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2.10. Homogeneous spaces 65 times (−1,82, −82), (−82,2, −41), (82,41,2), which do have points by a previous item in this list. How about all the other possible pairs (δ1,δ2)? Consider (−1,2, −2) and its homogeneous space: C(−1, 2) : 2Y 2 + X2 = 82, 2Y 2 + 2Z2 = 164. Let us show that there are solutions to C(−1, 2) over R, Q2 and Q41: • (Over R). The point (0, √ 41, √ 41) is a point on C(−1, 2) defined over R. • (Over Q41). Let Y0 = 1 and put f(X) = X2 − 80, g(Z) = Z2 − 81. By Hensel’s Lemma (see Appendix D.1 and Corol- lary D.1.2), it suffices to show that there are α0,β0 ∈ F41 such that f(α0) = g(β0) ≡ 0 mod 41 and f (α0), g (β0) ≡ 0 mod 41. The reader can check that the congruences α0 ≡ 11 mod 41 and β0 ≡ 9 mod 41 work. Thus, there are α, β ∈ Q41 such that f(α) = 0 = g(β). Hence, (X0,Y0,Z0) = (α,1,β) is a point on C(−1, 2) defined over Q41, as desired. • (Over Q2). Let X0 = 0 and put f(Y ) = Y 2 −41. Let α0 = 1. Then f(α0) = −40, f (α0) = 82 and 3 = ν2(−40) ν2(822) = ν2(22 · 412) = 2. Thus, by Hensel’s Lemma (Theorem D.1.1 see also Ex- ample D.1.4), there is α ∈ Q2 such that f(α) = 0, or α2 = 41. Hence, the point (X0,Y0,Z0) = (0,α,α) is a point on C(−1, 2) defined over Q2, as desired. One can also show that, in fact, C(−1, 2) has a point over Qp for all p ≥ 2. Therefore, we cannot deduce any contradictions working locally about whether C(−1, 2) has a point over Q. A computer search does not yield any Q-points on C(−1, 2). Therefore, our method breaks at this point, and we cannot determine whether there is a point on E(Q) that comes from C(−1, 2). It turns out that C(−1, 2) does not have rational points (but this is difficult to show). This type of space, a space that has solutions