$\newcommand{\CC}{\mathbb{C}}$
$\newcommand{\Spec}{\operatorname{Spec}}$
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?
Best Answer
$\newcommand{\Spec}{\operatorname{Spec}}$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 $$(-y^2-1)dx + \frac{2}{3}xydy = -x^3dx + \frac{2}{3}xydy = \frac{x}{3}(2ydy - 3x^2dx) = 0$$