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§8. Tate’s Theorem on Good Reduction 123

Let x = 3y so that y ≡ 7 (mod 8), and hence c 6 = 9c. This gives the equation

2
2
2
2
3c = y(y ± 12y + 4 · 12 ) = y((y ± 6) + 540).
2
Now y > 0 since y((y ± 6) + 540) is positive. If p is unequal to 3 and divides y,it
2
does so to an even power. Also the relation for 3c shows that if 3 divides y, then 27
2
divides 3c . In this case let y = 3z and c = 3d which leads to
2
2
d = z(z ± 4z + 64)
2
2
from the relation for 3c . If an odd prime p divides z, then p |z and z is a square. But
y ≡ 7 (mod 8) implies z ≡ 5 (mod 8) which contradicts the fact that z is a square.
This proves the theorem.
(8.2) Example. The following curve
√
5 + 29
2
3
2
y + xy + ε y = x , where ε =
2
√
over K = Q( 29) was shown by Tate to have good reduction at all places of K,see
Serre [1972, p. 320]. The norm of ε is −1, and the group of totally positive units in
2
R = Z[ε] is generated by ε . The reader can check that =−ε 10 and verify the
good reduction properties.
(8.3) Example (Unpublished example of R. Oort).
2
3
y + xy = x − εx
√ √
where ε = 32 + 5 41 the fundamental unit in Z[(1/2) + (1/2) 41]. Here = ε 4
√
and the curve has good reduction at all places of Q( 41).