔭 ∈ X and k some extension field of κ(𝔭). We note that the polynomial g is reducible or of degree strictly smaller than d over k if and only if G is reducible over k. Moreover, G is reducible over k if and only if the reduction of V over k is reducible or not reduced over k.

By [EGAIV, Théorème 9.7.7], the set of prime ideals 𝔭 such that G is reducible over k is constructible. In order to show that this set is even Zariski closed, it is hence sufficient to show that it is closed under specialization. We will prove the following claim.

Claim 1: Let 𝔭 ⊂ 𝔓 be prime ideals in A. If the image of g in Q(A/𝔭)[x_{1}, …, x_{n}] is of degree strictly smaller than d or becomes reducible over a finite extension field of Q(A/𝔭), then the image of g in Q(A/𝔓)[x_{1}, …, x_{n}] is reducible or of degree smaller than d after taking a finite extension field. (Here, we write Q(B) for the quotient field of an integral domain B.)

Since we assumed A to be Noetherian, it is enough to prove the claim for the case that there is no other prime ideal between 𝔭 and 𝔓.

Step 2: If the reduction of g modulo 𝔭 is of degree strictly smaller than d then clearly the same is true for the reduction of g modulo 𝔓. We hence assume that the image of g in Q(A/𝔭)[x_{1}, …, x_{n}] is reducible over \overline{Q(A/𝔭)}.

Step 3: In replacing the ring A by A/𝔭, we may assume w.l.o.g. that A is an integral domain and 𝔭 = (0) and that 𝔓 is a minimal prime ideal. We now consider the localization A_{𝔓}, which is a Noetherian, one-dimensional, local integral domain with maximal ideal 𝔓A_{𝔓}. Moreover, A_{𝔓}/𝔓A_{𝔓} = Q(A/𝔓). It is hence sufficient to establish the following claim.

Claim 2: Let A be a one-dimensional, Noetherian, local integral domain with maximal ideal 𝔪 and assume that g, considered as a polynomial in Q(A)[x_{1}, …, x_{n}], is reducible over a finite extension field. Then the reduction of g in (A/𝔪)[x_{1}, …, x_{n}] is of degree strictly smaller than d or is reducible, possibly over a finite extension field of A/𝔪.

Step 4: Assume that Q(A) ⊂ L is a finite extension field, such that g factors over L. Let R be the integral closure of A in L. By the Going-up Theorem [Mat, Theorem 9.4 i)], there is a prime ideal 𝔫 ⊂ R of height one lying above 𝔪. We note that R is Noetherian by the theorem of Krull-Akizuki [Mat, Theorem 11.7]. Hence, R_{𝔫} is a Noetherian, one-dimensional, local domain, which is integrally closed in its fraction field. By [Mat, Theorem 11.2], R_{𝔫} is a discrete valuation ring. We have assumed that g factors over Q(R_{𝔫}) = Q(R) = L. Since R_{𝔫} is a discrete valuation ring, it is a unique factorization domain and Gauss’s lemma implies that g already factors over R_{𝔫}.

Step 5: As g is reducible in the polynomial ring R_{𝔫}[x_{1}, …, x_{n}], its reduction in R_{𝔫}/𝔫R_{𝔫}[x_{1}, …, x_{n}] is also reducible or of strictly smaller degree. We conclude the proof in noting that R_{𝔫}/𝔫R_{𝔫} = Q(R/𝔫) and that Q(A/𝔪) ⊂ Q(R/𝔫) is a finite extension field. □

In the following, we assume that both A and B are symmetric matrices and write (b_{ij})_{1≤i≤j≤5} for the coefficients determining the matrix B. Moreover, we set b′ := (b_{ij})_{1≤i≤j≤5,(i,j)≠(1,1)} for the vector consisting of all entries except for the first one b_{11}.

Lemma 3.3. Let A = Id_{5} be the identity matrix. Then there is a Zariski closed subset Z_{0} ⊊ 𝔸_{ℚ}^{14} with the following property. If b′ ∈ 𝔸^{14}(\overline{ℚ}) \ Z_{0}(\overline{ℚ}), then f(λ, 1; Id_{5}, B) is absolutely irreducible of degree five as a polynomial in the variables λ and b_{11}.

Proof. We apply Proposition 3.1 to the ring A = ℚ[b_{12}, …, b_{55}] and the polynomial f(λ, 1; Id_{5}, B) ∈ A[b_{11}, λ]. Let Z_{0} ⊂ 𝔸_{ℚ}^{14} be the Zariski closed subset, given by Proposition 3.1. It remains to show that Z_{0} is not equal to the whole 14-dimensional affine space. Calculations using magma show that the polynomial

g(λ, b_{11}, b_{12}) := det (λI_{5} + (\begin{matrix}b_{11} & b_{12} & 0 & 0 & 0 \\ b_{12} & b_{12} & b_{12} & 0 & 0 \\ 0 & b_{12} & b_{12} & b_{12} & 0 \\ 0 & 0 & b_{12} & b_{12} & b_{12} \\ 0 & 0 & 0 & b_{12} & b_{12} \end{matrix}))

defines an irreducible curve of genus 0 in the projective plane. Moreover, up to height 100 we find eight regular ℚ-rational points. Hence, the curve is geometrically irreducible and the polynomial g(λ, b_{11}, 1) is absolutely irreducible and of degree five. We conclude that Z_{0} ⊊ 𝔸_{ℚ}^{14} as desired. □

Since f(λ, 1; Id_{5}, B) is homogeneous of degree five, we obtain the following corollary.

Corollary 3.4. The polynomial f(λ, 1; Id_{5}, B) is absolutely irreducible in the 16 variables λ and (b_{ij})_{1≤i≤j≤5}. □

Now we show that similar statements hold when we replace the identity matrix I_{5} by some symmetric invertible matrix A ∈ M_{5×5}(ℚ).

Lemma 3.5. Let A ∈ M_{5×5}(ℚ) be a symmetric matrix and assume that det(A) ≠ 0. Then the polynomial f(λ, 1; A, B) is absolutely irreducible in the 16 variables λ and (b_{ij})_{1≤i≤j≤5}.

Proof. As we ask for absolute irreducibility, we may work over ℚ̄. Let T be an invertible matrix such that T^{t}AT = Id_{5}. Then we have

f(λ, 1; A, B) = det(λA + B) = det(λ(T^{−1})^{t}Id_{5}T^{−1} + B)

= det(T^{−1})^{2} det(λI_{5} + T^{t}BT). (3.1)

By Corollary 3.4, the polynomial det(λI_{5} + B) is absolutely irreducible, considered as a polynomial in 16 variables. Hence, the polynomial on the right hand side of (3.1) is absolutely irreducible in λ and (b_{ij})_{1≤i≤j≤5}, as the linear variable substitution B ↦ T^{t}BT does not affect absolute irreducibility. □

For the case A = Id_{5}, Lemma 3.3 shows that the polynomial f(λ, 1; Id_{5}, B) is absolutely irreducible in λ and b_{11}, for almost all b′. We now provide a comparable statement for a general matrix A. For the proof, we use Lemma 3.5 together with a Bertini-type theorem for absolute irreducibility to deduce that there is a curve in the corresponding family that is absolutely irreducible. This will suffice for another application of Proposition 3.1.

Proposition 3.6. Let A ∈ M_{5×5}(ℚ) be a symmetric matrix with det(A) ≠ 0. Then there exist a tuple v′ := (v_{ij})_{1≤i≤j≤5,(i,j)≠(1,1)} ∈ ℤ^{14} and a Zariski closed subset Z_{A} ⊊ 𝔸_{ℚ}^{14} with the following property.

Define the polynomial

h_{A}(λ, b_{11}, c′) := det (λA + (\begin{matrix}b_{11} & c_{12} + v_{12}b_{11} & … & c_{15} + v_{15}b_{11} \\ c_{12} + v_{12}b_{11} & c_{22} + v_{22}b_{11} & … & c_{25} + v_{25}b_{11} \\ ⋮ & ... & & ⋮ \\ c_{15} + v_{15}b_{11} & c_{25} + v_{25}b_{11} & … & c_{55} + v_{55}b_{11} \end{matrix})) .

Then h_{A}(λ, b_{11}, c′) is absolutely irreducible as a polynomial in λ and b_{11}, for each c′ := (c_{ij})_{1≤i≤j≤5,(i,j)≠(1,1)} ∈ 𝔸^{14}(ℚ̄) \ Z_{A}(ℚ̄).

Proof. By Lemma 3.5, the polynomial f(λ, 1; A, B) is absolutely irreducible in the 16 variables λ and (b_{ij})_{1≤i≤j≤5}. Consider the affine variety X ⊂ 𝔸_{ℚ}^{16}, given by f(λ, 1; A, B) = 0, which is then geometrically integral. Define the projection

\begin{matrix}ϕ_{12} : X & → 𝔸_{ℚ}^{2} \\ (λ, (b_{ij})) & ↦ (b_{11}, b_{12}), \end{matrix}

and note that this is a dominant map. Then an application of [Jo, Théorème 6.3.3) and 4)] shows that the intersection

\begin{aligned}X^{(2)} : f(λ, 1; A, B) & = 0, \\ u_{12} + v_{12}b_{11} + w_{12}b_{12} & = 0 \end{aligned}

is geometrically integral for almost all triples (u_{12}, v_{12}, w_{12}) ∈ 𝔸^{3} in the sense that the exceptional set is contained in a Zariski closed subset of 𝔸^{3}. By homogeneity, the same holds after normalizing w_{12} = −1, say. Hence, there are integers u_{12} and v_{12} such that X^{(2)} is geometrically integral. We now apply the same argument to to find integers u_{13} and v_{13} such that the intersection

X^{(3)} = X^{(2)} ∩ {b_{13} = u_{13} + v_{13}b_{11}}

is geometrically integral. We repeat this process in total 14 times to obtain tuples (u_{ij})_{1≤i≤j≤5,(i,j)≠(1,1)}, (v_{ij})_{1≤i≤j≤5,(i,j)≠(1,1)} ∈ ℤ^{14} such that the affine curve given by

det (λA + (\begin{matrix}b_{11} & u_{12} + v_{12}b_{11} & … & u_{15} + v_{15}b_{11} \\ u_{12} + v_{12}b_{11} & u_{22} + v_{22}b_{11} & … & u_{25} + v_{25}b_{11} \\ ⋮ & ... & & ⋮ \\ u_{15} + v_{15}b_{11} & … & … & u_{55} + v_{55}b_{11} \end{matrix})) = 0

is geometrically integral. (Note that the maps ϕ_{1j} are all dominant since the polynomial f(λ, 1; A, B) contains the term det(A)λ^{5} and one can therefore always solve for λ.) Hence, h_{A}(λ, b_{11}, u) is absolutely irreducible as a polynomial in λ and b_{11}. An application of Proposition 3.1 provides us with a Zariski closed subset Z_{A} ⊊ 𝔸_{ℚ}^{14} with the desired property. □

The results so far are enough to deal with the case of Brauer classes that can occur due to f(λ, μ; A, B) having a rational root in λ, μ. We now need to provide similar irreducibility results for a resolvent that detects when the polynomial f(λ, μ; A, B) splits off a factor of degree two. Recall that we have defined f(λ, μ; A, B) = det(λA + μB). We can write f(λ, μ) in the form

f(λ, μ) = ∑_{l=0}^{5} p_{l}(A, B)μ^{5−l}λ^{l},

with p_{l}(A, B) a polynomial of degree l in the coefficients of the matrix A and of degree 5 − l in the coefficients of the matrix B. Moreover, p_{0}(A, B) = det(B) and p_{5}(A, B) = det(A). Assume that det(A) ≠ 0 and that A and B are fixed. We rewrite f(λ, μ) as

f(λ, μ) = det(A) ∑_{l=0}^{5} \frac{p_{l}(A, B)}{det(A)}μ^{5−l}λ^{l}.  (3.2)

Over some algebraic closure of ℚ, we then have

f(λ, μ) = det(A) ∏_{i=1}^{5}(λ + α_{i}μ).  (3.3)

We now define the resolvent

Φ(z; A, B) := det(A)^{4} ∏_{1≤i<j≤5}(z + α_{i} + α_{j}).

Lemma 3.7. a) The resolvent Φ(z; A, B) has the form

Φ(z; A, B) = ∑_{k=0}^{10} q_{k}(A, B)z^{k},

with q_{k}(A, B) ∈ ℤ[A, B] polynomials in the coefficients of the two matrices A, B.

b) The leading coefficient satisfies q_{10}(A, B) = det(A)^{4} and, for 0 ≤ k ≤ 10, the polynomial q_{k}(A, B) is homogeneous of degree 10 − k in the coefficients of B and homogeneous of degree 10 + k in the coefficients of A.

c) If f(λ, μ; A, B) splits off a factor of degree two over the rationals then Φ(z; A, B) has a rational root z ∈ ℚ.

Proof. Using the notation above, we can first write

Φ(z; A, B) = det(A)^{4} ∑_{k=0}^{10} q̃_{k}(α)z^{k},

with q̃_{k}(α) polynomials in the roots α_{i}, for 1 ≤ i ≤ 5. Note that q̃_{k}(α) are symmetric polynomials in the variables α_{i}. Hence, they can be expressed in terms of the following elementary symmetric polynomials

\begin{matrix}E_{1} := ∑_{i=1}^{5} α_{i}, E_{2} := ∑_{i<j} α_{i}α_{j}, E_{3} := ∑_{i<j<k} α_{i}α_{j}α_{k}, \\ E_{4} := ∑_{i<j<k<l} α_{i}α_{j}α_{k}α_{l}, E_{5} := ∏_{i=1}^{5} α_{i}. \end{matrix}

We compare equation (3.2) with equation (3.3) and obtain

det(A) ∑_{l=0}^{5} \frac{p_{l}(A, B)}{det(A)}μ^{5−l}λ^{l} = det(A) ∏_{i=1}^{5}(λ + α_{i}μ).

Hence

E_{i} = \frac{p_{5−i}(A, B)}{det(A)}, for 1 ≤ i ≤ 5.

By a short calculation in magma, we can express each of the polynomials q̃_{k}(α) in the elementary symmetric polynomials E_{i}, for 1 ≤ i ≤ 5. We find that they are of total degree at most four in this representation. Consequently, all the

q_{k}(A, B) := det(A)^{4}q̃_{k}(α), 0 ≤ k ≤ 10,

are polynomials in ℤ[A, B], where q_{k}(A, B) is homogeneous of degree 10 − k in the coefficients of B and homogeneous of degree 10 + k in the coefficients of A. This completes part a) and b) of the proof of the lemma. Part c) follows directly from the construction of the resolvent polynomial Φ(z; A, B). □

We next need an statement analogous to Lemma 3.6, for the resolvent Φ(z; A, B). The strategy of proof is the same as that for the polynomial f(λ, μ) itself.

Proposition 3.8. Let A ∈ M_{5×5}(ℚ) be a symmetric matrix with det(A) ≠ 0. Then there exist a tuple t′ := (t_{ij})_{1≤i≤j≤5,(i,j)≠(1,1)} ∈ ℤ^{14} and a Zariski closed subset V_{A} ⊊ 𝔸_{ℚ}^{14} with the following property.

The expression Φ(z; A, (b_{11}, c′ + b_{11}t′)) is absolutely irreducible as a polynomial in the variables z and b_{11} if c′ := (c_{ij})_{1≤i≤j≤5,(i,j)≠(1,1)} ∈ 𝔸^{14}(ℚ̄) \ V_{A}(ℚ̄).

Proof. First, a short calculation using magma shows that the polynomial

Φ(z; Id_{5}, (b_{11}, c′ + b_{11}t′))

is absolutely irreducible in z, b_{11}, if we set t′ := 0 and c_{ij} := 1, for |i − j| ≤ 1, and c_{ij} := 0, otherwise. Indeed, in this case, the homogenization of the polynomial Φ(z; Id_{5}, (b_{11}, c′ + b_{11}t′)) defines an irreducible curve of genus 3 with two regular rational points of height less than 100.

Since Φ(z; Id_{5}, (b_{11}, c′ + b_{11}t′)) is absolutely irreducible of degree 10 in z and b_{11} for these special choices of c′ and t′, we deduce that Φ(z; Id_{5}, B) is absolutely irreducible in the 16 variables z, (b_{ij})_{1≤i≤j≤5}.

Now let A ∈ M_{5×5}(ℚ) be an arbitrary symmetric matrix with det(A) ≠ 0. Over ℚ̄, we can diagonalize A with an invertible matrix C such that

C^{t}AC = Id_{5}.

Then we have

f(λ, μ; A, B) = det(C)^{−2} det(λC^{t}AC + μC^{t}BC)
= det(C)^{−2}f(λ, μ; Id_{5}, C^{t}BC).

For the corresponding resolvents, we hence get the relation

Φ(z; A, B) = det(C)^{−8}Φ(z; Id_{5}, C^{t}BC).

We conclude that Φ(z; A, B) is an absolutely irreducible polynomial in the 16 variables z, (b_{ij})_{1≤i≤j≤5}. The rest of the proof proceeds in exactly the same way as the proof of Lemma 3.6. Note that, for A fixed with det(A) ≠ 0, the polynomial Φ(z; A, B) always has degree 10. □

4. Proof of Theorem 1.1

In this section, we prove Theorem 1.1. Let A ∈ M_{5×5}(ℤ) be fixed and assume that det(A) ≠ 0. Let S^{(1)}(P) be the number of symmetric matrices B ∈ M_{5×5}(ℤ) with |b_{ij}| ≤ P, for all 1 ≤ i ≤ j ≤ 5, such that f(λ, μ; A, B) has a rational root in λ, μ. Similarly, let S^{(2)}(P) be the number of such matrices B such that Φ(z; A, B) has a rational root in z.

By Theorem 2.2, and the construction of the resolvent polynomial Φ(z; A, B), we can bound N_{2}(P) by

N_{2}(P) ≪ S^{(1)}(P) + S^{(2)}(P).  (4.1)

If f(λ, μ; A, B) has a rational root in λ, μ, then, by homogeneity, it also has a solution λ, μ ∈ ℤ^{2} with gcd(λ, μ) = 1. Since the coefficient of λ^{5} in f(λ, μ) is equal to det(A), we deduce that μ| det(A). For a divisor d| det(A), we define S_{d}^{(1)}(P) to be the number of symmetric matrices B ∈ M_{5×5}(ℤ) with coefficients bounded by P such that f(λ, d; A, B) has an integer root in λ. Then we can bound

S^{(1)}(P) ≤ ∑_{d| det(A)} S_{d}^{(1)}(P),  (4.2)

where the sum over d runs over all positive and negative divisors of det(A).

Similarly, if Φ(z; A, B) has a rational root, say z = \frac{r}{s}, with r, s ∈ ℤ^{2} and gcd(r, s) = 1 then s| det(A)^{4}. For any divisor d| det(A)^{4}, we let S_{d}^{(2)}(P) be the number of symmetric matrices B ∈ M_{5×5}(ℤ) with coefficients bounded by P such that d^{10}Φ(d^{−1}z; A, B) (which is again a polynomial with integer coefficients) has an integer root in z. Hence, we have the upper bound

S^{(2)}(P) ≤ ∑_{d| det(A)^{4}} S_{d}^{(2)}(P),

where again the sum is over all positive and negative divisors d of det(A)^{4}.

With d| det(A) fixed, we now proceed to bound S_{d}^{(1)}(P). We recall the notation

f(λ, μ; A, B) = ∑_{l=0}^{5} p_{l}(A, B)μ^{5−l}λ^{l},

with p_{l}(A, B) polynomials of degree 5−l in the coefficients of B. If B is a matrix with coefficients bounded by P then p_{l}(A, B) ≪ P^{5−l}. As soon as f(λ, d; A, B) has a complex root λ_{0}, then we deduce that there is a positive constant C_{0} ≥ 1 such that |λ_{0}| ≤ C_{0}P, by Theorem (27,3) in [Ma] (see also Lemma 1 in [Di]). Let v′ and Z_{A} be as in Lemma 3.6. Then f(λ, 1; A, (b_{11}, c′+b_{11}v′)) is absolutely irreducible in λ, b_{11} if c′ ∉ Z_{A}. By a linear change of variables, we deduce that

f(λ, d; A, (b_{11}, c′ + b_{11}v′)) = f(λ, 1; A, (db_{11}, dc′ + db_{11}v′))

is absolutely irreducible in the two variables λ, b_{11} if c′ ∉ d^{−1}Z_{A}. Let R_{d}^{(1)}(P) be the number of vectors (b_{11}, c′) ∈ (ℤ ∩ [−P, P])^{15} such that the polynomial f(λ, d; A, (b_{11}, c′ + b_{11}v′)) has an integer root in λ. Then there is a positive constant C_{1} such that

S_{d}^{(1)}(P) ≤ R_{d}^{(1)}(C_{1}P).  (4.3)

For c′ ∈ ℤ^{14}, we define

T_{d}^{(1)}(c′; P) := ♯{(b_{11}, λ) ∈ (ℤ ∩ [−P, P])^{2} : f(λ, d; A, (b_{11}, c′ + b_{11}v′)) = 0}.

Let

E_{d}^{(1)}(P) := ♯{c′ ∈ (ℤ ∩ [−P, P])^{14} : c′ ∈ d^{−1}Z_{A}}.

Then we can bound R_{d}^{(1)}(C_{1}P) by

R_{d}^{(1)}(C_{1}P) ≪ ∑_{\begin{matrix}c′∈(ℤ∩[−C_{1}P,C_{1}P])^{14} \\ c′∉d^{−1}Z_{A} \end{matrix}} T_{d}^{(1)}(c′; C_{2}P) + P E_{d}^{(1)}(C_{1}P),

for a sufficiently large constant C_{2}. An application of Theorem 2.3 gives the bound

T_{d}^{(1)}(c′; C_{2}P) ≪_{ε} P^{1/5+ε},

with an implied constant that is independent of c′. On the other hand, as Z_{A} ⊊ 𝔸_{ℚ}^{14} is Zariski closed, we have

E_{d}^{(1)}(C_{2}P) ≪ P^{13}.

Together these estimates lead to the bound

R_{d}^{(1)}(C_{1}P) ≪_{ε} P^{14+1/5+ε},

for any ε > 0. Here, the implied constant may depend on d and C_{1}, but is independent of P. Since A is considered fixed, we can trivially perform the summation in equation (4.2), and together with equation (4.3), we obtain the bound

S^{(1)}(P) ≪_{ε} P^{14+1/5+ε}.

The bound for S^{(2)}(P) is obtained in the same way as the bound for S^{(1)}(P), with Proposition 3.8 in place of Proposition 3.6. In fact, since Φ(z; A, B) is typically irreducible of degree 10, one obtains the bound

S^{(2)}(P) ≪_{ε} P^{14+1/10+ε},

for any ε > 0.

Remark 4.1. Let A be as above. As the polynomials f(λ, 1; A, B) and Φ(z; A, B) are irreducible in ℚ̄(b_{ij})_{1≤i≤j≤5}[λ] and ℚ̄(b_{ij})_{1≤i≤j≤5}[z], respectively, the polynomial interpretation of thin sets as in section 9.1 of [Se] shows that the matrices B counted by N_{2}(P) indeed form a thin subset.

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