Elliptic Curve and Differential Form Determine Weierstrass Equation

Question

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?

Answer 1

In my opinion, the statement, that an elliptic curve together with a non-zero holomorphic differential form determines a Weierstrass equation, unfolds its full meaning when considering a family of elliptic curves $E\to S$ over a base scheme $S$ which is not necessarily a field. In this case the sheaf $\Omega_{E/S}$ does not always admit a global section. However, if it does, then one can construct a "Weierstrass equation of $E$ over $S$" from it.

This procedure is described in Section 2.2 of the book "Arithmetic Moduli of Elliptic Curves" by Katz and Masur. Of course, you can apply this in particular to the case $S=\textrm{Spec}(F)$ where $F$ is your favorite a field.

Answer 2

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.

References:
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.