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