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.


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

\[ Y^2 = X^3 - 2^4 3^3 \]