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§2. Normal Forms for Cubic Curves 67

solid lines. By construction these three cubics go through the nine points. Since the
given cubic C goes through the first eight points, by Theorem 2(3.3) we deduce that
the ninth point T is also on C and thus (P + Q)R = P(Q + R).
When P, Q,and R are not all distinct, using the commutative law, we see that
one has only to check the case P + (P + Q) = (P + P) + Q. This comes from
a limiting position of the above diagram or can be proved using a special direct
argument. Hence P + Q is an abelian group law, and this proves the theorem.
With this theorem we complete the formulation of the notion of an elliptic curve
given in (5.1) of the Introduction as a nonsingular cubic plane curve, with a chosen
point, and the group law given by the chord-tangent composition. Observe that any
projective transformation preserving the zero point will also be a group homomor-
phism since it preserves the chord-tangent construction.
Let E be an elliptic curve over a field k. For each field extension K of k the

points E(K) form an abelian group, and if K → K is a k-morphism of fields, then

the induced function E(K) → E(K ) is a group morphism. Thus E is a functor from
the category of fields over k to the category of abelian groups.

Exercises
1. Verify completely the assertions in (1.1).
2. Carry out the details of the proof of the associative law for the ases P + (P + Q) =
(P + P) + Q.

§2. Normal Forms for Cubic Curves
In the examples of Chapter 1 we always chose a cubic equation in normal form and
mentioned that any elliptic curve could be transformed into normal form. This can
be done in two ways. Transforming a flex, which exists by 2(4.8), to infinity with a
tangent line, the line at infinity gives the normal form after suitable rescaling of the
variables. For the reader with more background we sketch how it can be derived from
the Riemann–Roch theorem. Here the background reference is Hartshorne [1977],
Chapter 4.
(2.1) Remark. Any flex on a nonsingular cubic curve can be transformed to 0:0:1
in such a way that the tangent line to the transformed curve is w = 0, the line at
infinity. In this case the cubic equation has the form in the w, x-plane
2
f (w, x, 1) = w + f 2 (w, x) + f 3 (w, x) = w + a 1 wx + a 3 w + f 3 (w, x)
since w must divide f 2 (w, x) from the remarks following 2(4.7). Now normalize the
3
coefficient of x in f 3 (w, x) to −1, and we have the normal form f (w, x, y) = 0,
namely
2
3
2
3
2
2
0 = wy + a 1 wxy + a 3 w y − x − a 2 wx − a 4 w x − a 6 w .
Away from the origin in the elliptic curve, we set w = 1 and consider the differential
0 = df (1, x, y) = f x (1, x, y)dx + f y (1, x, y) dy.