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:

  • \(b_2 = a_1^2 + 4 a_2\)

  • \(b_4 = 2a_4 + a_1 a_3\)

  • \(b_6 = a_3^2 + 4 a_6\)

  • \(b_8 = a_1^2 a_6 + 4a_2 a_6 - a_1 a_3 a_4 + a_2 a_3^2 - a_4^2\)

  • \(c_4 = b_2^2 - 24 b_4\)

  • \(\Delta = -b_2^2 b_8 - 8b_4^3 - 27b_6^2 + 9b_2 b_4 b_6\). 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

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

Ben Lynn 💡