EVERYWHERE LOCAL SOLUBILITY FOR HYPERSURFACES IN PRODUCTS OF PROJECTIVE SPACES

TOM FISHER, WEI HO, AND JENNIFER PARK

Abstract. We prove that a positive proportion of hypersurfaces in products of projective spaces over ℚ are everywhere locally soluble, for almost all multidegrees and dimensions, as a generalization of a theorem of Poonen and Voloch [PV04]. We also study the specific case of genus 1 curves in ℙ^{1} × ℙ^{1} defined over ℚ, represented as bidegree (2, 2)-forms, and show that the proportion of everywhere locally soluble such curves is approximately 87.4%. As in the case of plane cubics [BCF16], the proportion of these curves in ℙ^{1} × ℙ^{1} soluble over ℚ_{p} is a rational function of p for each finite prime p. Finally, we include some experimental data on the Hasse principle for these curves.

1. Introduction

Let V be a variety defined over ℚ. The study of rational points on V often involves determining the local points V (ℚ_{ν}) for completions ℚ_{ν} of ℚ. We say that V is (globally) soluble if the set V (ℚ) of rational points is nonempty, and V is everywhere locally soluble if the set V (ℚ_{ν}) is nonempty for all places ν ≤ ∞ of ℚ.

Poonen and Voloch [PV04] show that a positive proportion of all hypersurfaces in ℙ^{n} of fixed degree d are everywhere locally soluble for n, d ≥ 2 and (n, d) ≠ (2, 2). In particular, they prove that this proportion (as a limit) is exactly the product c = ∏_{ν} c_{ν} > 0 of local factors c_{ν}, where c_{ν} is the proportion of hypersurfaces that have a ℚ_{ν}-point. In [BBL16], Bright, Browning, and Loughran generalized this theorem to other families of varieties. Poonen and Voloch also conjecture that the proportion of globally soluble hypersurfaces is c when 2 ≤ d ≤ n, and is 0 when d > n + 1. This conjecture implies that the Hasse principle is true for 100% and for 0% of the everywhere locally soluble hypersurfaces in the two cases, respectively. Browning, Le Boudec, and Sawin [BLS19] have recently proved this conjecture for n ≥ d + 2, i.e., that the Hasse principle is true 100% of the time for Fano hypersurfaces of degree d in ℙ^{n} with n ≥ d + 2.

However, Poonen and Voloch [PV04] do not make any conjectures about local versus global solubility in the boundary (Calabi-Yau) case where d = n + 1; this case is perhaps the most interesting and the most mysterious. The simplest of these cases, that of plane cubics (d = 3 and n = 2), has been studied by Bhargava, Cremona, and the first author [BCF16], in which the authors explicitly compute the proportion c of everywhere locally soluble plane cubics; the same authors [BCF] also study the proportion of everywhere locally soluble hyperelliptic curves. In addition, Bhargava [Bha14] shows that a positive proportion of plane cubics fail the Hasse principle and a positive proportion satisfy the Hasse principle.

In this paper, we are interested in answering the analogous questions and computing the explicit proportion of everywhere local soluble curves defined in products of projective spaces. We also study a specific family of genus one curves over ℚ (analogous to the plane cubics studied in [BCF16]) and determine the proportion of such curves that are everywhere locally soluble. In forthcoming work, Bhargava and the second author show that the Hasse principle fails for a positive proportion of curves in this family [BH19a]. In this paper, we include some data on the success and failure of the Hasse principle for randomly selected sets of these curves with bounded coefficients.

Date: October 24, 2019.

In Section 2, we begin by proving the analogue of the theorem of Poonen and Voloch for hypersurfaces in products of projective spaces:

Theorem 1.1. Let d_{1}, …, d_{k}, n_{1}, …, n_{k} be positive integers. The proportion of multidegree (d_{1}, …, d_{k}) hypersurfaces over ℚ in ℙ^{n_{1}} × ⋯ × ℙ^{n_{k}} that are everywhere locally soluble tends to a real number c > 0, where c is the product ∏_{ν} c_{ν} over all places ν and c_{ν} is the proportion of such multidegree homogeneous polynomials with a nontrivial zero over ℚ_{ν}, as long as one of the following holds:

(i) k = 1 and n_{1}, d_{1} ≥ 2 but not (n_{1}, d_{1}) = (2, 2); (ii) k = 2 and if n_{1} = n_{2} = 1, then d_{1}, d_{2} ≥ 2; (iii) k ≥ 3.

We also predict the success or failure of the Hasse principle for families of hypersurfaces in products of projective spaces in many cases, depending on the multidegree of the hypersurfaces:

Conjecture 1.2. For any positive integers d_{i}, n_{i} for i ∈ {1, …, k}, consider the family of multidegree (d_{1}, …, d_{k})-hypersurfaces in ℙ^{n_{1}} × ⋯ × ℙ^{n_{k}} over ℚ. Then

(i) If d_{i} > n_{i} + 1 for all i = 1, …, k, then 100% of everywhere locally soluble (d_{1}, …, d_{k})- hypersurfaces fail the Hasse principle.

(ii) If d_{i} < n_{i} + 1 for some i = 1, …, k, except if (k, n_{1}, n_{2}, d_{1}, d_{2}) = (2, 1, 2, ≥ 2, 2) or (2, 2, 1, 2, ≥ 2), then 100% of everywhere locally soluble (d_{1}, …, d_{k})-hypersurfaces satisfy the Hasse principle. With these conditions, we have that the proportion of soluble hypersurfaces equals the product of local factors c = ∏_{ν} c_{ν}.

Note that the numerical conditions in Conjecture 1.2(ii) are significantly more general than just requiring the hypersurface to be Fano, which is equivalent to having d_{i} < n_{i} + 1 for all i (not just some i). Still, as in the original Poonen–Voloch heuristic, there are some remaining boundary cases, namely when d_{i} ≥ n_{i} + 1 for all i with equality for at least one i. In this paper, we compute the explicit probability of everywhere local solubility for the simplest such case with k > 1, namely the family of bidegree (2, 2) curves in ℙ^{1} × ℙ^{1}. Explicitly, these are the curves defined by the bihomogeneous bidegree (2, 2)-polynomials of the form

F (X_{0}, X_{1}, Y_{0}, Y_{1}) = a_{00}X_{0}^{2}Y_{0}^{2} + a_{01}X_{0}^{2}Y_{0}Y_{1} + a_{02}X_{0}^{2}Y_{1}^{2}
+ a_{10}X_{0}X_{1}Y_{0}^{2} + a_{11}X_{0}X_{1}Y_{0}Y_{1} + a_{12}X_{0}X_{1}Y_{1}^{2}  (1.3)
+ a_{20}X_{1}^{2}Y_{0}^{2} + a_{21}X_{1}^{2}Y_{0}Y_{1} + a_{22}X_{1}^{2}Y_{1}^{2}

where a_{ij} ∈ ℚ for i, j ∈ {0, 1, 2}. Smooth curves of this form have genus one. We show:

Theorem 1.4. For a finite prime p, the probability a bidegree (2, 2)-form with coefficients in ℤ_{p} is soluble over ℚ_{p} is

ρ(p) = 1 − \frac{p(p − 1)(p^{2} − 1)f(p)}{8(p^{8} − 1)(p^{9} − 1)}

where f(p) = 4p^{11} − 4p^{10} + 4p^{9} − p^{8} + 5p^{7} − 2p^{6} + 5p^{5} − p^{4} + 2p^{3} − 2p^{2} + 6p − 2.

The probability a (2, 2)-form is soluble over the reals (when we choose the coefficients uniformly at random in [−B, …, B] with B large) appears to be about 96.46%, based on a Monte Carlo simulation. Note that the probability of solubility over ℝ is greater than 7/8, since for insolubility a_{00}, a_{02}, a_{20}, a_{22} must all have the same sign.

Theorem 1.4 gives ∏_{p<∞} ρ(p) ≈ 0.90592. Combining this with the estimate of the previous paragraph, we have

∏_{p≤∞} ρ(p) ≈ 0.8739,

thereby giving the following theorem:

Theorem 1.5. The probability that a random (2, 2)-form with coefficients in ℚ is p-locally soluble for all p < ∞ is approximately 90.592%; assuming the truth of the Monte Carlo experiment, the probability that a random (2, 2)-form with coefficients in ℚ is everywhere locally soluble is approximately 87.39%.

A heuristic similar to that in [Bha14] suggests that of the (2, 2)-forms that are everywhere locally soluble, exactly 1/4 should be globally soluble. In Section 6, we report on an experiment to test this prediction numerically.

Acknowledgments. Much of this work was completed during a workshop organized by Alexander Betts, Tim Dokchitser, Vladimir Dokchitser and Celine Maistret, a trimester organized by David Harari, Emmanuel Peyre, and Alexei Skorobogatov, and a workshop organized by Michael Stoll; we thank the organizers as well as Baskerville Hall, Institut Henri Poincaré, and Franken-Akademie Schloss Schney, respectively, for their hospitality during those periods. We also thank Bhargav Bhatt, John Cremona, David Harari, Max Lieblich, Daniel Loughran, and Bjorn Poonen for helpful conversations and comments.

WH was partially supported by NSF grants DMS-1701437 and DMS-1844763 and the Sloan Foundation. JP was partially supported by NSF grant DMS-1902199.

2. Generalization of Poonen–Voloch’s theorem and conjecture

In this section, we generalize the main result and conjectures of [PV04] to hypersurfaces in products of projective spaces. The main idea of the proof of the theorem is identical to that of

[PV04] but relies on a more general combinatorial inequality. We work over the field ℚ for this section.

Fix an integer k ≥ 1 and positive integers n_{i}, d_{i} for 1 ≤ i ≤ k. We consider multidegree (d_{1}, ⋯, d_{k}) hypersurfaces in the product space 𝒫 := ℙ^{n_{1}} × ⋯ × ℙ^{n_{k}}. In particular, let ℤ[{x_{ij}}]_{d} denote the set of multihomogeneous polynomials in ℤ[{x_{ij} : 1 ≤ i ≤ k, 0 ≤ j ≤ n_{i}}] of multidegree d = (d_{1}, …, d_{k}). There are m_{i} := \binom{n_{i}+d_{i}}{d_{i}} monomials of degree d_{i} in (n_{i} + 1) variables, so the total number of monomials of multidegree d in ℤ[{x_{ij}}]_{d} is m = ∏_{i=1}^{k} m_{i}.

Define the height h(f) of f ∈ ℤ[{x_{ij}}]_{d} to be the maximum of the absolute values of the coefficients of f. Let M_{ℚ} be the set of places of ℚ. We define the following counts for H > 0 and ν ∈ M_{ℚ}:

N_{tot}(H) := #{f ∈ ℤ[{x_{ij}}]_{d} : h(f) ≤ H} = (2⌊H⌋ + 1)^{m}
N(H) := #{f ∈ ℤ[{x_{ij}}]_{d} : h(f) ≤ H and ∃ x = {x_{ij}} ∈ ∏_{i=1}^{k} ℤ^{n_{i}+1} \ {0} with f(x) = 0}

N_{ν}(H) := #{f ∈ ℤ[{x_{ij}}]_{d} : h(f) ≤ H and ∃ x = {x_{ij}} ∈ ∏_{i=1}^{k} ℚ_{ν}^{n_{i}+1} \ {0} with f(x) = 0}

N_{loc}(H) := #{f ∈ ℤ[{x_{ij}}]_{d} : h(f) ≤ H and ∀ ν ∈ M_{ℚ}, ∃ x ∈ ∏_{i=1}^{k} ℚ_{ν}^{n_{i}+1} \ {0} with f(x) = 0}.

In other words, N_{tot}(H) is the total number of multidegree d polynomials in 𝒫 of height at most H, and N(H), N_{ν}(H), and N_{loc}(H) are the number of such polynomials that are globally soluble, soluble over ℚ_{ν}, and everywhere locally soluble, respectively. The limits

lim_{H→∞} \frac{N(H)}{N_{tot}(H)} and lim_{H→∞} \frac{N_{loc}(H)}{N_{tot}(H)},

if they exist, will be called the proportion of globally soluble and everywhere locally soluble multidegree d hypersurfaces, respectively. Note that for any place ν, the local proportion lim_{H→∞} N_{ν}(H)/N_{tot}(H) exists. Indeed, as explained in Remark 2.3 of [PV04], after normalizing the Haar measure (or Lebesgue measure for ν = ∞) on the space ℤ_{ν}^{m} of the multihomogeneous polynomials of degree d in the variables {x_{ij} : 1 ≤ i ≤ k, 0 ≤ j ≤ n_{i}}, we see that this local proportion is the measure of the ν-adically closed subset of ℤ_{ν}^{m} corresponding to the multihomogeneous polynomials with a nontrivial zero over ℚ_{ν}. We restate Theorem 1.1 using this language:

Theorem 2.1. Let c_{ν} = lim_{H→∞} N_{ν}(H)/N_{tot}(H). As H → ∞, the proportion N_{loc}(H)/N_{tot}(H) tends to c := ∏_{ν} c_{ν} > 0, if one of the following conditions holds:

(i) k = 1 and n_{1}, d_{1} ≥ 2 but not (n_{1}, d_{1}) = (2, 2); (ii) k = 2, and if n_{1} = n_{2} = 1, then d_{1}, d_{2} ≥ 2; (iii) k ≥ 3.

Proof. If f is absolutely irreducible modulo p, then for sufficiently large p, the Lang-Weil estimate guarantees a smooth point on the hypersurface f = 0 modulo p, which may be lifted to a ℚ_{p}-point by Hensel’s lemma. By Lemmas 20 and 21 of [PS99], it suffices to show that the space of reducible polynomials is of codimension at least 2 in the space of all polynomials in ℤ[{x_{ij}}]_{d}. This follows from Lemma 2.2 below, which shows that the product of the projective spaces of multidegree (r_{1}, ⋯, r_{k}) polynomials and of multidegree (d_{1} − r_{1}, ⋯, d_{k} − r_{k}) polynomials has dimension at most m − 3, where m − 1 is the dimension of the projective space of multidegree (d_{1}, ⋯, d_{k}) polynomials. □

Lemma 2.2. Fix a positive integer k. Given positive integers n_{i}, d_{i}, r_{i} for 1 ≤ i ≤ k such that 0 ≤ r_{i} ≤ d_{i} for all i but 0 < ∑_{i} r_{i} < ∑_{i} d_{i}, we have

∏_{i=1}^{k} \binom{n_{i} + r_{i}}{n_{i}} + ∏_{i=1}^{k} \binom{n_{i} + d_{i} − r_{i}}{n_{i}} < ∏_{i=1}^{k} \binom{n_{i} + d_{i}}{n_{i}}  (2.3)

if one of the following conditions holds:

(i) k = 1 and n_{1}, d_{1} ≥ 2 but not (n_{1}, d_{1}) = (2, 2); (ii) k = 2, and if n_{1} = n_{2} = 1, then d_{1}, d_{2} ≥ 2; (iii) k ≥ 3.

Proof. We define

S = {(A_{1}, …, A_{k}) : A_{i} ⊂ {1, 2, …, n_{i} + d_{i}} with |A_{i}| = n_{i}}

and subsets

S_{1} = {(A_{1}, …, A_{k}) ∈ S : A_{i} ∩ {1, …, r_{i}} = ∅ for all 1 ≤ i ≤ k},
S_{2} = {(A_{1}, …, A_{k}) ∈ S : A_{i} ∩ {r_{i} + 1, …, d_{i}} = ∅ for all 1 ≤ i ≤ k}.

Then |S_{1} ∩ S_{2}| = 1. To prove |S_{1}| + |S_{2}| < |S| it suffices to show that |S \ (S_{1} ∪ S_{2})| ≥ 2. (i) If k = 1 then by choosing A_{1} with

|A_{1} ∩ {1, …, r_{1}}| = 1 and |A_{1} ∩ {r_{1} + 1, …, d_{1}}| = 1

we have |S \ (S_{1} ∪ S_{2})| ≥ r_{1}(d_{1} − r_{1})\binom{n_{1}}{2}, which is at least 2 under the stated hypotheses. (ii) If k = 2 then by choosing (A_{1}, A_{2}) with

|A_{1} ∩ {1, …, r_{1}}| = δ,  |A_{1} ∩ {r_{1} + 1, …, d_{1}}| = 1 − δ,
|A_{2} ∩ {1, …, r_{2}}| = 1 − δ,  |A_{2} ∩ {r_{2} + 1, …, d_{2}}| = δ,

for δ = 0, 1, we have |S \ (S_{1} ∪ S_{2})| ≥ (r_{1}(d_{2} − r_{2}) + r_{2}(d_{1} − r_{1}))n_{1}n_{2}, which is at least 2 under the stated hypotheses.

(iii) If k ≥ 3 then |S \ (S_{1} ∪ S_{2})| ≥ ∑_{i≠j} r_{i}(d_{j} − r_{j}) ≥ 2. □

We also discuss the analogue of the conjecture of Poonen–Voloch in our setting:

Conjecture 2.4. For any positive integers d_{i}, n_{i} for i ∈ {1, …, k}, consider the family of multidegree (d_{1}, …, d_{k})-hypersurfaces in ℙ^{n_{1}} × ⋯ × ℙ^{n_{k}} over ℚ. Then

(i) If d_{i} > n_{i} + 1 for all i = 1, …, k, then 100% of everywhere locally soluble (d_{1}, …, d_{k})- hypersurfaces fail the Hasse principle.

(ii) If d_{i} < n_{i} + 1 for some i = 1, …, k, except if (k, n_{1}, n_{2}, d_{1}, d_{2}) = (2, 1, 2, ≥ 2, 2) or (2, 2, 1, 2, ≥ 2), then 100% of everywhere locally soluble (d_{1}, …, d_{k})-hypersurfaces satisfy the Hasse principle.

Combining Conjecture 2.4(ii) with Theorem 2.1 implies that the proportion of globally soluble hypersurfaces in 𝒫 is a product of local densities for many cases, i.e., the last part of Conjecture 1.2: If d_{i} < n_{i} + 1 for some i = 1, …, k, except for the cases (2, 1, 2, ≥ 2, 2), or (2, 2, 1, 2, ≥ 2), we have that

lim_{H→∞} \frac{N(H)}{N_{tot}(H)} = c = ∏_{ν} c_{ν}.  (2.5)

Remark 2.6. If d_{i} < n_{i} + 1 for some i, and yet the hypotheses of Theorem 2.1 are not satisfied then either some d_{i} = 1 or (k, n_{1}, d_{1}) = (1, 2, 2). In the first case both sides of (2.5) are 1. In the second case concerning plane conics, see, e.g., Theorem 2 of [BCF^{+}16], which states that c_{p} = 1 − \frac{p}{2(p+1)^{2}} for p a prime; one concludes immediately that both sides of (2.5) are 0 in this case.

Motivation for Conjecture 2.4(i). Fix a point a = (a_{1}, …, a_{k}) ∈ ℤ^{n_{1}+1} × ⋯ × ℤ^{n_{k}+1} with each a_{i} ≠ 0 and having coprime coordinates, and consider the set of multidegree (d_{1}, …, d_{k})-polynomials vanishing on a. The set of these polynomials form a hyperplane in ℤ^{m}, and to count the multidegree (d_{1}, …, d_{k}) polynomials up to height H vanishing on a is to count the set of integral points contained in this hyperplane, whose coordinates are bounded by H. As in [PV04], the number of integral points of height at most H in this hyperplane is given by

\frac{c(a)H^{m−1}}{ϕ(a)} + O(H^{m−2}),

where ϕ(a) denotes the covolume of the lattice of integer points on the hyperplane of polynomials vanishing on a, and c(a) denotes the (m − 1)-dimensional volume of the part of the hyperplane inside [−1, 1]^{m}.

Ignoring the error term, we get that

N(H) ≤ H^{m−1} ∑_{a} \frac{c(a)}{ϕ(a)},

where the sum ranges over a ∈ ℤ^{n_{1}+1} × ⋯ × ℤ^{n_{k}+1}, excluding the zero vectors in each of the components. Since c(a) is bounded by definition, it remains to understand the convergence of ∑_{a} \frac{1}{ϕ(a)}. Lemma 3.1 of [PV04] shows that ϕ(a) equals the norm of the vector b formed by the monomials of degree (d_{1}, …, d_{k}) in the coordinates of (a_{1}, …, a_{k}) (the coprime condition in that lemma is not needed). Let ϕ_{i}(a_{i}) denote the norm of the vector formed by plugging in the coordinates of a_{i} into each of the degree d_{i} monomials in n_{i} + 1 variables. Because ϕ(a) = ∏_{i=1}^{k} ϕ_{i}(a_{i}), we have

∑_{a∈(ℤ^{n_{1}+1}\\{0\})×···×(ℤ^{n_{k}+1}\\{0\})} \frac{1}{ϕ(a)} = ( ∑_{a_{1}∈ℤ^{n_{1}+1}\\{0\}} \frac{1}{ϕ_{1}(a_{1})}) · · · · · ( ∑_{a_{k}∈ℤ^{n_{k}+1}\\{0\}} \frac{1}{ϕ_{k}(a_{k})}) . (2.7)

We claim that each of the sums ∑_{a_{i}∈ℤ^{n_{i}+1}\\{0\}} \frac{1}{ϕ_{i}(a_{i})} converges.

We decompose ℤ^{n_{i}+1} \ {0} = T_{n_{i}+1} ∪ T_{n_{i}} ∪ ⋯ ∪ T_{1}, where T_{j} consists of the vectors in ℤ^{n_{i}+1} with exactly j nonzero coordinates. Let a_{i} = (a_{i0}, …, a_{in_{i}}). Then by applying the AM-GM inequality to the definition of ϕ_{i}(a_{i}), we have

ϕ_{i}(a_{i}) ≥ m_{i}^{1/2} |∏_{ℓ=0}^{n_{i}} a_{iℓ}|^{\frac{d_{i}}{n_{i}+1}} .  (2.8)

In fact, for a_{i} ∈ T_{j}, the product over ℓ on the right side of (2.8) may be taken to be the product of only the j nonzero a_{iℓ} if we replace m_{i} with m_{ij} := \binom{j+d_{i}}{d_{i}} and n_{i} by j. We thus have

∑_{a∈ℤ^{n_{i}+1}\\{0\}} \frac{1}{ϕ_{i}(a_{i})} = ∑_{j=1}^{n_{i}+1} ∑_{a_{i}∈T_{j}} \frac{1}{ϕ_{i}(a_{i})}

≤ ∑_{j=1}^{n_{i}+1} ∑_{a_{i}∈T_{j}} m_{ij}^{−1/2} | ∏_{ℓ: a_{iℓ}≠0} a_{iℓ}|^{− \frac{d_{i}}{j}} .

The above sum converges if d_{i} > j for all j = 1, …, n_{i} + 1, so it converges if d_{i} > n_{i} + 1. Thus, the original quantity (2.7) converges if d_{i} > n_{i} + 1 for all i = 1, …, k.

This heuristic therefore predicts N(H) ∼ O(H^{m−1}). Combining this with the estimate N_{tot}(H) ∼ (2H)^{m} gives Conjecture 2.4(i).

Motivation for Conjecture 2.4(ii). We may restrict ourselves to only consider polynomials f for which f = 0 defines a smooth geometrically integral hypersurface X in 𝒫 (see also [PV04, Remark 2.1]). We use a well known conjecture of Colliot-Thélène (see, e.g., [CT03]):

Conjecture 2.9 (Colliot-Thélène). Let X be a smooth proper geometrically integral variety over a number field. If X is (geometrically) rationally connected, then the Brauer-Manin obstruction to the Hasse principle for X is the only obstruction.

In particular, we show in Proposition 2.10 that if X is a multidegree (d_{1}, …, d_{k})-hypersurface in 𝒫 = ℙ^{n_{1}} × ⋯ × ℙ^{n_{k}} with d_{i} < n_{i} + 1 for some i, then X is rationally connected. We will show in Proposition 2.11 below that the Brauer-Manin obstruction is vacuous for smooth complete intersections (of dimension at least 3) in 𝒫 over number fields. Thus, conditional on Conjecture 2.9, we see that the Hasse principle is satisfied for these X of dimension ≥ 3.

In almost all cases with d_{i} < n_{i} + 1 for some i where X has dimension 1 or 2, either some d_{i} = 1 (and so X has a rational point) or X is a quadric hypersurface (and so is known to satisfy the Hasse principle). For the case of bidegree (2, d_{2})-surfaces in ℙ^{2} × ℙ^{1} with d_{2} ≥ 2, however, it is possible for the Hasse principle to not be satisfied (e.g., the famous example of Iskovskikh [Isk71] where d_{2} = 4), so we exclude this case. The other nontrivial case is handled in the proof of [PV04, Proposition 3.4], which shows, under Conjecture 2.9, that the Hasse principle holds for a density 1 set of cubic surfaces in ℙ^{2}.

Proposition 2.10. Fix an integer k ≥ 1 and positive integers d_{i}, n_{i} for i ∈ {1, …, k}. Let X be a smooth integral multidegree (d_{1}, …, d_{k})-hypersurface in ℙ^{n_{1}} ×⋯×ℙ^{n_{k}} over ℂ. Then X is rationally connected if and only if d_{i} < n_{i} + 1 for some i.

Proof. One direction is easy: if d_{i} > n_{i} for all i, then H^{0}(X, ω_{X}) is nonzero, hence X is not rationally connected. For the reverse direction, without loss of generality, suppose d_{1} < n_{1} + 1. Then consider the projection π : X → ℙ^{n_{2}} ×⋯×ℙ^{n_{k}} (note that this codomain is a point if k = 1). The fibers of this projection π are degree d_{1} hypersurfaces in ℙ^{n_{1}}; the general ones are smooth and, by assumption, Fano and thus rationally connected [Cam92,KMM92].

We apply a theorem of Graber–Harris–Starr [GHS03], which states that if X → Y is a dominant morphism of complex varieties where the general fiber and Y are both rationally connected, then X is rationally connected as well. Here the projection map π is surjective by a dimension count, and the codomain ℙ^{n_{2}} × ⋯ × ℙ^{n_{k}} is rationally connected, so X itself is rationally connected. □

In the following proposition, the notation Br Y refers to the cohomological Brauer group H_{ét}^{2}(Y, 𝔾_{m}) for a scheme Y.

Proposition 2.11. Let K be a number field. Let X be a smooth complete intersection in a product 𝒫 = ℙ^{n_{1}} × ⋯ × ℙ^{n_{k}} satisfying dim X ≥ 3. Then the natural map Br K → Br X is an isomorphism, hence the Brauer-Manin obstruction for X is vacuous.

Proof. We need only slightly modify the proof in [PV04, Appendix A and B] for the same result where X is a smooth complete intersection in a single projective space ℙ^{n} (see also [Sch16, Proposition 2.6] for a smooth complete intersection in the product of two projective spaces). We summarize the argument here. We use the low degree exact sequence from the Leray spectral sequence

0 → Pic X → (Pic X\overline{K})^{G_{K}} → Br K → ker(Br X → Br X\overline{K}) → H^{1}(K, Pic X\overline{K})  (2.12)

where G_{K} = Gal(\overline{K}/K). We need to check that Pic X → (Pic X\overline{K})^{G_{K}} is an isomorphism, H^{1}(K, Pic X\overline{K}) = 0, and Br X\overline{K} = 0.

The restriction map Pic 𝒫\overline{K} → Pic X\overline{K} is an isomorphism since X is a smooth complete intersections of dimension at least 3 [Har70, Corollary 3.3]. Then since Pic 𝒫 → Pic 𝒫\overline{K} is an isomorphism, we find that the injections

Pic X ↪ (Pic X\overline{K})^{G_{K}} ↪ Pic X\overline{K}

are isomorphisms and H^{1}(K, Pic X\overline{K}) = 0.

To show that Br X\overline{K} = 0 here, we use the Kummer sequence to obtain, for any prime ℓ,

\xymatrix{0 \ar[r] & (Pic 𝒫\overline{K})/ℓ \ar[r] \ar[d] & H^{2}(𝒫\overline{K}, ℤ/ℓℤ) \ar[d]^{ψ} \\ 0 \ar[r] & (Pic X\overline{K})/ℓ \ar[r] & H^{2}(X\overline{K}, ℤ/ℓℤ) \ar[r] & (Br X\overline{K})[ℓ] \ar[r] & 0.}

The top horizontal injection is in fact an isomorphism, since both groups are rank k over ℤ/ℓℤ (using the Kunneth formula to compute H^{2}(𝒫, ℤ/ℓℤ)). By a version of Weak Lefschetz (see, e.g., [PV04, Corollary B.5] with V = 𝒫), the vertical map ψ is an isomorphism, so we have (Br X\overline{K})[ℓ] = 0 for all ℓ. Since Br X\overline{K} is torsion, it is in fact 0.

Thus, from (2.12), the map Br K → Br X is an isomorphism, so no elements of Br X obstruct rational points on X. □

3. Counting polynomials over finite fields

Our strategy for proving Theorems 1.4 and 1.5 can be viewed as an extension of Hensel’s lemma. If the reduction of the (2, 2)-form mod p has a smooth 𝔽_{p}-point, then Hensel’s lemma implies that this can be lifted to a ℚ_{p}-point; conversely, if the reduction of the (2, 2)-form has no 𝔽_{p}-points, then clearly it has no ℚ_{p}-points. The p-adic solubility of many (2, 2)-forms can be determined in this way, so it is crucial to understand (2, 2)-forms over finite fields. We work over a finite field 𝔽_{q}, where q is a power of a prime p, for this section, since it is no extra work to do so, though we will only need these results over 𝔽_{p} for the later sections.

3.1. Preliminaries. Many (2, 2)-forms modulo p contain binary quadratic forms as one of its factors, so we collect some relevant calculations here. These are easy to check.

Lemma 3.1. Of the q^{2} monic quadratic polynomials f ∈ 𝔽_{q}[X],

• q(q − 1)/2 have two distinct roots in 𝔽_{q};

• (q^{2} − q)/2 have conjugate roots in 𝔽_{q^{2}};

• q have a double root in 𝔽_{q}.

Lemma 3.2. Of the q^{3} binary quadratic forms f ∈ 𝔽_{q}[X, Y],

• (q − 1)(q + 1)q/2 have two distinct roots in 𝔽_{q};

• (q − 1)(q^{2} − q)/2 have two conjugate roots in 𝔽_{q^{2}};

• (q − 1)(q + 1) have a double root in 𝔽_{q};

• 1 is the zero form.

3.2. Reducible (2, 2)-forms. Now we look at the bihomogeneous polynomials of bidegree (d_{1}, d_{2}) in 𝔽_{q}[X_{0}, X_{1}; Y_{0}, Y_{1}] for 0 ≤ d_{0}, d_{1} ≤ 2, starting with the irreducible (d_{1}, d_{2})-forms with (d_{1}, d_{2}) ≠ (2, 2). These correspond to (d_{1}, d_{2})-curves in ℙ^{1} × ℙ^{1}, where we assume that (X_{0}, X_{1}) corresponds to the coordinates in the first factor of ℙ^{1} and (Y_{0}, Y_{1}) corresponds to the coordinates in the second ℙ^{1}.

Lemma 3.3. The number of irreducible bihomogeneous polynomials in 𝔽_{q}[X_{0}, X_{1}; Y_{0}, Y_{1}] (with bidegrees as indicated) are as follows:

Bidegree Number of forms up to scaling by 𝔽_{q}^{×}
(1, 0)  m_{10} = q + 1
(2, 0)  m_{20} = (q^{2} − q)/2
(1, 1)  m_{11} = q^{3} − q
(2, 1)  m_{21} = q^{5} − q^{3}

Proof. There are q^{2} − 1 nonzero (1, 0)-forms, and so q + 1 up to scaling. The number of irreducible (2, 0)-forms is taken from Lemma 3.2, taking into account scaling. Each irreducible (1, 1)-form is the graph of a Mobius map, and so m_{11} = | PGL_{2}(𝔽_{q})| = q^{3} − q. Finally we compute

m_{21} = (q^{6} − 1)/(q − 1) − m_{10}m_{11} − m_{10}m_{20} − m_{10}^{2}(m_{10} + 1)/2 = q^{5} − q^{3}.  □

We now consider all the different ways in which a (2, 2)-form can factor. We use the notation (a_{1}, b_{1})^{e_{1}} ⋯ (a_{r}, b_{r})^{e_{r}} to denote the bidegrees of the irreducible factors, with multiplicity. For example, the factorization type (1, 0)^{2}(0, 1)(0, 1) indicates that the (2, 2)-form factors as a product F_{10}^{2}F_{01}G_{01}, where F_{10}, F_{01}, and G_{01} are irreducible polynomials in 𝔽_{q}[X_{0}, X_{1}; Y_{0}, Y_{1}] with bidegrees (1, 0), (0, 1), and (0, 1), respectively, and that F_{01} is not an 𝔽_{q}^{×}-multiple of G_{01}.

Lemma 3.4. The number of reducible (2, 2)-forms over 𝔽_{q} with each factorization type are as follows. Moreover, the curve C ⊂ ℙ^{1} × ℙ^{1} defined by such a form either always has a smooth 𝔽_{q}-point, or never has a smooth 𝔽_{q}-point, as indicated in the right hand column.

Factorization type  Number of forms up to scaling by 𝔽_{q}^{×} Smooth point?
(1, 1)(1, 1)  \binom{m_{11}}{2} = (q^{3} − q)(q^{3} − q − 1)/2  yes
(2, 1)(0, 1) or (1, 2)(1, 0)  2m_{21}m_{10} = 2q^{3}(q + 1)^{2}(q − 1)  yes
(1, 1)(1, 0)(0, 1)  m_{11}m_{10}^{2} = q(q + 1)^{3}(q − 1)  yes
(1, 0)(1, 0)(0, 1)(0, 1)  \binom{m_{10}}{2}^{2} = q^{2}(q + 1)^{2}/4  yes
(2, 0)(0, 1)(0, 1) or (0, 2)(1, 0)(1, 0) 2m_{20}\binom{m_{10}}{2} = q^{2}(q + 1)(q − 1)/2  yes
(2, 0)(0, 2) m_{20}^{2} = q^{2}(q − 1)^{2}/4 no
(1, 0)^{2}(0, 1)(0, 1) or (0, 1)^{2}(1, 0)(1, 0) 2m_{10}\binom{m_{10}}{2} = q(q + 1)^{2}  yes
(2, 0)(0, 1)^{2} or (0, 2)(1, 0)^{2}  2m_{20}m_{10} = q(q + 1)(q − 1)  no
(1, 1)^{2}  m_{11} = q(q + 1)(q − 1)  no
(1, 0)^{2}(0, 1)^{2}  m_{10}^{2} = (q + 1)^{2}  no

Proof. The counts all follow from Lemma 3.3 as indicated. In each of the cases listed as having smooth 𝔽_{q}-points, there is an irreducible factor of multiplicity 1 with bidegree (1, 0), (0, 1), (1, 1), (2, 1) or (1, 2). Each such factor defines a smooth curve of genus 0 which, by projection to one of the factors, is isomorphic to ℙ^{1}. Moreover a case-by-case analysis shows that this curve meets the other components of C in at most 2 points. Since #ℙ^{1}(𝔽_{q}) = q + 1 > 2 this shows that C has a smooth 𝔽_{q}-point. In the remaining cases each irreducible factor is either repeated or has bidegree (2, 0) or (0, 2). So in these cases there are no smooth 𝔽_{q}-points. □

3.3. Irreducible (2, 2)-forms. We now consider the irreducible (2, 2)-forms over 𝔽_{q}. We distinguish between those that are absolutely irreducible (i.e., do not factor over \overline{𝔽}_{q}) and those that factor over 𝔽_{q^{2}} as the product of a bidegree (1, 1)-form and its conjugate. In the latter case, we say the form has factorization type (1, 1)\overline{(1, 1)}.

Lemma 3.5. Let C ⊂ ℙ^{1} × ℙ^{1} be a curve defined by an absolutely irreducible (2, 2)-form F ∈ 𝔽_{q}[X_{0}, X_{1}; Y_{0}, Y_{1}]. Then C has a smooth 𝔽_{q}-point.

Proof. If C is smooth then it has genus 1, and the lemma follows by the Hasse-Weil bound. If C is singular, then it has geometric genus 0. The normalization is a smooth genus 0 curve, in fact ℙ^{1} itself (e.g., by the Hasse-Weil bound), and thus has (q + 1) 𝔽_{q}-points. Since the preimage of the singular point is at most length 2, the curve C must have at least one smooth 𝔽_{q}-point. □

Remark 3.6. The numbers of irreducible (2, 2)-forms are as follows:

Factorization type  Number of forms up to scaling by 𝔽_{q}^{×}
smooth  q^{4}(q + 1)^{2}(q − 1)^{2}
absolutely irreducible yet singular  q^{3}(q + 1)^{2}(q − 1)^{2}
(1, 1)\overline{(1, 1)}  (q^{3} − q)(q^{3} + q − 1)/2

In the case (1, 1)\overline{(1, 1)} the (1, 1)-forms are irreducible over 𝔽_{q^{2}}, since the original (2, 2)-form was irreducible over 𝔽_{q}. Each is therefore the graph of a Mobius map, and so the count in the last row is (# PGL_{2}(𝔽_{q^{2}}) − # PGL_{2}(𝔽_{q}))/2. We omit the details of the other two counts since these are not needed for the proof of Theorem 1.4. However, as a check on our calculations, we note that these counts, together with those in Lemma 3.4, do indeed add up to (q^{9} − 1)/(q − 1).

If F has factorization type (1, 1)\overline{(1, 1)} then C is geometrically the union of two rational curves. We subdivide into cases according as these meet in

(i) a pair of points defined over 𝔽_{q}, (ii) a pair of conjugate points defined over 𝔽_{q^{2}}, (iii) a single point defined over 𝔽_{q}.

Lemma 3.7. The number of (2, 2)-forms over 𝔽_{q} (up to scaling by 𝔽_{q}^{×}) in cases (i), (ii) and (iii) above are, respectively,

n_{11} = q^{3}(q + 1)^{2}(q − 1)/4,
n_{12} = q^{2}(q + 1)(q − 1)^{2}(q − 2)/4,
n_{13} = q(q + 1)^{2}(q − 1)^{2}/2.

Proof.

(i) The singular points can be any pair of points in ℙ^{1}(𝔽_{q}) × ℙ^{1}(𝔽_{q}) that remain distinct under both projection maps. This last condition comes from the fact, noted in Remark 3.6, that each (1, 1)-form is the graph of a Mobius map. We make a change of coordinates to move the singular points to ((0 : 1), (0 : 1)) and ((1 : 0), (1 : 0)). By hypothesis F factors as the product of two (1, 1)-forms over 𝔽_{q^{2}}, and these are now linear combinations of X_{0}Y_{1} and X_{1}Y_{0}. Therefore F = f(X_{0}Y_{1}, X_{1}Y_{0}) for some irreducible binary quadratic form f. We compute n_{11} as the product of the (q^{2} − q)/2 choices for f (up to scaling), and the q^{2}(q + 1)^{2}/2 choices for the (unordered) pair of singular points.

(ii) We write 𝔽_{q^{2}} = 𝔽_{q}(α) where α^{2} + rα + s = 0 for some r, s ∈ 𝔽_{q}. As noted in (i), the singular points remain distinct under both projection maps. After a change of coordinates, defined over 𝔽_{q}, we may therefore assume that the singular points are ((α : 1), (α : 1)) and its Galois conjugate. Then F = f(X_{0}Y_{0} + rX_{0}Y_{1} + sX_{1}Y_{1}, X_{0}Y_{1} − X_{1}Y_{0}) for some irreducible binary quadratic form f. There are (q^{2} − q)/2 choices for f (up to scaling), but one of these gives F = (X_{0}^{2} + rX_{0}X_{1} + sX_{1}^{2})(Y_{0}^{2} + rY_{0}Y_{1} + sY_{1}^{2}). Therefore n_{12} is the product of (q + 1)(q − 2)/2 and the (q^{2} − q)^{2}/2 choices for the singular points.

(iii) We make a change of coordinates to move the singular point to ((0 : 1), (0 : 1)). Then F is the product of αX_{0}Y_{0} + βX_{0}Y_{1} + γX_{1}Y_{0}, and its Galois conjugate, for some α, β, γ ∈ 𝔽_{q^{2}} with β, γ ≠ 0. If β and γ are a basis for 𝔽_{q^{2}} as an 𝔽_{q}-vector space, then by substitutions of the form X_{1} ← X_{1} + λX_{0} and Y_{1} ← Y_{1} + μY_{0} with λ, μ ∈ 𝔽_{q} we may reduce to the case α = 0. But then the two components also meet at ((1 : 0), (1 : 0)), which cannot happen in case (iii). Therefore F = f(X_{0}Y_{0}, X_{0}Y_{1} + cX_{1}Y_{0}) for some irreducible binary quadratic form f and constant c ∈ 𝔽_{q}^{×}. We compute n_{13} as the product of the (q^{2} − q)/2 choices for f (up to scaling), the q − 1 choices for c, and the (q + 1)^{2} choices for the singular point.

As a final check, we note that n_{11} + n_{12} + n_{13} = (q^{3} − q)(q^{3} + q − 1)/2. □

4. Local solubility for bidegree (2, 2)-forms

Fix a prime p, and consider the space of all (2, 2)-forms F ∈ ℤ_{p}[X_{0}, X_{1}; Y_{0}, Y_{1}] with its natural product Haar measure when viewed as a copy of ℤ_{p}^{9}. In this section, we determine the density of ℚ_{p}-soluble forms in this space.

In order to determine the solubility of a given form F, it will often suffice to look at its reduction mod p, denoted \overline{F} ∈ 𝔽_{p}[X_{0}, X_{1}; Y_{0}, Y_{1}], and look for a smooth 𝔽_{p}-point on the curve \overline{C} defined by \overline{F}, so that we may apply Hensel’s lemma. As seen in the tables of Section 3 (where we now take q = p), it is easy to see that most of the factorization types for \overline{F} have a smooth 𝔽_{p}-point on \overline{C}. According to the results of Section 3, only five cases require further consideration, which we analyze in Section 4.3.

4.1. Preliminaries. Let v(a) = v_{p}(a) denote the p-adic valuation of a ∈ ℚ_{p}.

We will often keep track of the valuations of the coefficients of the (2, 2)-form (1.3) as a 3 × 3 table:

\begin{matrix}v(a_{00}) & v(a_{01}) & v(a_{02}) \\ v(a_{10}) & v(a_{11}) & v(a_{12}) \\ v(a_{20}) & v(a_{21}) & v(a_{22}), \end{matrix}  (4.1)

where each entry is the valuation of the coefficient of the corresponding monomial term in

\begin{matrix}X_{0}^{2}Y_{0}^{2} & X_{0}^{2}Y_{0}Y_{1} & X_{0}^{2}Y_{1}^{2} \\ X_{0}X_{1}Y_{0}^{2} & X_{0}X_{1}Y_{0}Y_{1} & X_{0}X_{1}Y_{1}^{2} \\ X_{1}^{2}Y_{0}^{2} & X_{1}^{2}Y_{0}Y_{1} & X_{1}^{2}Y_{1}^{2}. \end{matrix}  (4.2)

4.2. A useful lemma. Before analyzing the cases where Hensel’s lemma does not directly apply, we note that the following lemma will be used several times in the next section, and is, in some sense, typical of the arguments we use. In fact, we use it once in the analysis of Case 3 and twice in the analysis of Case 5 (in Lemmas 4.11 and 4.15).

Lemma 4.3. Let a_{00}, a_{01}, a_{02}, a_{10}, a_{11}, a_{20} be any fixed elements of ℤ_{p} satisfying

v(a_{00}) ≥ 2, v(a_{01}) ≥ 2, v(a_{02}) = 1,
v(a_{10}) ≥ 1, v(a_{11}) ≥ 1,
v(a_{20}) = 0.

Let

𝒮 := { ∑_{i,j=0}^{2} a_{ij}X_{0}^{2−i}X_{1}^{i}Y_{0}^{2−j}Y_{1}^{j} : a_{12} ∈ pℤ_{p}, a_{21} ∈ ℤ_{p}, a_{22} ∈ ℤ_{p}} .

Then the proportion of the polynomials in 𝒮 that have ℚ_{p}-solutions for which p ∤ X_{1}Y_{1} is 1/2.

Proof. Let σ(a_{00}, a_{01}, a_{02}, a_{10}, a_{11}, a_{20}) be the desired probability (that a polynomial in 𝒮 has a ℚ_{p}- solution with p ∤ X_{1}Y_{1}), and let τ(a_{00}, a_{01}, a_{02}, a_{10}, a_{11}, a_{20}) be the corresponding probability when 𝒮 is replaced by its subset

𝒯 := { ∑_{i,j=0}^{2} a_{ij}X_{0}^{2−i}X_{1}^{i}Y_{0}^{2−j}Y_{1}^{j} : a_{12}, a_{21}, a_{22} ∈ pℤ_{p}} .

If F ∈ 𝒮, then F (0, 1; Y_{0}, 1) reduces mod p to a quadratic polynomial in Y_{0}, i.e., the coefficient of Y_{0}^{2} is nonzero mod p. Furthermore, all such quadratic polynomials occur with equal probability. By Lemma 3.1, this quadratic splits into distinct factors over 𝔽_{p} with probability \frac{1}{2}(1 − \frac{1}{p}). In this case, the point ((0 : 1), (α : 1)), where α is one of the roots of the quadratic, is a smooth 𝔽_{p}-point, so by Hensel’s lemma, the curve defined by F = 0 has a ℚ_{p}-point of the form ((0 : 1), (α̃ : 1)) for some lift α̃ ∈ ℤ_{p}. If the quadratic is instead irreducible, as happens with probability \frac{1}{2}(1 − \frac{1}{p}), then there are no ℚ_{p}-points with p ∤ X_{1}. It remains to consider the case F ≡ a_{20}X_{1}^{2}(Y_{0} − cY_{1})^{2} (mod p), for some 0 ≤ c ≤ p − 1. Transforming F by the substitution Y_{0} ← Y_{0} + cY_{1} we find

σ (\begin{matrix}a_{00} & a_{01} & a_{02} \\ a_{10} & a_{11} \\ a_{20} \end{matrix}) = \frac{1}{2}(1 − \frac{1}{p}) + \frac{1}{p^{2}} ∑_{c=0}^{p−1} τ (\begin{matrix}a_{00} & 2ca_{00} + a_{01} & c^{2}a_{00} + ca_{01} + a_{02} \\ a_{10} & 2ca_{10} + a_{11} \\ a_{20} \end{matrix}) . (4.4)

We note in particular that the arguments of τ satisfy the conditions in the statement of the lemma.

If F ∈ 𝒯 then F ≡ a_{20}X_{1}^{2}Y_{0}^{2} (mod p). For a solution with p ∤ X_{1} we need p | Y_{0}. This suggests making the substitution Y_{0} ← pY_{0}. Dividing through by p, and then swapping the X’s and Y ’s we find

τ (\begin{matrix}a_{00} & a_{01} & a_{02} \\ a_{10} & a_{11} \\ a_{20} \end{matrix}) = σ (\begin{matrix}pa_{00} & pa_{10} & pa_{20} \\ a_{01} & a_{11} \\ p^{−1}a_{02} \end{matrix}) .  (4.5)

Using (4.4) and (4.5) to solve for σ and τ, we find that σ = τ = 1/2. □

4.3. The cases without smooth 𝔽_{p}-points. As remarked before, these are the factorization types of the (2, 2)-form F over 𝔽_{p} which do not immediately yield a smooth 𝔽_{p}-point in the reduction of the corresponding curve.

Case number Factorization type  Number of forms up to scaling by 𝔽_{p}^{×}
1  (1, 1)\overline{(1, 1)}  n_{1} := (p^{3} − p)(p^{3} + p − 1)/2
2  (2, 0)(0, 2)  n_{2} := p^{2}(p − 1)^{2}/4
3  (2, 0)(0, 1)^{2} or (0, 2)(1, 0)^{2} n_{3} := p(p + 1)(p − 1)
4  (1, 1)^{2}  n_{4} := p(p + 1)(p − 1)
5  (1, 0)^{2}(0, 1)^{2}  n_{5} := (p + 1)^{2}

Let n_{0} = (p^{9} − 1)/(p − 1) − (n_{1} + n_{2} + n_{3} + n_{4} + n_{5}) be the number of forms lying in none of the 5 cases, and let ξ_{i} be the probability of solubility in case i. Then the overall probability of solubility is

ρ = \frac{n_{0} + n_{1}ξ_{1} + n_{2}ξ_{2} + n_{3}ξ_{3} + n_{4}ξ_{4} + n_{5}ξ_{5}}{(p^{9} − 1)/(p − 1)}

In this section we compute ξ_{1}, …, ξ_{5} and hence obtain the final answer stated in Theorem 1.4. In the context of computing ξ_{5}, it is helpful to make the following definition for ξ′_{i}, and to compute the ξ′_{i} alongside the ξ_{i}:

Definition 4.6. For 1 ≤ i ≤ 4, we let ξ′_{i} be the probability of solubility given the following conditions: we are in case i, the point ((0 : 1), (0 : 1)) is a singular point on the reduction mod p, and v(a_{22}) ≥ 2. We write ξ′_{1j} for 1 ≤ j ≤ 3 for the same probability in cases 1(i), 1(ii), and 1(iii), as defined below. We define ξ′_{5} in the same way, except that we require that the singular point ((0 : 1), (0 : 1)) is not the point where the two lines meet.

We compute the values of ξ_{i} and ξ′_{i} in the next sections.

4.3.1. Case 1. In this case the reduction of our curve mod p is geometrically the union of two rational curves. We subdivide into the cases (i), (ii), (iii) as defined immediately before Lemma 3.7, and note that n_{1} = n_{11} + n_{12} + n_{13}. Writing ξ_{11}, ξ_{12}, ξ_{13} for the probabilities of solubility in cases 1(i), 1(ii) and 1(iii), respectively, we have

ξ_{1} = (n_{11}ξ_{11} + n_{12}ξ_{12} + n_{13}ξ_{13})/n_{1}.

Case 1(i). In this case, the two components meet at a pair of points defined over 𝔽_{p}. We begin by computing ξ_{11}. As in the proof of Lemma 3.7(i) we may assume that \overline{F} = f(X_{0}Y_{1}, X_{1}Y_{0}) for some irreducible binary quadratic form f. The only 𝔽_{p}-points on the reduction are the singular points ((0 : 1), (0 : 1)) and ((1 : 0), (1 : 0)). We must decide if they lift to ℚ_{p}-points. Let α be the probability that the singular point ((0 : 1), (0 : 1)) lifts.

Since ((0 : 1), (0 : 1)) is a singular point on the curve defined by \overline{F}, we deduce that all the valuations v(a_{12}), v(a_{21}), v(a_{22}) are ≥ 1. If v(a_{22}) = 1, then the singular point ((0 : 1), (0 : 1)) does not lift. Otherwise (with probability 1/p), we have v(a_{22}) ≥ 2. Then the valuations of the coefficients of F satisfy

\begin{aligned}≥ 1 & ≥ 1 & = 0 \\ ≥ 1 & ≥ 0 & ≥ 1 \\ = 0 & ≥ 1 & ≥ 2 \end{aligned}

where the equalities follow from f being irreducible. Making the substitutions X_{0} ← pX_{0}, Y_{0} ← pY_{0} and dividing through by p^{2}, we obtain a (2, 2)-form G(X_{0}, X_{1}, Y_{0}, Y_{1}) whose coefficients b_{ij} have valuations satisfying

\begin{aligned}≥ 3 & ≥ 2 & = 0 \\ ≥ 2 & ≥ 0 & ≥ 0 \\ = 0 & ≥ 0 & ≥ 0. \end{aligned}

We now investigate whether \overline{G} ∈ 𝔽_{p}[X_{0}, X_{1}; Y_{0}, Y_{1}] is absolutely irreducible. Define a ternary quadratic form Q(X_{0}, Y_{0}, Z_{0}) by \overline{G}(X_{0}, 1, Y_{0}, 1) = Q(X_{0}, Y_{0}, 1), so that the zero-set of Q in 𝔸_{X_{0},Y_{0}}^{2} ⊂ ℙ^{2} coincides with the zero-set of \overline{G} in 𝔸_{X_{0}}^{1} × 𝔸_{Y_{0}}^{1} ⊂ ℙ^{1} × ℙ^{1}. Then the curve defined by Q (and thus, the curve defined by \overline{G}) is geometrically irreducible if and only if the discriminant of Q is nonzero, equivalently

b_{22} disc(f) − f(b_{21}, −b_{12}) ≢ 0 (mod p).

(This argument still works in characteristic 2 provided that the formula for the discriminant of a ternary quadratic form is scaled by appropriate powers of 2.)

If the curve defined by \overline{G} is geometrically irreducible, then the argument in Lemma 3.5 shows that it has a smooth 𝔽_{p}-point. Otherwise (with probability 1/p), the reduction mod p is geometrically the union of two rational curves meeting at ((1 : 0), (1 : 0)) and an 𝔽_{p}-point of the form ((λ : 1), (μ : 1)). The two rational curves are not defined over 𝔽_{p}, since the binary quadratic form f is irreducible. We make the substitutions X_{0} ← X_{0} +λX_{1} and Y_{0} ← Y_{0} +μY_{1} to move the second point of intersection to ((0 : 1), (0 : 1)), and start over again considering whether this singular point lifts. The probability that it lifts is again α. We thus obtain the recursive formula

α = \frac{1}{p} ((1 − \frac{1}{p}) + \frac{1}{p} α) ,

and so α = 1/(p + 1).

We are interested in the probability that at least one of the singular points lifts. Since these events depend on different coefficients of the (2, 2)-form they are independent. Therefore

ξ_{11} = 1 − (1 − \frac{1}{p + 1})^{2} = \frac{2p + 1}{(p + 1)^{2}} .

A small modification of this argument (as required by Definition 4.6) gives

ξ′_{11} = 1 − (1 − \frac{1}{p + 1}) (1 − \frac{p}{p + 1}) = \frac{p^{2} + p + 1}{(p + 1)^{2}}.

Case 1(ii). The two components meet at a pair of conjugate points defined over 𝔽_{p^{2}}. There are no 𝔽_{p}-points on the reduction. Therefore ξ_{12} = 0 and ξ′_{12} is not defined.

Case 1(iii). The two components meet at a single point defined over 𝔽_{p}. As in the proof of Lemma 3.7(iii) we may assume that \overline{F} = f(X_{0}Y_{0}, X_{0}Y_{1}+X_{1}Y_{0}) for some irreducible binary quadratic form f. The only 𝔽_{p}-point on the reduction is the singular point ((0 : 1), (0 : 1)). If this lifts to a ℚ_{p}-point then we must have v(a_{22}) ≥ 2. Since this is exactly the condition in Definition 4.6 it follows that ξ_{13} = (1/p)ξ′_{13}.

We show in Section 5, using results from [BCF], that

ξ_{13} = \frac{2p^{10} + 3p^{9} − p^{5} + 2p^{4} − 2p^{2} − 3p − 1}{2(p + 1)^{2}(p^{9} − 1)}.

As noted in the last paragraph, we have ξ′_{13} = pξ_{13}.

4.3.2. Case 2. This is the case (2, 0)(0, 2). There are no 𝔽_{p}-points on the reduction. Therefore ξ_{2} = 0 and ξ′_{2} is not defined.

4.3.3. Case 3. This is the case (2, 0)(0, 1)^{2} or (0, 2)(1, 0)^{2}. We may assume without loss of generality that \overline{F} = f(X_{0}, X_{1})Y_{0}^{2} for some irreducible binary quadratic form f. The coefficients satisfy

\begin{aligned}= 0 & ≥ 1 & ≥ 1 \\ ≥ 0 & ≥ 1 & ≥ 1 \\ = 0 & ≥ 1 & ≥ 1 \end{aligned}

where the equalities follow from f being irreducible. Making the substitution Y_{0} ← pY_{0} and dividing through by p gives

\begin{aligned}= 1 & ≥ 1 & ≥ 0 \\ ≥ 1 & ≥ 1 & ≥ 0 \\ = 1 & ≥ 1 & ≥ 0 \end{aligned}

The reduction mod p is now g(X_{0}, X_{1})Y_{1}^{2} for some binary quadratic form g. If g is irreducible, splits, or has repeated roots, then the probability of solubility is 0, 1, or 1/2, respectively. In the last of these cases, we are using Lemma 4.3: more specifically, we assume the double root is at (X_{0} : X_{1}) = (0 : 1), make the substitution X_{0} ← pX_{0}, divide through by p, and then apply the lemma. Note that the lemma applies as v(a_{20}) = 0, and there are no solutions with p | X_{1}Y_{1} in view of the substitutions we made to reach this situation.

If g is identically zero, then we divide through by p, to obtain a (2, 2)-form H satisfying the line condition, by which we mean that H(X_{0}, X_{1}; 1, 0) (mod p) is an irreducible binary quadratic form. Writing δ_{line} for the probability of solubility in this case, we have

ξ_{3} = \frac{1}{p^{3}} (\frac{p^{3} − p}{2} · 1 + \frac{p(p − 1)^{2}}{2} · 0 + (p^{2} − 1) · \frac{1}{2} + δ_{line})  (4.7)

and

ξ′_{3} = \frac{1}{p^{2}} (p(p − 1) · 1 + (p − 1) · \frac{1}{2} + δ_{line}) .  (4.8)

Lemma 4.9. There are (up to scaling by 𝔽_{p}^{×}) exactly p^{7}(p − 1)/2 forms over 𝔽_{p} satisfying the line condition. The numbers of these in Cases 1 to 3 are

r_{11} = p^{3}(p + 1)(p − 1)^{2}/4
r_{12} = p^{2}(p + 1)(p − 1)^{2}(p − 2)/4
r_{13} = p^{2}(p + 1)(p − 1)^{2}/2
r_{2} = p^{2}(p − 1)^{2}/4
r_{3} = p^{2}(p − 1)/2

There are none in Cases 4 and 5.

Proof. It is easy to check that forms in Cases 1(ii) and 2 always satisfy the line condition, and those in Cases 4 and 5 never satisfy the line condition. By double counting pairs consisting of (2, 2)-forms and (0, 1)-forms (both up to scalars) that meet in a pair of conjugate points over 𝔽_{p^{2}}, we find that (p+1)r_{11} = (p−1)n_{11} and (p+1)r_{13} = pn_{13}. In Case 3 we must count the forms f(X_{0}, X_{1})g(Y_{0}, Y_{1})^{2} where f is an irreducible binary quadratic form, and g is a linear form with g(1, 0) ≠ 0. We find that r_{3} is the product of the (p^{2} − p)/2 choices for f and the p choices for g. □

Let r_{0} = p^{7}(p − 1)/2 − (r_{11} + r_{12} + r_{13} + r_{2} + r_{3}) be the number of forms satisfying the line condition not in Cases 1 to 3. Then

δ_{line} = \frac{r_{0} + r_{11}ξ_{11} + r_{12}ξ_{12} + r_{13}ξ_{13} + r_{2}ξ_{2} + r_{3}ξ_{3}}{p^{7}(p − 1)/2}  (4.10)

Using the values of ξ_{11}, ξ_{12}, ξ_{13} and ξ_{2} already computed, we can now solve (4.7) and (4.10) for ξ_{3} and δ_{line}. We then use (4.8) to compute ξ′_{3}. We find that

ξ_{3} = \frac{p^{10} + 2p^{9} + p^{6} − 2p^{5} + 2p^{3} + p^{2} − 3p − 2}{2(p + 1)(p^{9} − 1)}

and

ξ′_{3} = \frac{2p^{10} + p^{9} + p^{7} − 2p^{6} + 2p^{4} + p^{3} − 2p^{2} − 2p − 1}{2(p + 1)(p^{9} − 1)}.

4.3.4. Case 4. This is the case (1, 1)^{2}. By a change of coordinates we may assume

F ≡ (X_{0}Y_{1} − X_{1}Y_{0})^{2} (mod p).

We show in Section 5, using results from [BCF], that

ξ_{4} = \frac{5p^{10} + 8p^{9} + p^{8} − p^{7} + 2p^{6} − 3p^{5} + 4p^{3} − 10p − 6}{8(p + 1)(p^{9} − 1)},

and

ξ′_{4} = \frac{4p^{10} + 3p^{9} − p^{7} + 2p^{6} − 2p^{5} + 2p^{3} − p^{2} − 5p − 2}{4(p + 1)(p^{9} − 1)}.

4.3.5. Case 5. This is the case (1, 0)^{2}(0, 1)^{2}. By a change of coordinates we may assume

F ≡ X_{0}^{2}Y_{0}^{2} (mod p).

The coefficients of F have valuations satisfying

\begin{aligned}= 0 & ≥ 1 & ≥ 1 \\ ≥ 1 & ≥ 1 & ≥ 1 \\ ≥ 1 & ≥ 1 & ≥ 1. \end{aligned}

Let Q and Q′ be the binary quadratic forms over 𝔽_{p} determined by the last row and column, i.e.,

Q(Y_{0}, Y_{1}) = \frac{1}{p}F (0, 1; Y_{0}, Y_{1}) (mod p)
Q′(X_{0}, X_{1}) = \frac{1}{p}F (X_{0}, X_{1}; 0, 1) (mod p)

Note that these forms have the same last coefficient c ∈ 𝔽_{p} as they share one entry in the coefficient matrix corresponding to X_{1}^{2}Y_{1}^{2}. Writing ξ_{51} and ξ_{52} for the probabilities of solubility in the cases c ≠ 0 and c = 0, respectively, we have

ξ_{5} = (1 − 1/p)ξ_{51} + (1/p)ξ_{52}.

Lemma 4.11. We have ξ_{51} = 3/4.

Proof. The coefficients satisfy

\begin{aligned}= 0 & ≥ 1 & ≥ 1 \\ ≥ 1 & ≥ 1 & ≥ 1 \\ ≥ 1 & ≥ 1 & = 1. \end{aligned}

The reduction mod p is the union of two double lines, meeting at a single point. Any ℚ_{p}-point has p | X_{0} or p | Y_{0}, but not both since v(a_{22}) = 1. In other words, any ℚ_{p}-point must reduce to lie on exactly one of the lines.

To investigate whether there are solutions with p | X_{0} we make the substitution X_{0} ← pX_{0} and divide by p to get

\begin{aligned}= 1 & ≥ 2 & ≥ 2 \\ ≥ 1 & ≥ 1 & ≥ 1 \\ ≥ 0 & ≥ 0 & = 0. \end{aligned}

We then apply Lemma 4.3 (with Y_{0} ↔ Y_{1}). The probability of a solution with p | X_{0} and the probability of a solution with p | Y_{0} are each 1/2. The lemma also implies that these two events are independent of each other, so the probability of insolubility of these polynomials is 1/4. Hence, the probability of solubility is 3/4. □

Definition 4.12. Let δ_{1} and δ_{2} be the probabilities of solubility in the cases

\begin{aligned}≥ 1 & ≥ 1 & ≥ 0 \\ ≥ 1 & ≥ 0 & ≥ 0 \\ = 0 & ≥ 0 & ≥ 0 \end{aligned}  \begin{aligned}≥ 1 & ≥ 1 & = 0 \\ ≥ 1 & ≥ 0 & ≥ 0 \\ = 0 & ≥ 0 & ≥ 0 \end{aligned}

(The subscript is the number of equalities in the matrix.) Let δ_{1}^{∗} and δ_{2}^{∗} be the probabilities when we change the top left ≥ 1 to = 1. Let ε_{1} and ε_{2} be the probabilities when we change the top left ≥ 1 to ≥ 2.

Clearly we have

\begin{matrix}δ_{1} = (1 − 1/p)δ_{1}^{∗} + (1/p)ε_{1} \\ δ_{2} = (1 − 1/p)δ_{2}^{∗} + (1/p)ε_{2} \end{matrix}  (4.13)

Lemma 4.14. We have

ξ_{52} = (1 − \frac{1}{p^{2}} ) + \frac{1}{p^{2}} (\frac{1}{p} (1 − \frac{1}{p})^{2} δ_{2}^{∗} + 2 \frac{1}{p} (1 − \frac{1}{p}) δ_{1}^{∗} + (\frac{1}{p})^{2} ε_{1})

Proof. If at least one of the forms Q and Q′ has distinct roots in 𝔽_{p} then the (2, 2)-form is soluble over ℚ_{p}. This happens with probability 1 − 1/p^{2}. Otherwise (with probability 1/p^{2}) the coefficients satisfy

\begin{aligned}= 0 & ≥ 1 & ≥ 1 \\ ≥ 1 & ≥ 1 & ≥ 2 \\ ≥ 1 & ≥ 2 & ≥ 2 \end{aligned}

(The bottom right ≥ 2 comes from the assumption that c = 0, and the two adjacent ≥ 2 entries arise from assuming that neither Q nor Q′ has distinct roots in 𝔽_{p}.) We split into 3 cases:

(i) Suppose v(a_{02}) = v(a_{20}) = 1. If v(a_{22}) = 2 then the (2, 2)-form is insoluble over ℚ_{p}. Otherwise, we find by substituting X_{0} ← pX_{0}, Y_{0} ← pY_{0}, and dividing through by p^{3}, that the probability of solubility is δ_{2}^{∗}.

(ii) Suppose v(a_{02}) = 1 and v(a_{20}) ≥ 2. We find by substituting X_{0} ← pX_{0}, and dividing through by p^{2}, that the probability of solubility is δ_{1}^{∗}. The case where v(a_{02}) ≥ 2 and v(a_{20}) = 1 works in exactly the same way via the substitution Y_{0} ← pY_{0}.

(iii) Suppose v(a_{02}) ≥ 2 and v(a_{20}) ≥ 2. Via either of the substitutions in (ii), the probability

of solubility is ε_{1}. Combining these gives the desired expression for ξ_{52}. □

Lemma 4.15. We have

ξ′_{5} = (1 − \frac{1}{p}) + \frac{1}{p} (1 − \frac{1}{p}) \frac{3}{4} + (\frac{1}{p})^{2} (1 − \frac{1}{p}) + (\frac{1}{p})^{3} δ_{1}.

Proof. According to Definition 4.6, we may suppose the coefficients of F satisfy

\begin{aligned}= 0 & ≥ 1 & ≥ 2 \\ ≥ 1 & ≥ 1 & ≥ 1 \\ ≥ 1 & ≥ 1 & ≥ 1 \end{aligned}

If v(a_{12}) = 1, then Q′ has distinct roots in 𝔽_{p} and so F is soluble over ℚ_{p}. Otherwise (with probability 1/p), we have v(a_{12}) ≥ 2.

If v(a_{22}) = 1, then by an argument similar to Lemma 4.11, the probability of solubility is 3/4. (The solutions with p | X_{0} are analysed in exactly the same way as before, whereas to analyse those with p | Y_{0} we substitute Y_{0} ← pY_{0} and then X_{1} ← pX_{1}.)

Otherwise (with probability 1/p), we have v(a_{22}) ≥ 2. If v(a_{21}) = 1, then Q has distinct roots in 𝔽_{p} and so F is soluble over ℚ_{p}. Otherwise (with probability 1/p), we have v(a_{21}) ≥ 2. We find by substituting Y_{0} ← pY_{0} and then dividing through by p^{2} that the probability of solubility is δ_{1}. □

Lemma 4.16. There are p^{5} possibilities for \overline{F} (up to scaling by 𝔽_{p}^{×}) satisfying the conditions in the definition of δ_{1}. The number of these in Cases 1 to 5 are

s_{11} = p^{3}(p − 1)/2,
s_{12} = 0,
s_{13} = p(p − 1)^{2}/2,
s_{2} = 0,
s_{3} = p(p − 1)/2,
s_{4} = p(p − 1)
s_{5} = p.

In Cases 1(i), 1(iii), and 4, these forms also satisfy the conditions in the definition of δ_{2}. In Cases 3 and 5, they do not.

Proof. The conditions in the definition of δ_{1} are that \overline{F} = 0 is singular at ((1 : 0), (1 : 0)) but does not contain the line Y_{1} = 0. We have (p + 1)^{2}s_{11} = 2n_{11} and (p + 1)^{2}s_{13} = n_{13}. In Cases 1(ii) and 2, there are no 𝔽_{p}-points so s_{12} = s_{2} = 0. In Cases 3, 4, and 5, we count the forms X_{1}^{2}f(Y_{0}, Y_{1}), (αX_{1}Y_{0} + βX_{0}Y_{1} + γX_{1}Y_{1})^{2}, and X_{1}^{2}g(Y_{0}, Y_{1})^{2} where f is an irreducible binary quadratic form, α, β, γ ∈ 𝔽_{p}^{×} with αβ ≠ 0, and g is a linear form with g(1, 0) ≠ 0. Finally, it is only in Cases 3 and 5 that the reduction mod p contains the line X_{1} = 0. □

For the final computation of ξ_{5}, let s_{0} = p^{5} − (s_{11} + s_{13} + s_{3} + s_{4} + s_{5}) and t_{0} = p^{4}(p − 1) − (s_{11} + s_{13} + s_{4}) so that

δ_{1} = \frac{s_{0} + s_{11}ξ_{11} + s_{13}ξ_{13} + s_{3}ξ_{3} + s_{4}ξ_{4} + s_{5}ξ_{5}}{p^{5}}
δ_{2} = \frac{t_{0} + s_{11}ξ_{11} + s_{13}ξ_{13} + s_{4}ξ_{4}}{p^{4}(p − 1)} .

Replacing each ξ by ξ′ (see Definition 4.6) we have

ε_{1} = \frac{s_{0} + s_{11}ξ′_{11} + s_{13}ξ′_{13} + s_{3}ξ′_{3} + s_{4}ξ′_{4} + s_{5}ξ′_{5}}{p^{5}}
ε_{2} = \frac{t_{0} + s_{11}ξ′_{11} + s_{13}ξ′_{13} + s_{4}ξ′_{4}}{p^{4}(p − 1)} .

Putting together all the equations derived in this section, together with the previously computed ξ’s, we are now able to solve for ξ_{5}. We find ξ_{5} = f(p)/g(p) where

f(p) = 6p^{18} + 8p^{17} + 2p^{16} − 8p^{15} + 16p^{14} − 12p^{13} − 4p^{12} + 3p^{11} + 9p^{10} − 35p^{9}
+ 8p^{8} − 11p^{7} + 3p^{6} − p^{5} + 8p^{4} − 6p^{3} − 4p^{2} + 10p + 8,
g(p) = 8(p + 1)(p^{9} − 1)(p^{8} − 1).

5. Relation to binary quartics

We compute some of the probabilities required in Section 4 by reducing them to probabilities already computed in [BCF]. The basic idea is that a (2, 2)-form determines a binary quartic form, by writing the (2, 2)-form as a binary quadratic form in Y_{0}, Y_{1} and taking the discriminant. However, since we also want results in the case p = 2, we will in fact work with generalised binary quartics, defined as follows.

Definition 5.1. A generalised binary quartic (G_{2}, G_{4}) is a pair of binary forms of degrees 2 and 4. A generalised binary quartic (G_{2}, G_{4}) is soluble over a field K if for some X_{0}, X_{1}, Z ∈ K with (X_{0}, X_{1}) ≠ (0, 0) we have Z^{2} + G_{2}(X_{0}, X_{1})Z = G_{4}(X_{0}, X_{1}).

We write ℤ_{p}[X_{0}, X_{1}] = ⊕_{d}ℤ_{p}[X_{0}, X_{1}]_{d} and ℤ_{p}[X_{0}, X_{1}; Y_{0}, Y_{1}] = ⊕_{d,e}ℤ_{p}[X_{0}, X_{1}; Y_{0}, Y_{1}]_{de} for the gradings of these rings by degree d and by bidegree (d, e), respectively.

Lemma 5.2. Let ℓ, a ∈ 𝔽_{p} such that Z^{2} + ℓZ − a is irreducible over 𝔽_{p}, and let

𝒮 := {(G_{2}, G_{4}) ∈ ℤ_{p}[X_{0}, X_{1}]_{2} × ℤ_{p}[X_{0}, X_{1}]_{4} : G_{2} ≡ ℓX_{0}^{2} (mod p) and G_{4} ≡ aX_{0}^{4} (mod p)},
𝒯 := {(G_{2}, G_{4}) ∈ ℤ_{p}[X_{0}, X_{1}]_{2} × ℤ_{p}[X_{0}, X_{1}]_{4} : G_{2} ≡ 0 (mod p) and G_{4} ≡ 0 (mod p)},

and 𝒯^{∗} := {(G_{2}, G_{4}) ∈ 𝒯 : G_{4}(0, 1) ≢ 0 (mod p^{2})}. Then the proportions of generalised binary quartics in 𝒮, 𝒯 and 𝒯^{∗} that are soluble over ℚ_{p} are, respectively,

σ = \frac{2p^{10} + 3p^{9} − p^{5} + 2p^{4} − 2p^{2} − 3p − 1}{2(p + 1)^{2}(p^{9} − 1)},

τ = \frac{5p^{10} + 8p^{9} + p^{8} − p^{7} + 2p^{6} − 3p^{5} + 4p^{3} − 10p − 6}{8(p + 1)(p^{9} − 1)},

and

τ^{∗} = \frac{5p^{10} + 5p^{9} − p^{7} + 3p^{6} − 4p^{5} + 4p^{3} − 8p − 4}{8(p + 1)(p^{9} − 1)}.

Proof. These probabilities were computed in [BCF]. The probability σ was computed in Lemma 2.12, where it was denoted α_{4}^{−}. The probabilities τ and τ^{∗} were computed in Section 2.3, where they were denoted σ_{4} and β_{4}^{0}. □

We define a map

Φ :  ℤ_{p}[X_{0}, X_{1}; Y_{0}, Y_{1}]_{22} → ℤ_{p}[X_{0}, X_{1}]_{2} × ℤ_{p}[X_{0}, X_{1}]_{4}
F_{0}Y_{0}^{2} + F_{1}Y_{0}Y_{1} + F_{2}Y_{1}^{2} ↦ (F_{1}, −F_{0}F_{2}).

It is easy to check that a (2, 2)-form F is soluble over ℚ_{p} if and only if the generalised binary quartic Φ(F) is soluble over ℚ_{p}. We write Φ_{p} for the corresponding map on forms with coefficients in 𝔽_{p}. The cases (i) and (ii) in the following lemma relate to Cases 1(iii) and 4 in Section 4.3.

Lemma 5.3. Let \overline{F} ∈ 𝔽_{p}[X_{0}, X_{1}; Y_{0}, Y_{1}]_{22} take one of the following forms:

(i) \overline{F} = f(X_{0}Y_{0}, X_{0}Y_{1} + X_{1}Y_{0}) where f is an irreducible binary quadratic form,
(ii) \overline{F} = (X_{0}Y_{1} − X_{1}Y_{0})^{2}.

Then Φ restricts to a measure-preserving map

\{F ∈ ℤ_{p}[X_{0}, X_{1}; Y_{0}, Y_{1}]_{22} : F ≡ \overline{F} (mod p)\}
→ \{G ∈ ℤ_{p}[X_{0}, X_{1}]_{2} × ℤ_{p}[X_{0}, X_{1}]_{4} : G ≡ Φ_{p}(\overline{F}) (mod p)\}.

Proof. The proof comes down to showing that the derivative of Φ_{p} at \overline{F} is a surjective linear map 𝔽_{p}^{9} → 𝔽_{p}^{8}. In cases (i) and (ii), this linear map is given by

F_{0}Y_{0}^{2} + F_{1}Y_{0}Y_{1} + F_{2}Y_{1}^{2} ↦ (F_{1}, −f(0, 1)X_{0}^{2}F_{0} − f(X_{0}, X_{1})F_{2}), and
F_{0}Y_{0}^{2} + F_{1}Y_{0}Y_{1} + F_{2}Y_{1}^{2} ↦ (F_{1}, −X_{0}^{2}F_{0} − X_{1}^{2}F_{2}),

respectively, which are both surjective. □

Case 1(iii). We use these lemmas to compute ξ_{13}, that is, the probability of solubility where F mod p is of the form indicated in Lemma 5.3(i). Let f have coefficients a, b, c. Then Φ_{p}(\overline{F}) is the generalised binary quartic with equation

(Z + cX_{0}X_{1})^{2} + bX_{0}^{2}(Z + cX_{0}X_{1}) + acX_{0}^{4} = 0.

From this we see that ξ_{13} = σ as defined in Lemma 5.2.

Case 4. We use these lemmas to compute ξ_{4}, that is, the probability of solubility where F mod p is of form indicated in Lemma 5.3(ii). Since Φ_{p}(\overline{F}) is identically zero, we see that ξ_{4} = τ as defined in Lemma 5.2. To compute ξ′_{4} we must consider (2, 2)-forms F that additionally satisfy v(a_{22}) ≥ 2. Under the measure preserving map in Lemma 5.3(ii) these are mapped to 𝒯 \ 𝒯^{∗}. Therefore τ = (1 − 1/p)τ^{∗} + (1/p)ξ′_{4} and so

ξ′_{4} = pτ − (p − 1)τ^{∗} = \frac{4p^{10} + 3p^{9} − p^{7} + 2p^{6} − 2p^{5} + 2p^{3} − p^{2} − 5p − 2}{4(p + 1)(p^{9} − 1)}.

Remark 5.4. The same approach could be used to compute ξ_{11}, and indeed our answer agrees with [BCF, Lemma 2.11].

6. Connections to the Hasse principle

In Theorem 1.5, we determined that the proportion of (2, 2)-forms that are everywhere locally soluble is c ≈ 0.8739. A heuristic, similar to [Bha14, Conjectures 6 and 7] predicts that the proportion of everywhere locally soluble (2, 2)-forms that are globally soluble is \frac{1}{4}, i.e., in the notation of §2, that lim_{H→∞} \frac{N(H)}{N_{loc}(H)} = \frac{1}{4} and thus lim_{H→∞} \frac{N(H)}{N_{tot}(H)} = \frac{1}{4}c.

In this section we report on some experiments to test this conjecture numerically. A similar study in the case of plane cubics was made in [Fis15]. With the one exception noted below, all computations were performed using Magma [BCP97] and the data may be found at [FHP19].

6.1. Experiments and results. For each H ∈ {10, 30, 100, 300, 1000}, we chose 1000 (2, 2)-forms (i.e., polynomials of the form (1.3)), with coefficients chosen uniformly at random from [−H, H] ∩ ℤ. The numbers of these that were soluble or everywhere locally soluble (ELS) were as follows:

initial range improved range
H  #soluble  #soluble  #ELS
10 [753, 755]  753  885
30 [640, 652]  642  885
100 [536, 582]  549  875
300 [378, 502]  [432, 433]  867
1000 [275, 464]  [357, 464]  879

The second column gives our initial estimate for the number of soluble (2, 2)-forms out of the 1000. The lower bound was obtained by searching for rational solutions, with the assistance of 4-descent in Magma. The upper bound was obtained by computing the Cassels-Tate pairing on the 2-Selmer group of the Jacobian. For the improved estimates in the third column we used a range of methods, described more fully below, that are conditional on standard conjectures and sometimes were only practical for H sufficiently small.

For the first 4 experiments, we also give the breakdown of these totals by the rank of the Jacobian elliptic curve E/ℚ. The annotations + and − in the case H = 300 indicate that we should add or subtract one if the remaining form whose solubility has not yet been decided turns out to be soluble.

rank E(ℚ) singular 0 1  2  3  4 5 6 Total
H = 10  #ELS  5  0 86 344 313 116 21 0 885
#soluble  5  0 16 285 310 116 21 0 753
H = 30  #ELS  0  0 122 310 291 129 29 4 885
#soluble  0  0 1 208 274 126 29 4 642
H = 100 #ELS  0  0 171 321 257 96 25 5 875
#soluble  0  0 0 205 221 93 25 5 549
H = 300 #ELS  0  0 230− 373 187+ 58 19 0 867
#soluble  0  0 0 210 151+ 52 19 0 432+

Although we can see from our first table that the proportion of everywhere locally soluble forms that are globally soluble is decreasing with H, this hardly amounts to strong evidence that the limit is 1/4. However, the prediction of 1/4 arises since, in the limit, it is expected that (i) 50% of the Jacobians have rank 1 and 50% have rank 2 (by, e.g., the Minimalist Conjecture), and (ii) the proportions of forms that are soluble in these two cases are 0 and 1/2, respectively (as explained below). Our second table therefore provides much stronger evidence for the conjecture, and indeed we see that the convergence in (ii) is happening much faster than that in (i).

6.2. The marked point. As we saw in Section 5, if F ∈ ℤ[X_{0}, X_{1}; Y_{0}, Y_{1}] is a (2, 2)-form, then it determines a pair of binary quartics. These binary quartics have the same discriminant, which is accordingly called the discriminant of F. We should expect a randomly chosen (2, 2)-form to have nonzero discriminant (and hence define a smooth curve). This was true in all our experiments, except for 5 cases with H = 10, which were all in any case soluble. From now on we assume that the discriminant is nonzero, and write E for the Jacobian of C = {F = 0} ⊂ ℙ^{1} × ℙ^{1}. Since the discriminant is a degree 12 polynomial in the coefficients of F, the conductor and discriminant of E each have size about H^{12}.

There are two maps C → ℙ^{1}, given by projection to each factor, and the difference of fibres is a nonzero marked point P_{0} ∈ E(ℚ). There is an explicit formula for P_{0} (see [BH16, Section 6.1.2] or [FR18, Lemma 2.1]) in terms of the coefficients of F. As might be predicted from this formula, we found in our experiments that P_{0} had canonical height at most log(2H^{2}). The torsion subgroup of E(ℚ) was trivial in all but 6 cases with H = 10, when it had order 2. In only one of these cases was P_{0} a torsion point.

We should expect that for a randomly chosen (2, 2)-form, the associated binary quartics should not have any rational roots (i.e., linear factors). This was true in all but 69, 9 and 1 of our examples with H = 10, 30 and 100. We should also expect that P_{0} ∉ 2E(ℚ), and this was true in all but 6 cases when H = 10, and 2 cases when H = 30. If P_{0} ∉ 2E(ℚ) and the associated binary quartics do not have any rational roots, then if rank E(ℚ) = 1, the (2, 2)-form is not soluble. We thus expect that the (2, 2)-form is not soluble in general if rank E(ℚ) = 1.

We now explain why half of the forms with rank 2 Jacobian are expected to be soluble. For an elliptic curve E of rank 2, we want to estimate the proportion of elements in Sel^{2}(E/ℚ) that are in the image of E(ℚ)/2E(ℚ). The average size of Sel^{2}(E/ℚ) is 6 in this family [BH19b], but two of the 2-Selmer group elements correspond to (2, 2)-forms for which one of the associated binary quartics has a rational root, which should only happen 0% of the time when we order by height. Now the size of E(ℚ)/2E(ℚ) is 4 for the 100% of elliptic curves E for which there is no 2-torsion, but we also subtract the same 2 elements corresponding to the (2, 2)-forms for which one of the associated binary quartics has a rational root. We thus predict that (4 − 2)/(6 − 2) = 1/2 of the forms are soluble.

Remark 6.1. Although one of the best methods for finding generators of large height on an elliptic curve E/ℚ is to use Heegner points, this only works for curves of rank 1. Since our elliptic curves all come with a known point of infinite order, this method was of no use to us. We instead relied almost exclusively on descent methods.

6.3. The initial estimates: computing ranks of elliptic curves. The curve defined by a (2, 2)- form is isomorphic to the curve defined by either of the associated binary quartics. Our interest is therefore in deciding the solubility of the genus one curves associated to binary quartics.

Let C/ℚ be a genus one curve defined by a binary quartic, and let E/ℚ be its Jacobian. If C is everywhere locally soluble, then it defines a class [C] in the 2-Selmer group Sel^{2}(E/ℚ). Moreover C(ℚ) ≠ ∅ if and only if [C] ∈ im(δ) where δ in the connecting map in the Kummer exact sequence

0 → E(ℚ)/2E(ℚ) →^{δ} Sel^{2}(E/ℚ) → �(E/ℚ)[2] → 0.

Given a point P ∈ E(ℚ) the Magma function GenusOneModel(2,P) computes a binary quartic representing δ(P). In conjunction with the function IsEquivalent for testing equivalence of binary quartics, this gives a convenient way of reducing the problem of deciding whether C(ℚ) ≠ ∅ to that of finding generators for E(ℚ).

An initial upper bound for the rank of E(ℚ) is obtained by 2-descent, that is, by computing the 2-Selmer group Sel^{2}(E/ℚ). This upper bound can sometimes be improved by computing the Cassels-Tate pairing. Let S_{n} be the image of the natural map Sel^{2^{n}}(E/ℚ) → Sel^{2}(E/ℚ). If ξ, η ∈ S_{n}, say with ξ′ ↦ ξ and η′ ↦ η, then there is an alternating pairing

⟨ , ⟩_{n} : S_{n} × S_{n} → 𝔽_{2} ; (ξ, η) ↦ ⟨ξ′, η⟩_{CT} = ⟨ξ, η′⟩_{CT}  (6.2)

whose kernel is S_{n+1}. We note the inclusions of 𝔽_{2}-vector spaces

im(δ) ⊂ … ⊂ S_{3} ⊂ S_{2} ⊂ S_{1} = Sel^{2}(E/ℚ).

The function CasselsTatePairing in Magma, written by S. Donnelly, computes the pairings

⟨ , ⟩_{CT} : Sel^{2}(E/ℚ) × Sel^{2}(E/ℚ) → 𝔽_{2},  (6.3)

and

⟨ , ⟩_{CT} : Sel^{2}(E/ℚ) × Sel^{4}(E/ℚ) → 𝔽_{2},  (6.4)

taking as input either a pair of binary quartics, or a binary quartic and a quadric intersection. (A variant of his method for computing (6.3) is described in [Fis16].) We may thus compute the pairing (6.2) for n = 1 and n = 2.

A lower bound for the rank of E(ℚ) may be obtained by searching for points either directly on E, or better on one of its 4-coverings as computed using FourDescent in Magma. In this way we obtained generators for a subgroup Γ ⊂ E(ℚ) of known points. In all cases where it is possible that rank Γ < rank E(ℚ) we searched up to height 10^{10} on the 4-coverings.

Our initial (unconditional) estimate on the number of (2, 2)-forms that are soluble was obtained as follows. First, if the curve C defined by our (2, 2)-form is not everywhere locally soluble, then it is obviously not soluble. Otherwise C (or more precisely one of the associated binary quartics) determines a class [C] ∈ Sel^{2}(E/ℚ). If [C] ∈ δ(Γ) then we know that C(ℚ) ≠ ∅, and indeed from the generators for Γ we may compute an explicit solution. Otherwise we look for [D] ∈ Sel^{2}(E/ℚ) with ⟨[C], [D]⟩_{CT} ≠ 0. If we succeed in finding a binary quartic D with these properties, then it is a witness to the fact that C(ℚ) = ∅.

6.4. The improved estimates. Both the improved estimates, and the second table (giving the breakdown by rank of the Jacobian) are conditional on the following two standard conjectures.

• Parity conjecture: This is the parity part of the Birch–Swinnerton-Dyer conjecture, i.e., the Mordell-Weil rank of an elliptic curve E/ℚ is even or odd according as its root number w(E/ℚ) is +1 or −1.

• Generalised Riemann Hypothesis (GRH) : This is needed for the class number calculations for 2-descent (but would be easy to remove for small H) and for the computation of analytic rank bounds.

For ease of exposition, we will assume (as is the case in all examples of interest) that E(ℚ) has trivial torsion subgroup. We write r_{2^{n}} = dim_{𝔽_{2}} S_{n} for the upper bound on the rank of E(ℚ) obtained by 2^{n}-descent. Thus we have

rank Γ ≤ rank E(ℚ) ≤ … ≤ r_{8} ≤ r_{4} ≤ r_{2}.

It is a theorem, original due to Monsky [Mon96], that w(E/ℚ) = (−1)^{r_{2}}.

We improve our lower bounds on the number of forms that are soluble by using the parity conjecture. Indeed, if rank Γ = r_{2^{n}} − 1 for some n, then we may conclude by the parity conjecture that im(δ) = S_{n}. We mainly used this idea with n = 1, when the conclusion is that binary quartics with Jacobian E satisfy the Hasse principle, but also used it with n = 2 in four examples with H = 1000.

In the two examples where we used the parity conjecture in the case H = 30, we were also able to find the missing generators using EightDescent in Magma [Sta05]. In particular on the curve {F = 0} ⊂ ℙ^{1} × ℙ^{1} where

F = 27X_{0}^{2}Y_{0}^{2} − X_{0}X_{1}Y_{0}^{2} + 17X_{1}^{2}Y_{0}^{2} − 27X_{0}^{2}Y_{0}Y_{1} + 15X_{0}X_{1}Y_{0}Y_{1} + 9X_{1}^{2}Y_{0}Y_{1}
− 25X_{0}^{2}Y_{1}^{2} − 12X_{0}X_{1}Y_{1}^{2} − 13X_{1}^{2}Y_{1}^{2},

we found the solution

X_{0} = 5998800628516423107297133082973646629266881508307007941326966876023,
X_{1} = 342294900150114936634770190317380320064921533929615189995360150770683,
Y_{0} = 246468494594162038245191010835877699291643209107952263886240062422805,
Y_{1} = −206172926328604047309514129427033995615708844556361901784128916991449.

This maps to a point on the Jacobian of canonical height ĥ ≈ 644.736, which is well beyond the range that could be found by 4-descent. Unfortunately it was not practical to run EightDescent in the experiments with H = 100, 300 and 1000, and so our improved lower bounds in those cases remain conditional on the parity conjecture.

The main method we used to improve the upper bounds on the number of forms that are soluble was to compute the Cassels-Tate pairing (6.4). In the experiments with H = 10, 30, 100 and 300, we were left with 0, 1, 2 and 5 examples where r_{2} = r_{4} = r_{8} = 3, yet (despite searching on all 4coverings up to height bound 10^{10}) we could only find one generator. The elliptic curves in question are recorded in the following table.

y^{2} = x^{3} − 385216x − 118546643  Δ = 2.0
y^{2} + xy = x^{3} − x^{2} − 21940631x − 10062163381  Δ = 2.6
y^{2} + xy = x^{3} − x^{2} − 130106786x − 418444299752  Δ = 2.5
y^{2} = x^{3} + x^{2} − 674939767x + 9768411280745  Δ = 3.5
y^{2} + xy + y = x^{3} + 1365438724x + 1088450102306  Δ = 3.0
y^{2} + xy = x^{3} + x^{2} + 13646956x + 36868880351052  Δ = 3.6
y^{2} + y = x^{3} + 463718380x − 1653282652263  Δ = 3.9
y^{2} + xy = x^{3} + x^{2} − 6811523942x + 180704627470189

In all but the last of these examples, we were able to prove that the rank is 1 by using Sage [Sag19] to compute an upper bound on the analytic rank. The parameter Δ we used for this calculation (see [Bob13]) is recorded in the right hand column. In the last example we obtained no rank bound better than 3, despite taking Δ = 4.0.

Unfortunately it was not practical to compute the pairing (6.4) in the experiment with H = 1000. So we are left with a large number of unresolved cases. Writing t = rank Γ for the number of generators known, there were 90 cases with (r_{2}, r_{4}, t) = (3, 3, 1), 14 cases with (r_{2}, r_{4}, t) = (4, 4, 2), and one each with (r_{2}, r_{4}, t) = (4, 4, 1), (5, 3, 1) and (5, 5, 3).

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University of Cambridge, DPMMS, Centre for Mathematical Sciences, Wilberforce Road, Cambridge CB3 0WB, UK

E-mail address: T.A.Fisher@dpmms.cam.ac.uk

University of Michigan, Department of Mathematics, Ann Arbor, MI 48109, USA

E-mail address: weiho@umich.edu

The Ohio State University, Department of Mathematics, Columbus, OH 43210, USA

E-mail address: park.2720@osu.edu