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122 5. Reduction mod p and Torsion Points

(7.5) Definition. An elliptic curve E over K has potential good reduction provided
there exists a finite extension L and an extension w of v to L such that E over L has
good reduction at the valuation w.

Be extending the ground field from K to L, the minimal equation of E can change
to one with good reduction, or, in other words, E can become isomorphic over L to

a curve E with good reduction at w.Inthe above case E was the Legendre or
Hessian form of the equation. In general the discriminant of E over K is different

from the discriminant of E over L and in this case of potential good reduction

v( ) > 0and w( ) = 0 while j(E) = j(E ). Since any curve E with good

reduction has v( j(E )) ≥ 0, it follows immediately that if E has potential good

reduction, then j(E) is a local integer, i.e., v( j(E)) ≥ 0. The above argument with
Legendre or Hessian forms of the equations shows that the converse holds, and this
is the theorem of Deuring.
(7.6) Theorem. An elliptic curve E defined over K has potential good reduction if
and only if j(E) is a local integer, i.e., v( j(E)) ≥ 0.
§8. Tate’s Theorem on Good Reduction over the Rational
Numbers

Tate’s theorem says that over the rational numbers there are no elliptic curves with
good reduction everywhere. We give a proof using formulas for coefficients follow-
ing Ogg [1966].
(8.1) Theorem. Every elliptic curve E over the rational numbers Q has bad reduc-
tion at some prime, i.e., cannot be equal to ±1.
Proof. Assume that =±1. Recall from 3(3.1) that

2 3 2
=−b b 8 − 8b − 27b − 9b 2 b 4 b 6
2 4 6
and
2
b 2 = a + 4a 2 , b 4 = a 1 a 3 + 2a 4 .
1
If a 1 is even, then 4|b 2 and 2|b 4 so that b 6 will have to be odd, and in fact, ±1 =
2
≡ 5b (mod 8). Since any square modulo 8 is conguent to 0, 1, 4 (mod 8), this is
6
impossible.
Now consider the case where a 1 is odd, and hence b 2 is also odd. Then the co-
2
efficient c 4 = b − 24b 4 ≡ 1 (mod 8). We write c 4 = x ± 12 from the relation
2
3
3
3
2
c − c = 12 =±12 , and thus
4 6
2
3
2
2
c = x(x ± 36x + 3 · 12 ) ≡ x (x + 4) (mod 8).
6
This means that x ≡ 5(mod8). Now3|x, for otherwise any p|x with p > 3, it would
2
follow that p |x and ±x would be a square. This would contradict x ≡ 5(mod8).