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
Purchased from American Mathematical Society for the exclusive use of nofirst nolast (email unknown) Copyright 2011 American Mathematical Society. Duplication prohibited. Please report unauthorized use to cust-serv@ams.org. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.