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§6. Computability of Torsion Points on Elliptic Curves 119

the above argument showing x = m shows again that x = m and h is even. This
proves the theorem.
The last step in our analysis of torsion points, which we now know must have
integer coordinates, is to obtain a divisibility requirement on the y-coordinate.
2
(6.2) Theorem. Let E be an elliptic curve over Q, and let y = f (x) be a Weier-
strass equation for E where f (x) has integer coefficients. If (x, y) is a torsion point
on E, then the integer y is zero or y divides D( f ), the discriminant of the cubic
polynomial f (x).

Proof. If y = 0, then (x, 0) is of order 2 and 0 divides the discriminant. Otherwise,
2(x, y) = (¯x, ¯y) unequal to 0 on E(Q). The tangent line to E at (x, y) has slope
f (x)/2y, and when its equation y = λx + β is substituted into the Weierstrass

3
2
2
equation y = f (x) = x + ax + bx + c, we obtain a cubic equation with x as
2

double root and ¯x as a single root. This equation has coefficient a − ( f (x)/2y) of
2
x , and hence the sum of the roots of the cubic in x is the negative of this coefficient,
so
2

f (x)
2x +¯x = a − .
2y

Since x, ¯x,and a are integers, it follows that f (x)/2y is an integer, and 2y divides
f (x).

By (4.2) in the Appendix to Chapter 2, we can write the discriminant D( f ) of

f (x) as a linear combination D( f ) = u(x) f (x) + v(x) f (x), where u(x), v(x) ∈

Z[x]. Since y = f (x) and y divides f (x) for the point (x, y) on E, we deduce that
y divides D( f ). This proves the theorem.
(6.3) Remark. The effectively computable method for finding E(Q) tors is to take
3
2
2
for E a Weierstrass equation y = f (x), where f (x) = x + ax + bx + c and a, b,
and c are integers. Consider the finite set of all divisors y 0 of D( f ), the discriminant
2
of f (x). Solve the cubic f (x) = y for integer solutions x 0 . Among these (x 0 , y 0 )
0
are all points of E(Q) tors which are unequal to 0.
3 2 3
(6.4) Example. For a cubic in the form x + ax + bx the discriminant is D(x +
2 2 2 2 3 2
ax + bx) = b (4b − a ).Inthe caseof E: y = x + x − x we have D =
−4 − 1 =−5. The divisors of 5 are +1, −1, +5, and −5. For y = 0wehavethe
point (0, 0), for y =+1 we have the points (1, 1) and (−1, 1), for y =−1we
have the points (1, −1) and (−1, −1), and for y = 5 we obtain the cubic equation
2
3
0 = x + x − x − 25. It has a root strictly between 2 and 3 which is not integral,
and, in fact, it is not rational. The same applies to y =−5. Thus we have six rational
torsion points on E(Q),and E(Q) tors is cyclic of order 6. Although not all points
need be torsion points in general.
(6.5) Remark. If (x, y) is a point E(Q) such that some multiple n(x, y) has nonin-
tegral coefficients, then (x, y) is not a torsion point. This applies to (0, 0) on E and
E in (5.6) by the calculations in 1(1.6) and 1(2.4).