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
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