MA342J: Introduction to Modular Forms
Tutorial 4, March 15 & 21
Transforming an elliptic curve into a Weierstrass form
Recall that an elliptic curve over the field k is a nonsingular projective plane
cubic curve CF (F ∈ k[X, Y, Z], homogeneous, irreducible, of degree 3) with a
point O ∈ CF (k).
Theorem. (i) For any a, b, c ∈ k the curve
Ea,b,c : Y 2 Z = X 3 + aX 2 Z + bXZ 2 + cZ 3 ,
O = (0 : 1 : 0) .
is an elliptic curve over k whenever the polynomial x3 + ax2 + bx + c has no
multiple roots in k.
(ii) Let E = (CF , O) be an elliptic curve over the field k. If the characteristics
of the field char(k) 6= 2, 31 then E is isomorphic over k to an elliptic curve of
the above form Ea,b,c . (This is called a Weierstrass form of E, though it is not
unique.)
Here is the algorithm to transform a cubic F (X, Y, Z) = 0 with a point
O = (X0 : Y0 : Z0 ) into a Weierstrass form (or to discover that this cubic is not
an elliptic curve).
The tangent line to CF at O must intersect CF in three points. Moreover,
the multiplicity of O itself is at least 2, therefore the third point (let us call it
P ) is defined over k. Hence there are two possibilities: P = O and P 6= O. We
will consider them below.
Step 0. Performing a linear change of coordinates in P2 (k) if necessary, we can
assume that O = (0 : 1 : 0). Then the curve is given by
F3 (X, Z) + F2 (X, Z)Y + F1 (X, Z)Y 2 = 0
(Fi is homogeneous of degree i). Moreover, since the point O is nonsingular F1 (X, Z) = c1 X + c2 Z is nonzero, and performing the change of
coordinates (X 0 : Y 0 : Z 0 ) = (X : Y : c1 X + c2 Z) we can assume that
F1 (X, Z) = Z.
Case 1: P = O.
Step 2. Check that P = O implies that F2 (X, Z) = a1 XZ + a2 Z 2 (no X 2
term), then change the coordinates (X 0 : Y 0 : Z 0 ) = (X : Y + a21 X +
a2
2 Z) to get equation of the form
b0 X 3 + b1 X 2 Z + b2 XZ 2 + b3 Z 3 + ZY 2 = 0 .
Step 3. Rescale the coordinates to get the Weierstrass form.
Case 2: P 6= O.
1 char(k) is the characteristics of the field k. It is either 0, e.g. char(Q) = char(R) =
char(C) = 0, or a prime number, e.g. char(Fp ) = p.
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Step 2. Observe that it is possible to make a linear change of coordinates so
that O = (0 : 1 : 0) and P = (0 : 0 : 1). In the affine coordinates
Y
x= X
Z , y = Z the point P has coordinates (0, 0) and the curve is
given by
f3 (x, y) + f2 (x, y) + f1 (x, y) = 0
(fi is homogeneous of degree i). Let t = xy . Dividing the last equation
by y we get
y 2 φ3 (t) + yφ2 (t) + φ1 (t) = 0
φi (t) := fi (t, 1)
and therefore
√
−φ2 (t) ± δ
,
y =
2φ3 (t)
δ = φ2 (t)2 − 4φ1 (t)φ3 (t) .
√
Approximately, we want x0 = t and y 0 = δ to be our new (affine)
coordinates since they determine x and y. Since φi (t) = fi (t, 1) is a
polynomial of degree i in t, we have
φ2 (t)2 − 4φ1 (t)φ3 (t) = b0 t4 + b1 t3 + b2 t2 + b3 t + b4
and equation of the curve in these new ”coordinates” is given by
(y 0 )2 = b0 (x0 )4 + b1 (x0 )3 + b2 (x0 )2 + b3 x0 + b4 .
Moreover, the condition δ(t) = 0 means that the line of the slope t
through the origin P is a tangent line to a curve to some point. One
of these points is O and it is easy to check that the corresponding
value is t = 0. It follows that actually b4 = 0. Therefore in the
coordinates
1
1
y
=
=
0
x
t
x
y0
2φ3 (t)y + φ2 (t)
=
=
= ...
(x0 )2
t2
x00 =
y 00
the curve has the equation
(y 00 )2 = b0 + b1 x00 + b2 (x00 )2 + b3 (x00 )3 .
Step 3. Rescale the coordinates y 00 = b23 y 000 , x00 = b3 x000 to get the Weierstrass
form.
Step 4. Write the final transformation in projective coordinates and check
that it is a bijection from CF to Ea,b . Observe that in the case
P = O this transformation was linear (and therefore was obviously
1-to-1) while here it is given by homogeneous polynomials of higher
degree.
Question 1: Write the curve
X3 + Y 3 = Z3
O = (1 : 0 : 1)
2
in a Weierstrass form.
Question 2: Let t ∈ Q be a parameter. Consider the cubic curve
(X + Y + Z)(XY + XZ + Y Z) = tXY Z .
One can easily find 6 rational points on this curve, namely
(1 : 0 : 0) , (0 : 1 : 0) , (0 : 0 : 1) ,
(1 : −1 : 0) , (1 : 0 : −1) , (0 : 1 : −1) .
Up to simple permutations of coordinates, there are then two elliptic curves
to be considered: with O = (0 : 0 : 1) and O = (0 : −1 : 1). Write the
Weierstrass form for every case. For which values of t the curves are not elliptic.
(Here the answer should be independent of the choise of O because the reason
would be that the above cubic curve is either singular or reducible.)
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