We use the invariant theory of binary quartics to give a new formula for the Cassels–Tate pairing on the 2-Selmer group of an elliptic curve. Unlike earlier methods, our formula does not require us to solve any conics. An important role in our construction is played by a certain K3 surface defined by a (2, 2, 2)-form.
publisher-imprint-nameSpringervolume-issue-count4issue-article-count28issue-toc-levels0issue-pricelist-year2022issue-copyright-holderSpringer Nature Switzerland AGissue-copyright-year2022article-contains-esmNoarticle-numbering-styleContentOnlyarticle-registration-date-year2022article-registration-date-month9article-registration-date-day2article-toc-levels0toc-levels0volume-typeRegularjournal-productArchiveJournalnumbering-styleContentOnlyarticle-collection-editorJonathan Bober, Andrew Booker, Min Lee, and Chloe Martindalearticle-grants-typeOpenChoicemetadata-grantOpenAccessabstract-grantOpenAccessbodypdf-grantOpenAccessbodyhtml-grantOpenAccessbibliography-grantOpenAccessesm-grantOpenAccessonline-firstfalsepdf-file-referenceBodyRef/PDF/40993_2022_Article_376.pdfpdf-typeTypesettarget-typeOnlinePDFissue-typeRegulararticle-typeOriginalPaperjournal-subject-primaryMathematicsjournal-subject-secondaryNumber Theoryjournal-subject-collectionMathematics and Statisticsopen-accesstrueIntroduction
Let E be an elliptic curve over a number field K. The Mordell-Weil theorem tells us that the abelian group E(K) is finitely generated, but there is no known algorithm guaranteed to compute its rank. Instead, for each integer n≥2\documentclass[12pt]{minimal}
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\begin{document}$$n \ge 2$$\end{document} there is an exact sequence of abelian groupsThe n-Selmer group S(n)(E/K)\documentclass[12pt]{minimal}
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\begin{document}$$S^{(n)}(E/K)$$\end{document} is finite and effectively computable. Computing S(n)(E/K)\documentclass[12pt]{minimal}
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\begin{document}$$S^{(n)}(E/K)$$\end{document} gives an upper bound for the rank of E(K), but this will be sharp only if the n-torsion of the Tate-Shafarevich group is trivial.
Cassels [4] showed that there is an alternating pairing⟨,⟩CT:S(n)(E/K)×S(n)(E/K)→Q/Z\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} \langle ~,~\rangle _{{{\text {CT}}}} : S^{(n)}(E/K) \times S^{(n)}(E/K) \rightarrow {{\mathbb {Q}}}/{{\mathbb {Z}}}\end{aligned}$$\end{document}whose kernel is the image of S(n2)(E/K)\documentclass[12pt]{minimal}
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\begin{document}$$S^{(n^2)}(E/K)$$\end{document}. By computing this pairing, our upper bound for the rank of E(K) improves from that obtained by n-descent to that obtained by n2\documentclass[12pt]{minimal}
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\begin{document}$$n^2$$\end{document}-descent. In view of the generalisation to abelian varieties, due to Tate, the pairing is known as the Cassels–Tate pairing.
Cassels [6] also described a method for computing the pairing in the case n=2\documentclass[12pt]{minimal}
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\begin{document}$$n=2$$\end{document}. His method involves solving conics over the field of definition of each 2-torsion point on E. More recently, Donnelly [10] found a method that only involves solving conics over K, and implemented this in Magma [3]. In this article we use the invariant theory of binary quartics to give a self-contained account of a version of his method that is relatively simple to implement.
Since this article was first written, Jiali Yan has written her PhD thesis [18], extending some of these ideas to Jacobians of genus 2 curves, and Bill Allombert has implemented our method for computing the pairing as part of the function ellrank in pari/gp [15]. I thank them both, and also Steve Donnelly and John Cremona, for useful discussions.
Binary quartics
A binary quartic over a field K is a homogeneous polynomial g∈K[x,z]\documentclass[12pt]{minimal}
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\begin{document}$$g \in K[x,z]$$\end{document} of degree 4. Binary quartics g1\documentclass[12pt]{minimal}
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\begin{document}$$g_1$$\end{document} and g2\documentclass[12pt]{minimal}
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\begin{document}$$g_2$$\end{document} are K-equivalent ifg2(x,z)=λ2g1(αx+γz,βx+δz)\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} g_2(x,z) = \lambda ^2 g_1 (\alpha x + \gamma z, \beta x + \delta z) \end{aligned}$$\end{document}for some λ,α,β,γ,δ∈K\documentclass[12pt]{minimal}
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\begin{document}$$\lambda , \alpha , \beta , \gamma , \delta \in K$$\end{document} with λ(αδ-βγ)≠0\documentclass[12pt]{minimal}
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\begin{document}$$\lambda (\alpha \delta - \beta \gamma ) \not = 0$$\end{document}. They are properlyK-equivalent if in addition λ(αδ-βγ)=±1\documentclass[12pt]{minimal}
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\begin{document}$$\lambda (\alpha \delta - \beta \gamma ) = \pm 1$$\end{document}. The invariants of the binary quarticg(x,z)=ax4+bx3z+cx2z2+dxz3+ez4\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} g(x,z) = a x^4 + b x^3 z + c x^2 z^2 + d x z^3 + e z^4 \end{aligned}$$\end{document}areI=12ae-3bd+c2,J=72ace-27ad2-27b2e+9bcd-2c3.\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} I&= 12 a e - 3 b d + c^2 , \\ J&= 72 a c e - 27 a d^2 - 27 b^2 e + 9 b c d - 2 c^3. \end{aligned}$$\end{document}The binary quartics g1\documentclass[12pt]{minimal}
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\begin{document}$$g_1$$\end{document} and g2\documentclass[12pt]{minimal}
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\begin{document}$$g_2$$\end{document} have invariants related by I(g2)=λ4(αδ-βγ)4I(g1)\documentclass[12pt]{minimal}
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\begin{document}$$I(g_2) = \lambda ^4 (\alpha \delta - \beta \gamma )^4 I(g_1)$$\end{document} and J(g2)=λ6(αδ-βγ)6J(g1)\documentclass[12pt]{minimal}
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\begin{document}$$J(g_2) = \lambda ^6 (\alpha \delta - \beta \gamma )^6 J(g_1)$$\end{document}. In particular, properly equivalent binary quartics have the same invariants. The discriminant is Δ=16(4I3-J2)/27\documentclass[12pt]{minimal}
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\begin{document}$$\Delta = 16(4 I^3 - J^2)/27$$\end{document}. We say that g is K-soluble if there exist x,z∈K\documentclass[12pt]{minimal}
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\begin{document}$$x,z \in K$$\end{document}, not both zero, such that g(x, z) is a square in K. The reason for this terminology is that if Δ(g)≠0\documentclass[12pt]{minimal}
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\begin{document}$$\Delta (g) \not =0$$\end{document} then there is a smooth projective curve C of genus one with affine equation y2=g(x,1)\documentclass[12pt]{minimal}
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\begin{document}$$y^2 = g(x,1)$$\end{document}, and we are asking that C(K)≠∅\documentclass[12pt]{minimal}
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\begin{document}$$C(K) \not = \emptyset $$\end{document}. As shown by Weil [17], the Jacobian of C is the elliptic curveEI,J:y2=x3-27Ix-27J.\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} E_{I,J} : \quad y^2 = x^3 - 27 I x - 27 J. \end{aligned}$$\end{document}Now let K be a number field, and MK\documentclass[12pt]{minimal}
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\begin{document}$$M_K$$\end{document} its set of places. A binary quartic over K is everywhere locally soluble if it is Kv\documentclass[12pt]{minimal}
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\begin{document}$$K_v$$\end{document}-soluble for all places v∈MK\documentclass[12pt]{minimal}
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\begin{document}$$v \in M_K$$\end{document}. We note that every elliptic curve over K can be written in the form (2) for some I,J∈K\documentclass[12pt]{minimal}
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\begin{document}$$I,J \in K$$\end{document} with 4I3-J2≠0\documentclass[12pt]{minimal}
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\begin{document}$$4 I^3 - J^2 \not =0$$\end{document}.
The case K=Q\documentclass[12pt]{minimal}
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\begin{document}$$K = {{\mathbb {Q}}}$$\end{document} is proved in [2], the only simplification in this case being that (since the only roots of unity in Q\documentclass[12pt]{minimal}
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\begin{document}$${{\mathbb {Q}}}$$\end{document} are ±1\documentclass[12pt]{minimal}
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\begin{document}$$\pm 1$$\end{document}) equivalent quartics with the same invariants are always properly equivalent (even in the cases where I=0\documentclass[12pt]{minimal}
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\begin{document}$$I=0$$\end{document} or J=0\documentclass[12pt]{minimal}
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\begin{document}$$J=0$$\end{document}). The general case is similar. □\documentclass[12pt]{minimal}
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\begin{document}$$\square $$\end{document}
Although Lemma 2.1 specifies S(2)(EI,J/K)\documentclass[12pt]{minimal}
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\begin{document}$$S^{(2)}(E_{I,J}/K)$$\end{document} as a set, the group law is not obvious. The following description is taken from [8, 9]. Let L be the étale algebra K[φ]\documentclass[12pt]{minimal}
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\begin{document}$$K[\varphi ]$$\end{document} where φ\documentclass[12pt]{minimal}
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\begin{document}$$\varphi $$\end{document} is a root of X3-3IX+J=0\documentclass[12pt]{minimal}
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\begin{document}$$X^3 - 3 I X + J = 0$$\end{document}. Then the binary quartic (1) has cubic invariantz(g)=4aφ+3b2-8ac3.\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} z(g) = \frac{4 a \varphi + 3 b^2 - 8 a c}{3}. \end{aligned}$$\end{document}By a change of coordinates (that is, replacing g by a properly equivalent quartic) we may assume that z(g) is a unit in L. The group law on S(2)(EI,J/K)\documentclass[12pt]{minimal}
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\begin{document}$$S^{(2)}(E_{I,J}/K)$$\end{document} is then given by multiplying the cubic invariants in L×/(L×)2\documentclass[12pt]{minimal}
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\begin{document}$$L^\times /(L^\times )^2$$\end{document}. The method for converting an element of L×/(L×)2\documentclass[12pt]{minimal}
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\begin{document}$$L^\times /(L^\times )^2$$\end{document} back to a binary quartic does, however, involve solving a conic over K.
Statement of results
In this section we state our new formula for the Cassels–Tate pairing on the 2-Selmer group of an elliptic curve. First we need some more invariant theory. The binary quartic (1) has Hessianh(x,z)=(3b2-8ac)x4+4(bc-6ad)x3z+2(2c2-24ae-3bd)x2z2+4(cd-6be)xz3+(3d2-8ce)z4.\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} h (x, z ) = (3&b^2 - 8ac)x^4 + 4(bc - 6ad)x^3 z + 2(2c^2 - 24ae - 3bd)x^2 z^2 \\ {}&+4(cd - 6be)xz^3 + (3d^2 - 8ce)z^4. \end{aligned}$$\end{document}There are exactly three linear combinations of g(x, z) and h(x, z) that are singular (i.e. have repeated roots). Following [9] this prompts us to putG(x,z)=134φg(x,z)+h(x,z),\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} G(x,z)&= \frac{1}{3} \left( 4 \varphi g(x,z) + h(x,z) \right) , \end{aligned}$$\end{document}H(x,z)=112∂2G∂x2+29(I-φ2)z2,\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} H(x,z)&= \frac{1}{12} \frac{\partial ^2 G}{\partial x^2} + \frac{2}{9} (I -\varphi ^2) z^2, \end{aligned}$$\end{document}so that G(1,0)G(x,z)=H(x,z)2\documentclass[12pt]{minimal}
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\begin{document}$$G(1,0) G(x,z) = H(x,z)^2$$\end{document}. We note that z(g)=G(1,0)=H(1,0)\documentclass[12pt]{minimal}
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\begin{document}$$z(g) = G(1,0) = H(1,0)$$\end{document}.
Theorem 3.1
Let I,J∈K\documentclass[12pt]{minimal}
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\begin{document}$$I,J \in K$$\end{document} with 4I3-J2≠0\documentclass[12pt]{minimal}
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\begin{document}$$4 I^3 - J^2 \not =0$$\end{document}. Let g1,g2,g3\documentclass[12pt]{minimal}
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\begin{document}$$g_1,g_2,g_3$$\end{document} be everywhere locally soluble binary quartics over K with invariants I and J. Let H1(x,z)\documentclass[12pt]{minimal}
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\begin{document}$$H_1(x,z)$$\end{document} be the binary quadratic form (4), with coefficients in L=K[φ]\documentclass[12pt]{minimal}
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\begin{document}$$L = K[\varphi ]$$\end{document}, associated to g1\documentclass[12pt]{minimal}
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\begin{document}$$g_1$$\end{document}. Suppose that z(g1)z(g2)z(g3)=m2\documentclass[12pt]{minimal}
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\begin{document}$$z(g_1) z(g_2) z(g_3) = m^2$$\end{document} for some m∈L×\documentclass[12pt]{minimal}
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\begin{document}$$m \in L^\times $$\end{document}, and writez(g2)z(g3)mH1(x,z)=α1(x,z)+β1(x,z)φ+γ1(x,z)φ2\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} \frac{z(g_2) z(g_3)}{m} H_1(x,z) = \alpha _1(x,z) + \beta _1(x,z) \varphi + \gamma _1(x,z) \varphi ^2 \end{aligned}$$\end{document}where α1,β1,γ1∈K[x,z]\documentclass[12pt]{minimal}
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\begin{document}$$\alpha _1,\beta _1,\gamma _1 \in K[x,z]$$\end{document}. For each v∈MK\documentclass[12pt]{minimal}
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\begin{document}$$v \in M_K$$\end{document} we choose xv,zv∈Kv\documentclass[12pt]{minimal}
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\begin{document}$$x_v,z_v \in K_v$$\end{document} with g1(xv,zv)\documentclass[12pt]{minimal}
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\begin{document}$$g_1(x_v,z_v)$$\end{document} a square in Kv\documentclass[12pt]{minimal}
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\begin{document}$$K_v$$\end{document} and γ1(xv,zv)≠0\documentclass[12pt]{minimal}
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\begin{document}$$g_2(1,0) \not =0$$\end{document} then the Cassels–Tate pairing on S(2)(EI,J/K)\documentclass[12pt]{minimal}
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\begin{document}$$S^{(2)}(E_{I,J}/K)$$\end{document} is given by⟨[g1],[g2]⟩CT=∏v∈MK(g2(1,0),γ1(xv,zv))v\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} \langle [g_1], [g_2] \rangle _{{{\text {CT}}}} = \prod _{v \in M_K} (g_2(1,0),\gamma _1(x_v,z_v))_v \end{aligned}$$\end{document}where (,)v:Kv×/(Kv×)2×Kv×/(Kv×)2→μ2\documentclass[12pt]{minimal}
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\begin{document}$$(~,~)_v : K_v^\times /(K_v^\times )^2 \times K_v^\times /(K_v^\times )^2 \rightarrow \mu _2$$\end{document} is the Hilbert norm residue symbol.
Remark 3.2
If we wish to compute the pairing starting only with g1\documentclass[12pt]{minimal}
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\begin{document}$$z(g_2)$$\end{document} are units in L, multiply these together, and then compute g3\documentclass[12pt]{minimal}
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\begin{document}$$g_3$$\end{document} by solving a conic over K. This conic is the same as the one that has to be solved in Donnelly’s method [10].
We show in Remark 8.3 that the binary quadratic form γ1\documentclass[12pt]{minimal}
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\begin{document}$$\gamma _1$$\end{document} is not identically zero. Therefore, by our assumption that g1\documentclass[12pt]{minimal}
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\begin{document}$$g_1$$\end{document} is everywhere locally soluble, it is always possible to choose xv,zv∈Kv\documentclass[12pt]{minimal}
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\begin{document}$$x_v, z_v \in K_v$$\end{document} with the stated properties.
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\begin{document}$$g_2(1,0) \not =0$$\end{document} is no limitation, since if g2(1,0)=0\documentclass[12pt]{minimal}
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\begin{document}$$[g_2] = 0$$\end{document} in the 2-Selmer group, which certainly implies the pairing is trivial.
By definition the Cassels–Tate pairing takes values in Q/Z\documentclass[12pt]{minimal}
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\begin{document}$${{\mathbb {Q}}}/{{\mathbb {Z}}}$$\end{document}. In our formula it takes values in μ2\documentclass[12pt]{minimal}
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\begin{document}$$\mu _2$$\end{document}. It should be understood that we have identified μ2=12Z/Z\documentclass[12pt]{minimal}
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\begin{document}$$\mu _2 = \frac{1}{2}{{\mathbb {Z}}}/{{\mathbb {Z}}}$$\end{document}.
Since ⟨,⟩CT\documentclass[12pt]{minimal}
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\begin{document}$$\langle ~,~\rangle _{{{\text {CT}}}}$$\end{document} is alternating and bilinear and [g1]+[g2]+[g3]=0\documentclass[12pt]{minimal}
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\begin{document}$$[g_1] + [g_2] + [g_3] = 0$$\end{document} we have ⟨[g1],[g2]⟩CT=⟨[g1],[g3]⟩CT\documentclass[12pt]{minimal}
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\begin{document}$$\langle [g_1], [g_2] \rangle _{{{\text {CT}}}} = \langle [g_1], [g_3] \rangle _{{{\text {CT}}}}$$\end{document}. So we may equally write g2(1,0)\documentclass[12pt]{minimal}
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\begin{document}$$g_2(1,0)$$\end{document} or g3(1,0)\documentclass[12pt]{minimal}
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\begin{document}$$g_3(1,0)$$\end{document} in (6). Notice however that the binary quartics g2\documentclass[12pt]{minimal}
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\begin{document}$$g_3$$\end{document} do both contribute to the pairing via (5). Moreover we must use the exact formulae for z(g1)\documentclass[12pt]{minimal}
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\begin{document}$$z(g_1)$$\end{document}, z(g2)\documentclass[12pt]{minimal}
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\begin{document}$$z(g_3)$$\end{document}, these being linear in φ\documentclass[12pt]{minimal}
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\begin{document}$$\varphi $$\end{document}. It is not enough just to know these quantities up to squares, since this would change the left hand side of (5).
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\begin{document}$$E(K)[2]=0$$\end{document} then m is uniquely determined up to sign. By the product formula for the Hilbert norm residue symbol this makes no difference to (6). If E(K)[2]≠0\documentclass[12pt]{minimal}
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\begin{document}$$E(K)[2] \not =0$$\end{document} then there are more choices for m, but it turns out (see the proof of Theorem 8.2) that we may use any one of these to compute the pairing.
Remark 3.3
The product over all places in Theorem 3.1 is a finite product. Indeed, outside an easily determined finite set of places, we have
v is a finite prime, with residue field of size at least 11.
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\begin{document}$$g_2(1,0)$$\end{document} is a v-adic unit and v∤2\documentclass[12pt]{minimal}
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\begin{document}$$v \not \mid 2$$\end{document}.
Under conditions (i) and (ii) we can pick our local point (by Hensel lifting a smooth point on the reduction that is not a root of γ1\documentclass[12pt]{minimal}
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\begin{document}$$\gamma _1$$\end{document}) such that γ1(xv,zv)\documentclass[12pt]{minimal}
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\begin{document}$$\gamma _1(x_v,z_v)$$\end{document} is a unit. It follows by (iii) that the local contribution at v is trivial.
Example 3.4
Let E/Q\documentclass[12pt]{minimal}
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\begin{document}$$E/{{\mathbb {Q}}}$$\end{document} be the elliptic curvey2+y=x3-x2-929x-10595\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} y^2 + y = x^3 - x^2 - 929 x - 10595 \end{aligned}$$\end{document}labelled 571a1 in [7]. A 2-descent shows that S(2)(E/Q)≅(Z/2Z)2\documentclass[12pt]{minimal}
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\begin{document}$$S^{(2)}(E/{{\mathbb {Q}}}) \cong ({{\mathbb {Z}}}/2{{\mathbb {Z}}})^2$$\end{document}, and its non-zero elements are represented byg1(x,z)=-11x4+68x3z-52x2z2-164xz3-64z4,g2(x,z)=-4x4-60x3z-232x2z2-52xz3-3z4,g3(x,z)=-31x4-78x3z+32x2z2+102xz3-53z4.\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} g_1(x,z)&= -11 x^4 + 68 x^3 z - 52 x^2 z^2 - 164 x z^3 - 64 z^4, \\ g_2(x,z)&= -4 x^4 - 60 x^3 z - 232 x^2 z^2 - 52 x z^3 - 3 z^4, \\ g_3(x,z)&= -31 x^4 - 78 x^3 z + 32 x^2 z^2 + 102 x z^3 - 53 z^4. \end{aligned}$$\end{document}Each of these binary quartics has invariants I=44608\documentclass[12pt]{minimal}
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\begin{document}$$I=44608$$\end{document} and J=18842960\documentclass[12pt]{minimal}
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\begin{document}$$J=18842960$$\end{document}, and discriminant Δ=-212·571\documentclass[12pt]{minimal}
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\begin{document}$$\Delta = -2^{12} \cdot 571$$\end{document}. By (5), with m=19(20φ2-8656φ+936032)\documentclass[12pt]{minimal}
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\begin{document}$$m = \frac{1}{9}(20 \varphi ^2 - 8656 \varphi + 936032)$$\end{document}, we getγ1(x,z)=49(5x2-16xz-12z2).\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned}\gamma _1(x,z) = \frac{4}{9} (5 x^2 - 16 x z - 12 z^2).\end{aligned}$$\end{document}For each odd prime p there is a smooth Fp\documentclass[12pt]{minimal}
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\begin{document}$${{\mathbb {F}}}_p$$\end{document}-point on the reduction of y2=g1(x,1)\documentclass[12pt]{minimal}
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\begin{document}$$y^2 = g_1(x,1)$$\end{document} mod p, whose x-coordinate is not a root of 5x2-16x-12=0\documentclass[12pt]{minimal}
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\begin{document}$$5 x^2 - 16 x - 12 = 0$$\end{document}. Indeed we checked this claim directly for p=3,5,7,11\documentclass[12pt]{minimal}
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\begin{document}$$p=3,5,7,11$$\end{document} and 571, and for all other primes it follows by Hasse’s bound. Therefore the odd primes make no contribution to (6).
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\begin{document}$$p=2$$\end{document} we write g1(x,1)=x4+4q(x)\documentclass[12pt]{minimal}
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\begin{document}$$g_1(x,1) = x^4 + 4 q(x)$$\end{document} whereq(x)=-3x4+17x3-13x2-41x-16.\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} q(x) = -3 x^4 + 17 x^3 - 13 x^2 - 41 x - 16. \end{aligned}$$\end{document}By Hensel’s lemma the equation q(x)=0\documentclass[12pt]{minimal}
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\begin{document}$$x = 2^4 + O(2^5)$$\end{document}. But then γ1(x,1)≡5mod(Q2×)2\documentclass[12pt]{minimal}
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\begin{document}$$(5,-1)_2=1$$\end{document} the contribution is again trivial. Finally, since g1(15,4)>0\documentclass[12pt]{minimal}
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\begin{document}$$\gamma _1(15,4) < 0$$\end{document}, there is a contribution from the real place. This shows that the Cassels–Tate pairing on S(2)(E/Q)\documentclass[12pt]{minimal}
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\begin{document}$$S^{(2)}(E/{{\mathbb {Q}}})$$\end{document} is non-trivial, and hence rankE(Q)=0\documentclass[12pt]{minimal}
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\begin{document}$${\text {rank}}E ({{\mathbb {Q}}}) = 0$$\end{document}.
The Cassels–Tate pairing
There are two standard definitions of the Cassels–Tate pairing (in the case of elliptic curves) called in [11, 16] the homogeneous space definition and the Weil pairing definition. Both definitions appear in Cassels’ original paper [4], although the method in [6] (see also [12]) is a variant of the Weil pairing definition. In this section we review the homogeneous space definition, and highlight its connection with the Brauer-Manin obstruction.
Let K be a field with separable closure K¯\documentclass[12pt]{minimal}
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\begin{document}$$H^i({{\text {Gal}}}({{\overline{K}}}/K),-)$$\end{document}. Let C/K be a smooth projective curve. We defineBr(C)=ker(H2(K,K¯(C)×)→H2(K,DivC)).\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} {\text {Br}}(C) = \ker \bigg ( H^2(K,{{\overline{K}}}(C)^\times ) \rightarrow H^2(K, {\text {Div}}C) \bigg ). \end{aligned}$$\end{document}It is shown in the Appendix to [14] that this is equivalent to the usual definition Br(C)=He´t2(C,Gm)\documentclass[12pt]{minimal}
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\begin{document}$${\text {Br}}(C) = H^2_{\acute{\textrm{e}}\text {t}}(C,{{\mathbb {G}}}_m)$$\end{document}. Identifying Br(K)=H2(K,K¯×)\documentclass[12pt]{minimal}
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\begin{document}$${\text {Br}}(K) = H^2(K,{{\overline{K}}}^\times )$$\end{document}, there is a natural mapBr(K)→Br(C).\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} {\text {Br}}(K) \rightarrow {\text {Br}}(C). \end{aligned}$$\end{document}We will need the following two facts, whose proofs we give below.
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\begin{document}$$P \in C(K)$$\end{document} there is an evaluation map Br(C)→Br(K);A↦A(P).\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} {\text {Br}}(C) \rightarrow {\text {Br}}(K) \, ; \,\, A \mapsto A(P). \end{aligned}$$\end{document} This is a group homomorphism, and a section to the map (8). Moreover the evaluation maps behave functorially with respect to all field extensions.
Suppose C is a smooth curve of genus one, with Jacobian elliptic curve E. If H3(K,K¯×)=0\documentclass[12pt]{minimal}
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\begin{document}$$H^3(K,{{\overline{K}}}^\times ) = 0$$\end{document} then there is an isomorphism ΨC:H1(K,E)⟨[C]⟩⟶∼Br(C)Br(K).\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} \Psi _C : \frac{H^1(K,E)}{\langle [C]\rangle } {\mathop {\longrightarrow }\limits ^{\sim }} \frac{{\text {Br}}(C)}{{\text {Br}}(K)}. \end{aligned}$$\end{document}
Now let E be an elliptic curve over a number field K. Let C and D be principal homogeneous spaces under E. Since H3(K,K¯×)=0\documentclass[12pt]{minimal}
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\begin{document}$$H^3(K,{{\overline{K}}}^\times ) = 0$$\end{document} for K a number field, we have ΨC([D])=AmodBr(K)\documentclass[12pt]{minimal}
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\begin{document}$$\Psi _C([D]) = A \mod {\text {Br}}(K)$$\end{document} for some A∈Br(C)\documentclass[12pt]{minimal}
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\begin{document}$$A \in {\text {Br}}(C)$$\end{document}. Now suppose that C and D are everywhere locally soluble. For each place v∈MK\documentclass[12pt]{minimal}
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\begin{document}$$v \in M_K$$\end{document} we pick a local point Pv∈C(Kv)\documentclass[12pt]{minimal}
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\begin{document}$$P_v \in C(K_v)$$\end{document}. The Cassels–Tate pairing is defined by⟨[C],[D]⟩CT=∑v∈MKinvv(A(Pv))\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} \langle [C], [D] \rangle _{{{\text {CT}}}} = \sum _{v \in M_K} {\text {inv}}_v (A(P_v)) \end{aligned}$$\end{document}where invv:Br(Kv)→Q/Z\documentclass[12pt]{minimal}
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\begin{document}$${\text {inv}}_v : {\text {Br}}(K_v) \rightarrow {{\mathbb {Q}}}/{{\mathbb {Z}}}$$\end{document} is the local invariant map. As this form of the definition makes clear, if ⟨[C],[D]⟩CT≠0\documentclass[12pt]{minimal}
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\begin{document}$$\langle [C], [D] \rangle _{{{\text {CT}}}} \not =0$$\end{document} then the genus one curve C is a counter-example to the Hasse Principle explained by the Brauer-Manin obstruction.
We check that the pairing is well defined, i.e. it does not depend on the choices of A and of the Pv\documentclass[12pt]{minimal}
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\begin{document}$$P_v$$\end{document}. By class field theory there is an exact sequence0⟶Br(K)⟶⨁v∈MKBr(Kv)⟶∑invvQ/Z⟶0.\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} 0 \longrightarrow {\text {Br}}(K) \longrightarrow \bigoplus _{v \in M_K} {\text {Br}}(K_v) {\mathop {\longrightarrow }\limits ^{\sum {\text {inv}}_v}} {{\mathbb {Q}}}/{{\mathbb {Z}}}\longrightarrow 0. \end{aligned}$$\end{document}It follows that if we change A by adding an element of Br(K)\documentclass[12pt]{minimal}
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\begin{document}$${\text {Br}}(K)$$\end{document} then the pairing (10) is unchanged. Next, since the class of D is trivial in H1(Kv,E)\documentclass[12pt]{minimal}
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\begin{document}$$H^1(K_v,E)$$\end{document}, the analogue of (9) over Kv\documentclass[12pt]{minimal}
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\begin{document}$$K_v$$\end{document} shows that the restriction of A to the Brauer group of C/Kv\documentclass[12pt]{minimal}
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\begin{document}$$C/K_v$$\end{document} is constant, i.e. it belongs to the image of Br(Kv)\documentclass[12pt]{minimal}
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\begin{document}$${\text {Br}}(K_v)$$\end{document}. Therefore the pairing (10) does not depend on the choice of local points Pv\documentclass[12pt]{minimal}
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\begin{document}$$P_v$$\end{document}.
We now prove the facts we quoted in (i) and (ii) above.
For P∈C(K)\documentclass[12pt]{minimal}
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\begin{document}$$P \in C(K)$$\end{document} there is a short exact sequence of Galois modules 0⟶OP×⟶K¯(C)×⟶ordPZ⟶0\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} 0 \longrightarrow {{\mathcal {O}}}_P^\times \longrightarrow {{\overline{K}}}(C)^\times {\mathop {\longrightarrow }\limits ^{{{\text {ord}}}_P}} {{\mathbb {Z}}}\longrightarrow 0 \end{aligned}$$\end{document} where OP\documentclass[12pt]{minimal}
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\begin{document}$${{\mathcal {O}}}_P$$\end{document} is the local ring at P. Taking Galois cohomology gives an exact sequence 0⟶H2(K,OP×)⟶H2(K,K¯(C)×)⟶ordPH2(K,Z).\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} 0 \longrightarrow H^2(K,{{\mathcal {O}}}_P^\times ) \longrightarrow H^2(K,{{\overline{K}}}(C)^\times ) {\mathop {\longrightarrow }\limits ^{{{\text {ord}}}_P}} H^2(K,{{\mathbb {Z}}}). \end{aligned}$$\end{document} It follows by (7) that each element of Br(C)\documentclass[12pt]{minimal}
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\begin{document}$${\text {Br}}(C)$$\end{document} can be represented by a cocycle taking values in OP×\documentclass[12pt]{minimal}
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\begin{document}$${{\mathcal {O}}}_P^\times $$\end{document}, and so can be evaluated at P.
There is an exact sequence of Galois modules 0→K¯×→K¯(C)×→DivC→PicC→0,\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} 0 \rightarrow {{\overline{K}}}^\times \rightarrow {{\overline{K}}}(C)^\times \rightarrow {\text {Div}}C \rightarrow {\text {Pic}}C \rightarrow 0, \end{aligned}$$\end{document} where DivC\documentclass[12pt]{minimal}
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\begin{document}$${\text {Div}}C$$\end{document} and PicC\documentclass[12pt]{minimal}
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\begin{document}$${\text {Pic}}C$$\end{document} are the divisor group and Picard group for C over K¯\documentclass[12pt]{minimal}
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\begin{document}$${{\overline{K}}}$$\end{document}. Splitting into short exact sequences, and taking Galois cohomology, gives the following exact sequences By Shapiro’s lemma and the fact that H1(K,Z)=0\documentclass[12pt]{minimal}
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\begin{document}$$H^1(K,{{\mathbb {Z}}})=0$$\end{document} we have H1(K,DivC)=0\documentclass[12pt]{minimal}
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\begin{document}$$H^1(K, {\text {Div}}C)=0$$\end{document}. It follows by (7) and a diagram chase that there is an exact sequence Br(K)→Br(C)→H1(K,PicC)→H3(K,K¯×).\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} {\text {Br}}(K) \rightarrow {\text {Br}}(C) \rightarrow H^1(K, {\text {Pic}}C) \rightarrow H^3(K,{{\overline{K}}}^\times ). \end{aligned}$$\end{document} In fact, had we started from the definition Br(C)=He´t2(C,Gm)\documentclass[12pt]{minimal}
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\begin{document}$${\text {Br}}(C) = H^2_{\acute{\textrm{e}}\text {t}}(C,{{\mathbb {G}}}_m)$$\end{document}, then (11) would follow from the Hochschild–Serre spectral sequence. If C is a smooth curve of genus one with Jacobian E, then taking Galois cohomology of the exact sequence 0⟶Pic0C⟶PicC⟶degZ⟶0\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} 0 \longrightarrow {\text {Pic}}^0 C \longrightarrow {\text {Pic}}C {\mathop {\longrightarrow }\limits ^{\deg }} {{\mathbb {Z}}}\longrightarrow 0 \end{aligned}$$\end{document} gives Z⟶δH1(K,E)⟶H1(K,PicC)⟶0\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} {{\mathbb {Z}}}{\mathop {\longrightarrow }\limits ^{\delta }} H^1(K,E) \longrightarrow H^1(K, {\text {Pic}}C) \longrightarrow 0 \end{aligned}$$\end{document} with δ(1)=[C]\documentclass[12pt]{minimal}
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\begin{document}$$\delta (1) = [C]$$\end{document}. If H3(K,K¯×)=0\documentclass[12pt]{minimal}
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\begin{document}$$H^3(K,{{\overline{K}}}^\times ) = 0$$\end{document} then from (11) and (12) we obtain the isomorphism ΨC\documentclass[12pt]{minimal}
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\begin{document}$$\Psi _C$$\end{document}.
Cyclic extensions
The definition of the map ΨC\documentclass[12pt]{minimal}
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\begin{document}$$\Psi _C$$\end{document} in the last section simplifies when we evaluate it on classes split by a cyclic extension L/K. Let G=Gal(L/K)\documentclass[12pt]{minimal}
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\begin{document}$$G = {{\text {Gal}}}(L/K)$$\end{document} be generated by σ\documentclass[12pt]{minimal}
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\begin{document}$$\sigma $$\end{document} of order n. We recall that for A a G-module, the Tate cohomology groups areH^0(G,A)=ker(Δ|A)im(N|A)andH^1(G,A)=ker(N|A)im(Δ|A)\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} {\widehat{H}}^0(G,A) = \frac{\ker (\Delta | A)}{{\text {im}}(N | A)} \quad \text { and } \quad {\widehat{H}}^1(G,A) = \frac{\ker (N | A)}{{\text {im}}(\Delta | A)} \end{aligned}$$\end{document}where Δ=1-σ\documentclass[12pt]{minimal}
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\begin{document}$$N = 1 + \sigma + \ldots + \sigma ^{n-1}$$\end{document} satisfy ΔN=NΔ=0\documentclass[12pt]{minimal}
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\begin{document}$${{\mathbb {Z}}}[G]$$\end{document}.
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\begin{document}$$b \in K^\times $$\end{document} there is a cyclic K-algebra with basis 1,v,…,vn-1\documentclass[12pt]{minimal}
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\begin{document}$$v x = \sigma (x) v$$\end{document} for all x∈L\documentclass[12pt]{minimal}
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\begin{document}$$x \in L$$\end{document}. We write (L/K, b) for the class of this algebra in Br(K)=H2(K,K¯×)\documentclass[12pt]{minimal}
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\begin{document}$${\text {Br}}(K)=H^2(K,{{\overline{K}}}^\times )$$\end{document}. Likewise if f∈K(C)×\documentclass[12pt]{minimal}
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\begin{document}$$f \in K(C)^\times $$\end{document} then (L/K, f) is an element of H2(K,K¯(C)×)\documentclass[12pt]{minimal}
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\begin{document}$$H^2(K,{{\overline{K}}}(C)^\times )$$\end{document}.
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\begin{document}$$\Psi _C$$\end{document} in Sect. 4. We start with the exact sequence of G-modules0→L×→L(C)×→DivLC→PicLC→0.\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} 0 \rightarrow L^\times \rightarrow L(C)^\times \rightarrow {\text {Div}}_L C \rightarrow {\text {Pic}}_L C \rightarrow 0. \end{aligned}$$\end{document}Splitting into short exact sequences, and taking Galois cohomology, gives a diagram as before. The connecting mapH^1(G,PicLC)→H^0(G,L(C)×/L×)\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} {\widehat{H}}^1(G,{\text {Pic}}_L C) \rightarrow {\widehat{H}}^0(G,L(C)^\times /L^\times ) \end{aligned}$$\end{document}is now given by Ξ↦f\documentclass[12pt]{minimal}
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\begin{document}$$\Xi \mapsto f$$\end{document}. Therefore ΨC(ξ)\documentclass[12pt]{minimal}
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\begin{document}$$\Psi _C(\xi )$$\end{document} is the image of f under the mapK(C)×NL/K(L(C)×)=H^0(G,L(C)×)≅H2(G,L(C)×)⟶infH2(K,K¯(C)×).\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} \frac{K(C)^\times }{N_{L/K}(L(C)^\times )} = {\widehat{H}}^0(G,L(C)^\times ) \cong H^2(G,L(C)^\times ) {\mathop {\longrightarrow }\limits ^\textrm{inf}} H^2(K,{{\overline{K}}}(C)^\times ). \end{aligned}$$\end{document}This is the cyclic algebra (L/K, f) as required. □\documentclass[12pt]{minimal}
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\begin{document}$$\square $$\end{document}
Pairs of binary quartics and (2, 2)-forms
Let C be a smooth curve of genus one. First suppose, as in Sect. 2, that C is defined by a binary quartic g. Then C→P1\documentclass[12pt]{minimal}
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\begin{document}$$Q \in C$$\end{document}.
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\begin{document}$$z_2$$\end{document}. Projecting C to either factor gives a double cover of P1\documentclass[12pt]{minimal}
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\begin{document}$${{\mathbb {P}}}^1$$\end{document}. The corresponding binary quartics are obtained by writing f as a binary quadratic form in one of the sets of variables, and taking its discriminant. We write pr1,pr2:C→P1\documentclass[12pt]{minimal}
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\begin{document}$${\text {pr}}_1, {\text {pr}}_2 : C \rightarrow {{\mathbb {P}}}^1$$\end{document} for the projection maps. Let H1,H2\documentclass[12pt]{minimal}
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\begin{document}$$H_1, H_2$$\end{document} and ι1,ι2\documentclass[12pt]{minimal}
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\begin{document}$$\iota _1, \iota _2$$\end{document} be the corresponding hyperplane sections and involutions.
If Q∈C\documentclass[12pt]{minimal}
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\begin{document}$$Q \in C$$\end{document} then ι2Q+ι1ι2Q∼H1\documentclass[12pt]{minimal}
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\begin{document}$$\iota _2 Q + \iota _1 \iota _2 Q \sim H_1$$\end{document} and Q+ι2Q∼H2\documentclass[12pt]{minimal}
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\begin{document}$$Q + \iota _2 Q \sim H_2$$\end{document}. Subtracting one from the other gives [ι1ι2Q-Q]=[H1-H2]\documentclass[12pt]{minimal}
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\begin{document}$$[\iota _1 \iota _2 Q - Q ] = [H_1 - H_2]$$\end{document} as required.
The specified rational function on C factors via pr1\documentclass[12pt]{minimal}
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\begin{document}$${\text {pr}}_1$$\end{document} and is therefore invariant under pull back by ι1\documentclass[12pt]{minimal}
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\begin{document}$$\iota _1$$\end{document}. It has a zero at each point in the support of H2\documentclass[12pt]{minimal}
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\begin{document}$$H_2$$\end{document}, and a double pole at each point in the support of H1\documentclass[12pt]{minimal}
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\begin{document}$$H_1$$\end{document}. Since there are no other poles, and the divisor has degree 0, it must therefore be as stated.
Lemma 6.1(i) is closely related to Poncelet’s Porism, as described in [13]. Our use of (2, 2)-forms was inspired by the treatment in [1].
We write disck(f)\documentclass[12pt]{minimal}
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\begin{document}$${\text {disc}}_k(f)$$\end{document} for the discriminant of f when it is viewed as a binary quadratic form in the kth set of variables.
Lemma 6.3
Let C={f(x1,z1;x2,z2)=0}⊂P1×P1\documentclass[12pt]{minimal}
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\begin{document}$$C = \{ f(x_1,z_1;x_2,z_2) = 0 \} \subset {{\mathbb {P}}}^1 \times {{\mathbb {P}}}^1$$\end{document} as above. Let a∈K×\documentclass[12pt]{minimal}
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\begin{document}$$a \in K^\times $$\end{document} and let C1\documentclass[12pt]{minimal}
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\begin{document}$$C_1$$\end{document}, C2\documentclass[12pt]{minimal}
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\begin{document}$$C_2$$\end{document} be the following quadratic twists of C.C1:ay2=disc2(f)C2:ay2=disc1(f)\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} C_1: \qquad a y^2&= {\text {disc}}_2(f) \\ C_2: \qquad a y^2&= {\text {disc}}_1(f) \end{aligned}$$\end{document}Then ΨC1([C2])=(K(a)/K,f(x1,z1;1,0)/z12)\documentclass[12pt]{minimal}
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\begin{document}$$\Psi _{C_1}([C_2]) = (K(\sqrt{a})/K, f(x_1,z_1;1,0)/z_1^2)$$\end{document}.
Proof
If a∈(K×)2\documentclass[12pt]{minimal}
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\begin{document}$$a \in (K^\times )^2$$\end{document} then C1\documentclass[12pt]{minimal}
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\begin{document}$$C_1$$\end{document} and C2\documentclass[12pt]{minimal}
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\begin{document}$$C_2$$\end{document} are isomorphic over K and so by (9) we have ΨC1([C2])=0\documentclass[12pt]{minimal}
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\begin{document}$$\Psi _{C_1}([C_2]) = 0$$\end{document}. We may therefore suppose that a∉(K×)2\documentclass[12pt]{minimal}
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\begin{document}$$a \not \in (K^\times )^2$$\end{document}. Let L=K(a)\documentclass[12pt]{minimal}
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\begin{document}$$L = K(\sqrt{a})$$\end{document} and G=Gal(L/K)={1,σ}\documentclass[12pt]{minimal}
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\begin{document}$$G = {{\text {Gal}}}(L/K) = \{1, \sigma \}$$\end{document}. We claim there is a divisor Ξ∈DivL0(C1)\documentclass[12pt]{minimal}
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\begin{document}$$\Xi \in {\text {Div}}_L^0(C_1)$$\end{document} such that
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\begin{document}$$C_2$$\end{document} is the twist of C1\documentclass[12pt]{minimal}
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\begin{document}$$C_1$$\end{document} by the class of Ξ\documentclass[12pt]{minimal}
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\begin{document}$$\Xi $$\end{document} in H^1(G,PicL0(C1))\documentclass[12pt]{minimal}
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\begin{document}$${\widehat{H}}^1(G,{\text {Pic}}_L^0(C_1))$$\end{document}, and
Let E/K be an elliptic curve. An n-covering of E is a pair (C,ν)\documentclass[12pt]{minimal}
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\begin{document}$$(C,\nu )$$\end{document} where C is a smooth curve of genus one, and ν:C→E\documentclass[12pt]{minimal}
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\begin{document}$$\nu : C \rightarrow E$$\end{document} is a morphism, such that, for some choice of isomorphism ψ:C→E\documentclass[12pt]{minimal}
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\begin{document}$$\psi : C \rightarrow E$$\end{document} defined over K¯\documentclass[12pt]{minimal}
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\begin{document}$${{\overline{K}}}$$\end{document}, there is a commutative diagramThe n-coverings of E are parametrised by H1(K,E[n])\documentclass[12pt]{minimal}
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\begin{document}$$H^1(K,E[n])$$\end{document}.
Suppose that C1\documentclass[12pt]{minimal}
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\begin{document}$$C_1$$\end{document}, C2\documentclass[12pt]{minimal}
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\begin{document}$$C_2$$\end{document}, C3\documentclass[12pt]{minimal}
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\begin{document}$$C_3$$\end{document} are 2-coverings of E that sum to zero in H1(K,E[2])\documentclass[12pt]{minimal}
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\begin{document}$$H^1(K,E[2])$$\end{document}. We pick isomorphisms ψi:Ci→E\documentclass[12pt]{minimal}
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\begin{document}$$\psi _i : C_i \rightarrow E$$\end{document} as above, and let εi=(σ↦σ(ψi)ψi-1)\documentclass[12pt]{minimal}
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\begin{document}$$\varepsilon _i = (\sigma \mapsto \sigma (\psi _i) \psi _i^{-1})$$\end{document} be the corresponding cocycle in Z1(Gal(K¯/K),E[2])\documentclass[12pt]{minimal}
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\begin{document}$$Z^1({{\text {Gal}}}({{\overline{K}}}/K),E[2])$$\end{document}. Our hypothesis is that ε1+ε2+ε3\documentclass[12pt]{minimal}
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\begin{document}$$\varepsilon _1 + \varepsilon _2 + \varepsilon _3$$\end{document} is a coboundary. However, by adjusting the choice of ψ3\documentclass[12pt]{minimal}
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\begin{document}$$\psi _3$$\end{document}, we may suppose that ε1+ε2+ε3=0\documentclass[12pt]{minimal}
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\begin{document}$$\varepsilon _1 + \varepsilon _2 + \varepsilon _3 = 0$$\end{document}. It may then be checked that the morphismμ:C1×C2×C3→E(P1,P2,P3)↦ψ1(P1)+ψ2(P2)+ψ3(P3)\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} \mu : C_1 \times C_2 \times C_3&\rightarrow E \\ (P_1,P_2,P_3)&\mapsto \psi _1(P_1) + \psi _2(P_2) + \psi _3(P_3) \end{aligned}$$\end{document}is defined over K.
Remark 7.1
We are still free to replace ψ3\documentclass[12pt]{minimal}
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\begin{document}$$\psi _3$$\end{document} by P↦ψ3(P)+T\documentclass[12pt]{minimal}
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\begin{document}$$P \mapsto \psi _3(P) + T$$\end{document} for T∈E(K)[2]\documentclass[12pt]{minimal}
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\begin{document}$$T \in E(K)[2]$$\end{document}, and for this reason there are #E(K)[2]\documentclass[12pt]{minimal}
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\begin{document}$$\# E(K)[2]$$\end{document} choices for the map μ\documentclass[12pt]{minimal}
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\begin{document}$$\mu $$\end{document}.
Suppose further that C1\documentclass[12pt]{minimal}
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\begin{document}$$C_1$$\end{document}, C2\documentclass[12pt]{minimal}
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\begin{document}$$C_2$$\end{document}, C3\documentclass[12pt]{minimal}
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\begin{document}$$C_3$$\end{document} are defined by binary quartics g1\documentclass[12pt]{minimal}
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\begin{document}$$g_1$$\end{document}, g2\documentclass[12pt]{minimal}
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\begin{document}$$g_2$$\end{document}, g3\documentclass[12pt]{minimal}
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\begin{document}$$g_3$$\end{document} with the same invariants I and J. Let π:C1×C2×C3→P1×P1×P1\documentclass[12pt]{minimal}
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\begin{document}$$\pi : C_1 \times C_2 \times C_3 \rightarrow {{\mathbb {P}}}^1 \times {{\mathbb {P}}}^1 \times {{\mathbb {P}}}^1$$\end{document} be the map that projects to the x-coordinates. Then S=π(μ-1(0E))\documentclass[12pt]{minimal}
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\begin{document}$$S = \pi (\mu ^{-1}(0_E))$$\end{document} is a surface in P1×P1×P1\documentclass[12pt]{minimal}
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\begin{document}$${{\mathbb {P}}}^1 \times {{\mathbb {P}}}^1 \times {{\mathbb {P}}}^1$$\end{document}. Geometrically it is the Kummer surface (E×E)/{±1}\documentclass[12pt]{minimal}
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\begin{document}$$(E \times E)/ \{\pm 1\}$$\end{document}.
We write disck(F)\documentclass[12pt]{minimal}
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\begin{document}$${\text {disc}}_k(F)$$\end{document} for the discriminant of a (2, 2, 2) form F when it is viewed as a binary quadratic form in the kth set of variables.
Proposition 7.2
The surface S⊂P1×P1×P1\documentclass[12pt]{minimal}
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\begin{document}$$S \subset {{\mathbb {P}}}^1 \times {{\mathbb {P}}}^1 \times {{\mathbb {P}}}^1$$\end{document} is defined by a (2, 2, 2)-form F. Moreover we may scale F so that it has coefficients in K, and for all permutations i, j, k of 1, 2, 3 we have disck(F)=gigj\documentclass[12pt]{minimal}
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\begin{document}$${\text {disc}}_k(F) = g_i g_j$$\end{document}.
Proof
We first consider the special case where C1=C2=C3=E\documentclass[12pt]{minimal}
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\begin{document}$$C_1 = C_2 = C_3 = E$$\end{document}. Suppose that P1,P2,P3∈E\documentclass[12pt]{minimal}
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\begin{document}$$P_1,P_2,P_3 \in E$$\end{document} satisfy P1+P2+P3=0E\documentclass[12pt]{minimal}
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\begin{document}$$P_1 + P_2 + P_3 = 0_E$$\end{document}. If we specify the x-coordinates of P1\documentclass[12pt]{minimal}
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\begin{document}$$P_1$$\end{document} and P2\documentclass[12pt]{minimal}
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\begin{document}$$P_2$$\end{document}, then in general this leaves two possibilities for the x-coordinate of P3\documentclass[12pt]{minimal}
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\begin{document}$$P_3$$\end{document}. The exceptional cases are when either P1\documentclass[12pt]{minimal}
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\begin{document}$$P_1$$\end{document} or P2\documentclass[12pt]{minimal}
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\begin{document}$$P_2$$\end{document} is a 2-torsion point.
Let Δi=ψi-1(E[2])\documentclass[12pt]{minimal}
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\begin{document}$$\Delta _i = \psi _i^{-1} (E[2])$$\end{document} be the set of ramification points for Ci→P1\documentclass[12pt]{minimal}
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\begin{document}$$C_i \rightarrow {{\mathbb {P}}}^1$$\end{document}. We identify Δi\documentclass[12pt]{minimal}
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\begin{document}$$\Delta _i$$\end{document} with its image in P1\documentclass[12pt]{minimal}
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\begin{document}$${{\mathbb {P}}}^1$$\end{document}, i.e., the set of roots of gi\documentclass[12pt]{minimal}
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\begin{document}$$g_i$$\end{document}. The observations in the last paragraph show that when we project onto the ith and jth factors, S→P1×P1\documentclass[12pt]{minimal}
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\begin{document}$$S \rightarrow {{\mathbb {P}}}^1 \times {{\mathbb {P}}}^1$$\end{document} is a double cover ramified over Δi×P1\documentclass[12pt]{minimal}
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\begin{document}$${{\mathbb {P}}}^1 \times \Delta _j$$\end{document}. This shows that S is defined by a (2, 2, 2)-form F. Moreover disck(F)=λkgigj\documentclass[12pt]{minimal}
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\begin{document}$${\text {disc}}_k(F) = \lambda _k g_i g_j$$\end{document} for some λ1,λ2,λ3∈K×\documentclass[12pt]{minimal}
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\begin{document}$$\lambda _1, \lambda _2, \lambda _3 \in K^\times $$\end{document}. We claim that (i) λ3∈(K×)2\documentclass[12pt]{minimal}
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\begin{document}$$\lambda _3 \in (K^\times )^2$$\end{document} and (ii) λ1=λ2=λ3\documentclass[12pt]{minimal}
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\begin{document}$$\lambda _1 = \lambda _2 = \lambda _3$$\end{document}. It is then clear we may rescale F so that λ1=λ2=λ3=1\documentclass[12pt]{minimal}
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\begin{document}$$\lambda _1 = \lambda _2 = \lambda _3=1$$\end{document}.
Let Ci\documentclass[12pt]{minimal}
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\begin{document}$$C_i$$\end{document} have equation yi2=gi(xi,zi)\documentclass[12pt]{minimal}
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\begin{document}$$y_i^2 = g_i(x_i,z_i)$$\end{document}. We note that K(S)⊂K(C1×C2)\documentclass[12pt]{minimal}
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\begin{document}$$K(S) \subset K(C_1 \times C_2)$$\end{document} is a quadratic extension of K(P1×P1)\documentclass[12pt]{minimal}
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\begin{document}$$K({{\mathbb {P}}}^1 \times {{\mathbb {P}}}^1)$$\end{document} with Kummer generator disc3(F)z14z24=λ3g1(x1,z1)g2(x2,z2)z14z24=λ3y1y2z12z222.\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} \frac{{\text {disc}}_3(F)}{z_1^4 z_2^4} = \frac{ \lambda _3 g_1(x_1,z_1) g_2(x_2,z_2)}{z_1^4 z_2^4} = \lambda _3 \left( \frac{ y_1 y_2 }{z_1^2 z_2^2} \right) ^2. \end{aligned}$$\end{document} Since this is a square in K(C1×C2)\documentclass[12pt]{minimal}
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\begin{document}$$K(C_1 \times C_2)$$\end{document} it follows that λ3∈(K×)2\documentclass[12pt]{minimal}
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\begin{document}$$\lambda _3 \in (K^\times )^2$$\end{document}.
Since g1,g2,g3\documentclass[12pt]{minimal}
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\begin{document}$$g_1,g_2,g_3$$\end{document} have the same invariants I and J, we may reduce by the action of SL2(K¯)×SL2(K¯)×SL2(K¯)\documentclass[12pt]{minimal}
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\begin{document}$${{\text {SL}}}_2({{\overline{K}}})\times {{\text {SL}}}_2({{\overline{K}}})\times {{\text {SL}}}_2({{\overline{K}}})$$\end{document} to the case g1(x,z)=g2(x,z)=g3(x,z)=x3z-13Ixz3-127Jz4.\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} g_1(x,z) = g_2(x,z) = g_3(x,z) = x^3 z - \tfrac{1}{3} I x z^3 - \tfrac{1}{27}J z^4. \end{aligned}$$\end{document} The result then follows by symmetry.
We put f(x1,z1;x2,z2)=F(x1,z1;x2,z2;1,0)\documentclass[12pt]{minimal}
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\begin{document}$$f(x_1,z_1;x_2,z_2) = F(x_1,z_1;x_2,z_2;1,0)$$\end{document}. By Proposition 7.2 we have disc1(f)=ag2(x2,z2)\documentclass[12pt]{minimal}
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\begin{document}$${\text {disc}}_1(f) = a g_2(x_2,z_2)$$\end{document} and disc2(f)=ag1(x1,z1)\documentclass[12pt]{minimal}
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\begin{document}$${\text {disc}}_2(f) = a g_1(x_1,z_1)$$\end{document}. The curves C1\documentclass[12pt]{minimal}
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\begin{document}$$C_1$$\end{document} and C2\documentclass[12pt]{minimal}
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\begin{document}$$C_2$$\end{document} are therefore isomorphic to those considered in Lemma 6.3. Applying Lemma 6.3 gives the result. □\documentclass[12pt]{minimal}
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\begin{document}$$\square $$\end{document}
Computing the (2, 2, 2)-forms
To complete the proof of Theorem 3.1 we must explain how to compute the (2, 2, 2)-form F. As before it is helpful to first consider the special case where C1=C2=C3=E\documentclass[12pt]{minimal}
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\begin{document}$$C_1 = C_2 = C_3 = E$$\end{document}.
Let E be the elliptic curve y2=x3+ax+b\documentclass[12pt]{minimal}
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\begin{document}$$y^2 = x^3 + a x + b$$\end{document}. We consider the mapswhere μ(P1,P2,P3)=P1+P2+P3\documentclass[12pt]{minimal}
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\begin{document}$$\mu (P_1,P_2,P_3) = P_1 + P_2 + P_3$$\end{document} and π\documentclass[12pt]{minimal}
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\begin{document}$$\pi $$\end{document} is the map taking the x-coordinate of each point. An equation for S=π(μ-1(0E))\documentclass[12pt]{minimal}
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\begin{document}$$S = \pi (\mu ^{-1} (0_E))$$\end{document} is computed as follows.
Let Pi=(xi,yi)\documentclass[12pt]{minimal}
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\begin{document}$$P_i=(x_i,y_i)$$\end{document} for i=1,2,3\documentclass[12pt]{minimal}
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\begin{document}$$i=1,2,3$$\end{document} be points on E with P1+P2+P3=0E\documentclass[12pt]{minimal}
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\begin{document}$$P_1+P_2+P_3 = 0_E$$\end{document}. These points lie on a line, say y=λx+ν\documentclass[12pt]{minimal}
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\begin{document}$$y = \lambda x + \nu $$\end{document}. Then as polynomials in x we havex3+ax+b-(λx+ν)2=(x-x1)(x-x2)(x-x3).\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} x^3 + a x + b - (\lambda x + \nu )^2 = (x-x_1)(x-x_2)(x-x_3). \end{aligned}$$\end{document}Comparing the coefficients of the powers of x we obtainλ2=s1,2λν=a-s2,ν2=b+s3,\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} \lambda ^2&= s_1, \\ 2 \lambda \nu&= a - s_2, \\ \nu ^2&= b + s_3, \end{aligned}$$\end{document}where s1,s2,s3\documentclass[12pt]{minimal}
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\begin{document}$$x_1,x_2,x_3$$\end{document}. Eliminating λ\documentclass[12pt]{minimal}
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\begin{document}$$\nu $$\end{document} gives the equation(a-s2)2-4s1(b+s3)=0.\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} (a-s_2)^2 - 4 s_1 (b+s_3) = 0. \end{aligned}$$\end{document}The required (2, 2, 2)-form F is obtained by homogenising this equation, i.e. we replace xi\documentclass[12pt]{minimal}
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\begin{document}$$x_i$$\end{document} by xi/zi\documentclass[12pt]{minimal}
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\begin{document}$$z_1^2 z_2^2 z_3^2$$\end{document}.
Remark 8.1
We have F(x1,1;x2,1;x3,1)=W0x32-W1x3+W2\documentclass[12pt]{minimal}
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\begin{document}$$F(x_1,1;x_2,1;x_3,1) = W_0 x_3^2 - W_1 x_3 + W_2$$\end{document} whereW0=(x1-x2)2,W1=2(x1x2+a)(x1+x2)+4b,W2=x12x22-2ax1x2-4b(x1+x2)+a2.\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} W_0&= (x_1 - x_2)^2, \\ W_1&= 2 (x_1 x_2 + a) (x_1+x_2) + 4b, \\ W_2&= x_1^2 x_2^2 - 2 a x_1 x_2 - 4 b (x_1+x_2) + a^2. \end{aligned}$$\end{document}These are the formulae used in [5, Chapter 17] to show that the height on an elliptic curve is a quadratic form.
We now turn to the general case. So let S⊂P1×P1×P1\documentclass[12pt]{minimal}
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\begin{document}$$S \subset {{\mathbb {P}}}^1 \times {{\mathbb {P}}}^1 \times {{\mathbb {P}}}^1$$\end{document} be as in Sect. 7. Let z(gi)\documentclass[12pt]{minimal}
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\begin{document}$$z(g_i)$$\end{document} be the cubic invariant, and write H1,H2,H3\documentclass[12pt]{minimal}
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\begin{document}$$H_1,H_2,H_3$$\end{document} for the binary quadratic forms (4) over L=K[φ]\documentclass[12pt]{minimal}
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\begin{document}$$L = K[\varphi ]$$\end{document} associated to g1,g2,g3\documentclass[12pt]{minimal}
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\begin{document}$$g_1,g_2,g_3$$\end{document}.
Theorem 8.2
If z(g1)z(g2)z(g3)=m2\documentclass[12pt]{minimal}
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\begin{document}$$z(g_1) z(g_2) z(g_3) = m^2$$\end{document} for some m∈L×\documentclass[12pt]{minimal}
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\begin{document}$$m \in L^\times $$\end{document}, andH1H2H3m=F0+F1φ+F2φ2\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} \frac{H_1 H_2 H_3}{m} = F_0 + F_1 \varphi + F_2 \varphi ^2 \end{aligned}$$\end{document}where F0,F1,F2\documentclass[12pt]{minimal}
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\begin{document}$$F_0,F_1,F_2$$\end{document} are (2, 2, 2)-forms defined over K, then S has equation F2=0\documentclass[12pt]{minimal}
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\begin{document}$$F_2 = 0$$\end{document}.
Proof
Let Pi=(xi:yi:zi)∈Ci\documentclass[12pt]{minimal}
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\begin{document}$$P_i= (x_i : y_i : z_i) \in C_i$$\end{document} for i=1,2,3\documentclass[12pt]{minimal}
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\begin{document}$$\mu (P_1,P_2,P_3) = 0_E$$\end{document}. Let Qi\documentclass[12pt]{minimal}
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\begin{document}$$Q_i$$\end{document} be the image of Pi\documentclass[12pt]{minimal}
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\begin{document}$$P_i$$\end{document} under the covering map Ci→E\documentclass[12pt]{minimal}
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\begin{document}$$C_i \rightarrow E$$\end{document}. By the formulae for the covering map coming from classical invariant theory (see for example [8, Proposition 4.2]), the x-coordinate of Qi\documentclass[12pt]{minimal}
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\begin{document}$$Q_i$$\end{document} isξi=3hi(xi,zi)4gi(xi,zi).\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} \xi _i = \frac{3 h_i(x_i,z_i)}{4 g_i(x_i,z_i)}. \end{aligned}$$\end{document}We recall from Sect. 3 thatz(gi)4φgi+hi3=Hi2.\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} z(g_i) \frac{ 4 \varphi g_i + h_i}{3} = H_i^2. \end{aligned}$$\end{document}By (14), (15) and the equation yi2=gi(xi,zi)\documentclass[12pt]{minimal}
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\begin{document}$$C_i$$\end{document} we haveξi+3φ=9Hi24z(gi)yi2\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} \xi _i + 3 \varphi = \frac{ 9 H_i^2 }{4 z(g_i) y_i^2} \end{aligned}$$\end{document}and hence∏i=13(ξi+3φ)=27H1H2H38my1y2y32.\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} \prod _{i=1}^3 (\xi _i + 3 \varphi ) = \left( \frac{27 H_1 H_2 H_3}{8 m y_1 y_2 y_3} \right) ^2. \end{aligned}$$\end{document}Since Q1+Q2+Q3=0E\documentclass[12pt]{minimal}
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\begin{document}$$Q_1 + Q_2 + Q_3 = 0_E$$\end{document} these points lie on a line, say y=λx+ν\documentclass[12pt]{minimal}
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\begin{document}$$y = \lambda x + \nu $$\end{document} for some λ,ν∈K\documentclass[12pt]{minimal}
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\begin{document}$$\lambda ,\nu \in K$$\end{document}. Then as a polynomial in x we havex3-27Ix-27J-(λx+ν)2=(x-ξ1)(x-ξ2)(x-ξ3).\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} x^3 - 27 I x - 27 J - ( \lambda x + \nu )^2 = (x-\xi _1)(x-\xi _2)(x-\xi _3). \end{aligned}$$\end{document}Putting x=-3φ\documentclass[12pt]{minimal}
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\begin{document}$$x = -3 \varphi $$\end{document} gives∏i=13(ξi+3φ)=(ν-3λφ)2.\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} \prod _{i=1}^3 (\xi _i + 3 \varphi ) = (\nu - 3 \lambda \varphi )^2. \end{aligned}$$\end{document}We first suppose E(K)[2]=0\documentclass[12pt]{minimal}
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\begin{document}$$E(K)[2] =0$$\end{document}. In this case L is a field, so comparing (16) and (17) we have27H1H2H38my1y2y3=±(ν-3λφ),\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned}\frac{27 H_1 H_2 H_3}{8 m y_1 y_2 y_3} = \pm (\nu - 3 \lambda \varphi ),\end{aligned}$$\end{document}in L(S). Taking the coefficient of φ2\documentclass[12pt]{minimal}
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\begin{document}$$\varphi ^2$$\end{document} we see that F2\documentclass[12pt]{minimal}
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\begin{document}$$F_2$$\end{document} vanishes on S. In general there are #E(K)[2]\documentclass[12pt]{minimal}
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\begin{document}$$\#E(K)[2]$$\end{document} choices for the square root, up to sign, and these correspond to the #E(K)[2]\documentclass[12pt]{minimal}
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\begin{document}$$\#E(K)[2]$$\end{document} choices in Remark 7.1.
It remains to check that F2\documentclass[12pt]{minimal}
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\begin{document}$$F_2$$\end{document} is not identically zero. For this we may work over an algebraically closed field. Then by a change of coordinates we may suppose that gi\documentclass[12pt]{minimal}
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\begin{document}$$g_i$$\end{document} and hi\documentclass[12pt]{minimal}
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\begin{document}$$h_i$$\end{document} are linear combinations of xi4+zi4\documentclass[12pt]{minimal}
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\begin{document}$$x_i^4 + z_i^4$$\end{document} and xi2zi2\documentclass[12pt]{minimal}
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\begin{document}$$x_i^2 z_i^2$$\end{document}. The singular quartics in this pencil are (xi2-zi2)2\documentclass[12pt]{minimal}
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\begin{document}$$(x_i^2 - z_i^2)^2$$\end{document}, (xi2+zi2)2\documentclass[12pt]{minimal}
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\begin{document}$$(x_i^2 + z_i^2)^2$$\end{document} and (xizi)2\documentclass[12pt]{minimal}
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\begin{document}$$(x_i z_i)^2$$\end{document}. Since L≅K×K×K\documentclass[12pt]{minimal}
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\begin{document}$$L \cong K \times K \times K$$\end{document} we may identify Hi\documentclass[12pt]{minimal}
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\begin{document}$$H_i$$\end{document} as a triple of binary quadratic forms. These are non-zero multiples of xi2-zi2\documentclass[12pt]{minimal}
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\begin{document}$$x_i^2 - z_i^2$$\end{document}, xi2+zi2\documentclass[12pt]{minimal}
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\begin{document}$$x_i^2 + z_i^2$$\end{document} and xizi\documentclass[12pt]{minimal}
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\begin{document}$$x_i z_i$$\end{document}, in this order if we made a suitable change of coordinates. (This last claim may be checked without any calculation if we use stereographic projection to identify the roots of the binary quadratic forms with the vertices of an octahedron, and then rotate the octahedron.) Therefore the space of (2, 2, 2)-forms spanned by F0,F1,F2\documentclass[12pt]{minimal}
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\begin{document}$$F_0,F_1,F_2$$\end{document} contains the forms(x12-z12)(x22-z22)(x32-z32),(x12+z12)(x22+z22)(x32+z32),x1z1x2z2x3z3.\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} (x_1^2 - z_1^2)(x_2^2 - z_2^2)(x_3^2 - z_3^2), \quad (x_1^2 + z_1^2)(x_2^2 + z_2^2)(x_3^2 + z_3^2), \quad x_1 z_1 x_2 z_2 x_3 z_3. \end{aligned}$$\end{document}Since these are linearly independent, it follows that F2\documentclass[12pt]{minimal}
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\begin{document}$$F_2$$\end{document} is non-zero. □\documentclass[12pt]{minimal}
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\begin{document}$$\square $$\end{document}
Proof of Theorem 3.1
Let F=F2\documentclass[12pt]{minimal}
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\begin{document}$$F = F_2$$\end{document} be the equation for S in Theorem 8.2. We specialise the last two sets of variables in (13) to (1, 0). Then comparing with (5) we have F(x,z;1,0;1,0)=γ1(x,z)\documentclass[12pt]{minimal}
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\begin{document}$$F(x,z;1,0;1,0) = \gamma _1(x,z)$$\end{document}. By Corollary 7.3 we haveΨC1([C2])=(K(a)/K,γ1(x,z)/z2),\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} \Psi _{C_1}([C_2]) = (K(\sqrt{a})/K,\gamma _1(x,z)/z^2), \end{aligned}$$\end{document}where a=g3(1,0)\documentclass[12pt]{minimal}
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\begin{document}$$a = g_3(1,0)$$\end{document}. Then by (10) we have⟨[C1],[C2]⟩CT=∑v∈MKinvv(Kv(a)/Kv,γ1(xv,zv)/zv2).\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} \langle [C_1],[C_2] \rangle _{{{\text {CT}}}} = \sum _{v \in M_K} {\text {inv}}_v (K_v(\sqrt{a})/K_v,\gamma _1(x_v,z_v)/z_v^2). \end{aligned}$$\end{document}Subject to identifying μ2=12Z/Z\documentclass[12pt]{minimal}
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\begin{document}$$\mu _2 = \frac{1}{2}{{\mathbb {Z}}}/{{\mathbb {Z}}}$$\end{document}, the Hilbert norm residue symbol is given by(a,b)v=invv(Kv(a)/Kv,b).\documentclass[12pt]{minimal}
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\begin{document}$$\begin{aligned} (a,b)_v = {\text {inv}}_v (K_v(\sqrt{a})/K_v, b). \end{aligned}$$\end{document}This gives the formula in Theorem 3.1, except that we have g3(1,0)\documentclass[12pt]{minimal}
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\begin{document}$$g_3(1,0)$$\end{document} in place of g2(1,0)\documentclass[12pt]{minimal}
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\begin{document}$$g_2(1,0)$$\end{document}. As noted in Remark 3.2(v), this change does not matter. □\documentclass[12pt]{minimal}
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\begin{document}$$\square $$\end{document}
Remark 8.3
To show that γ1(x,z)\documentclass[12pt]{minimal}
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\begin{document}$$\gamma _1(x,z)$$\end{document} is not identically zero we show more generally that F cannot be made to vanish by specialising two of the sets of variables. Indeed, by considering F as given in Remark 8.1, it suffices to show that the polynomials W0,W1,W2\documentclass[12pt]{minimal}
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\begin{document}$$W_0,W_1,W_2$$\end{document} never simultaneously vanish. This may be checked by setting x1=x2=x\documentclass[12pt]{minimal}
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\begin{document}$$x_1 = x_2 = x$$\end{document} and computing that the resultant of W1\documentclass[12pt]{minimal}
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\begin{document}$$W_1$$\end{document} and W2\documentclass[12pt]{minimal}
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\begin{document}$$W_2$$\end{document} is 28(4a3+27b2)2\documentclass[12pt]{minimal}
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\begin{document}$$2^8(4 a^3 + 27 b^2)^2$$\end{document}. This last expression is non-zero, by definition of an elliptic curve.
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