Lemma 15 ([The elliptic star]). Let J₀=E_1×E2×E3 be a product of elliptic curves with origins, and let Z₀ be the reduced union of the three coordinate elliptic axes through zero. Let K⊂J₀ be a finite subgroup of order at least four …

If axis i at g met axis j at h≠g , the coordinates of g and h outside i,j would agree. There is at least one such coordinate, contradicting injectivity there. This proves disjointness. Put V_i=H^0(E_i,D_i) . Formula gives V_i=2 and h^0(D_i(-q))=1 for every q∈E_i . Evaluation is therefore basepoint-free and defines a degree-two map φ_i:E_i→ℙ(V_i^*) , each of whose fibers has at most two distinct points. The marks (φ_1(g_1),φ_2(g_2),φ_3(g_3)), g=(g_1,g_2,g_3)∈K, have at most two indices in each coordinate fiber, since K→E_i is injective. The product formula for sections identifies a section of D_1 D_2 D_3 with a tensor in V_1⊗V_2⊗V_3 . Its restriction to axis i at g is contraction by evaluation at the other two coordinates. Vanishing on Z(g) is exactly. Lemma Lemma 14 ([Singular marked points of a trilinear form]) forbids this for four or more indices for a nonzero section. Finally, the ideal of a closed subscheme with this support is contained in the radical ideal of the reduced union, which proves the last assertion.