# seeing the differential dx/y on an elliptic curve as an element of the sheaf of differentials

$$\newcommand{\CC}{\mathbb{C}}$$ $$\newcommand{\Spec}{\operatorname{Spec}}$$

It's a well known fact that every elliptic curve (say, over a field $$k$$) has a global holomorphic nowhere vanishing differential. If the curve is given by $$y^2 = x^3 + ax + b$$, then it's computed in Silverman (p33, example 4.6), that the differential $dx/y is holomorphic and nonvanishing. Hence, the line bundle of holomorphic differentials should be trivial, corresponding to a free rank 1 sheaf of relative differentials. On the other hand, let $$E$$ be some elliptic curve over $$k = \CC$$, say given by the equation $$y^2 = x^3 - 1$$. Then the existence of a nowhere vanishing holomorphic differential should imply that the sheaf of relative differentials $$\Omega_{E/k}$$ is free of rank 1. In particular, we can restrict the sheaf $$\Omega_{E/k}$$ to the affine locus given by the ring $$R := \CC[x,y]/(y^2-x^3+1)$$ The restriction of $$\Omega_{E/k}$$ to $$U := \Spec R$$ is then just the sheaf associated to the $$R$$ module $$\Omega_{R/k} = (Rdx\oplus Rdy)/(2ydy - 3x^2dx)$$ The global sections of $$\Omega_{E/k}|_U$$ should then just be the elements of $$\Omega_{R/k}$$. In particular, $$dx/y$$ should be an element of the module $$\Omega_{R/k}$$. My question is: how do you see $$dx/y$$ as an element of $$\Omega_{R/k}$$? Am I missing some algebra trick? • The point of the global differential is that it changes forms when you change charts. Is that a chart where$\frac{dx}{y}$makes sense? May 5 '14 at 21:32 • I don't know, but$\Omega_{E/k}$is quasi-coherent, so over$U$it's just the sheaf associated to the$R$-module$\Omega_{R/k}$, which means that global sections are literally just elements of the module... In fact I'd be happy to even find an$R$-basis for$\Omega_{R/k}\$. May 5 '14 at 21:50

$$\newcommand{\Spec}{\operatorname{Spec}}$$So here's the answer. Since $$dx/y$$ is defined on $$D(y)$$, and over $$D(xy)$$, $$dx/y = 2dy/3x^2$$, which is defined on $$D(x)$$, and since $$D(x),D(y)$$ cover $$\Spec R$$, they must glue to a global section, ie, an element of $$\Omega_{R/k}$$.
Ie, we want to find an element (which we now know exists!) of $$\Omega_{R/k} = (Rdx\oplus Rdy)/(2ydy-3x^2dx)$$ which gets mapped to (under the localization map) to $$dy/x$$ on $$D(y)$$.
Ie, you want to find $$f,g\in R$$ such that $$fdx + gdy = dx/y$$ in $$\Omega_{R/k}|_{D(y)}$$, which is to say that there exists some $$t = y^k$$ such that $$t(y(fdx+gdy) - dx) = 0\qquad\text{in \Omega_{R/k}}$$ Equivalently, $$t((yf-1)dx + ygdy)$$ is a multiple of $$2ydy-3x^2dx$$. Using the relation $$y^2 = x^3-1$$, we may choose $$f = -y, g = \frac{2}{3}x, t = 1$$, and we see that $$(-y^2-1)dx + \frac{2}{3}xydy = -x^3dx + \frac{2}{3}xydy = \frac{x}{3}(2ydy - 3x^2dx) = 0$$