## The Weierstrass Form

Using Bezout’s Theorem, it can be shown that every irreducible cubic has a flex (a point where the tangent intersects the curve with multiplicity three) or a singular point (a point where there is no tangent because both partial derivatives are zero). [Reducible cubics consist of a line and a conic, which are easy to study.]

Irreducible cubics containing singular points can be affinely transformed into one of the following forms:

1. $$Y^2 = X^3$$

2. $$Y^2 = X^2(X-1)$$

3. $$Y^2 = X^2(X+1)$$

(I’m not sure if this is true for all characteristics.)

An irreducibe cubic with a flex can be affinely transformed into a Weierstrass equation:

$Y^2 + a_1 XY + a_3 Y = X^3 + a_2 X^2 + a_4 X + a_6$

We only consider cubic equations of this form. Define:

• $$b_2 = a_1^2 + 4 a_2$$

• $$b_4 = 2a_4 + a_1 a_3$$

• $$b_6 = a_3^2 + 4 a_6$$

• $$b_8 = a_1^2 a_6 + 4a_2 a_6 - a_1 a_3 a_4 + a_2 a_3^2 - a_4^2$$

• $$c_4 = b_2^2 - 24 b_4$$

• $$\Delta = -b_2^2 b_8 - 8b_4^3 - 27b_4^2 + 9b_2 b_4 b_6$$. This is the discriminant.

• $$j = c_4^3 / \Delta$$ for $$\Delta \ne 0$$. This is the $$j$$-invariant.

Fact: The discriminant is zero if and only if the curve is singular.

Fact: Isomorphic curves over some field $$K$$ have the same $$j$$-invariant. Two curves with the same $$j$$-invariant are isomorphic over $$\bar {K}$$.

This equation can be further simplified through another affine transformation. Let $$K$$ denote the field we are working in. If the $$\mathrm{char} K \ne 2$$, then completing the square on the left hand side (and performing an appropriate variable substitution) eliminates the $$XY$$ and $$Y$$ terms. In addition, if $$\mathrm{char} K \ne 3$$, then a similar trick eliminates the $$X^2$$ term (whereas if $$\mathrm{char} K = 3$$ we can eliminate either the $$X^2$$ or the $$X$$ term).

If $$\mathrm{char} K = 2$$ then one of the following two forms can be obtained:

1. $$Y^2 + XY = X^3 + a_2 X^2 + a_6$$ (the nonsupersingular case)

2. $$Y^2 + a_3 Y = X^3 + a_4 X + a_6$$ (the supersingular case)

In Weierstrass form, we see that for any given value of $$X$$, there are at most two values that $$Y$$ may take. If $$a_1 = a_3 = 0$$ (which is always the case for $$\mathrm{char} K \ne 2$$, we have that if $$(x,y)$$ is a point, then $$(x, -y)$$ is the other point with the same $$x$$-coordinate.

### Example

Consider the cubic Fermat curve

$X^3 + Y^3 = 1$

Assume $$\mathrm{char} K \ne 3$$ (otherwise the curve is the same as $$(X + Y)^3 = 1$$). An affine transformation takes it to its Weierstrass form:

$Y^2 - 9Y = X^3 - 27$

If $$\mathrm{char} K \ne 2$$ then we can further transform this to

$Y^2 = X^3 - 2^4 3^3$

Ben Lynn blynn@cs.stanford.edu 💡