I am reading Fermat's Last Theorem by Diamond, Darmon and Taylor and they state:

"An elliptic curve E over a field F is a proper smooth curve over F of genus one with a distinguished F-rational point. If $E/F$ is an elliptic curve and if $\omega$ is a non-zero holomorphic differential on E/F then E can be realised in the projective plane by an equation (called a Weierstrass equation) of the form $$Y^2Z + a_1XYZ + a_3Y Z^2 = X^3 + a_2X^2Z + a_4XZ^2 + a_6Z^3$$ such that the distinguished point is (0 : 1 : 0) (sometimes denoted $\infty$ because it corresponds to the “point at infinity” in the affine model obtained by setting $Z=1$) and $\omega =\frac{dx}{2y+a_1x+a_3}$."

My question is how does the choice of $\omega$ determine the Weierstrass for $E$? Why state this in terms of differential forms instead of the usual projective embedding?

  • $\begingroup$ At least in the complex case you can use the differential form to integrate along the first singular homology and obtain the lattice associated with the elliptic curve. Conversely if you have the lattice you get the differential form by passing dz to the quotient. Also, giving the lattice you can get the Weiertrass equation by the differential equation of the P-Weietrass function and given the Weietrass equation you can get the differential form by the formula that you wrote above. $\endgroup$ Feb 1, 2018 at 0:15
  • $\begingroup$ I see how this applies over $\mathbb{C}$, but how does one restrict this correspondence over a subfield $F\subseteq \mathbb{C}$, or even to finite fields $\endgroup$
    – Rdrr
    Feb 4, 2018 at 0:20

1 Answer 1


To ease notation, the Weierstrass equation is generally written using non-homogeneous coordinates $x = X/Z$ and $y = Y/Z$, then $$E:y^2 + a_1xy+a_3y=x^3+a_2x^2+a_4x+a6$$ If $char(\mathbb{K}) \neq 2$, then we can simplify the equation by completing the square. Thus the substitution $$y \to\frac{1}{2}(y-a_1x-a_3)$$ gives an equation of the form $$E:y^2 = 4x^3 + b_2x^2+2b_4 + b6$$ $$b_2 = a_1^2 + 4a_4, \quad\quad b_4 =2a_4+a_1a_3, \quad\quad b_6=a_3^2+4a_6$$ We also define quantities \begin{align} b_8 &= a_1^2a_6+4a_2a_6-a_1a_3a_4+a_2a_3^2-a_4^2\\ c_4 &= b_2^2 - 24b_4\\ c_6 &= -b_2^3 +36b_2b_4-216b_6\\ \Delta &= -b_2^2b_8 - 8b_4^3 - 27b_6^2+9b_2b_4b_6\\ j &=c_4^3/\Delta\\ \omega &=\frac{dx}{2y+a_1x+a_3} = \frac{dy}{3x^2 + 2a_2x+a_4-a_1y} \end{align} Where the quantity $\Delta$ is the discriminant of the Weierstrass equation, the quantity $j$ is the $j$-invariant of the elliptic curve, and $\omega$ is the invariant differential associated to the Weierstrass equation.

J. Silverman - The Arithmetic of Elliptic Curves 2-ed. s. $42$ (pdf)
Silverman and Tate's - Rational Points on Elliptic Curves
Ludwig Bauer - Weierstrass Equations (pdf)

PS: Maybe this is not a clear answer to your questions, but surely the references will be very helpful.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.