Generator for Kahler differentials of an affine elliptic curve Consider the affine (nonsingular) elliptic curve $A = \mathbb C[x,y]/(y^2-x^3+x)$. Since the cotangent bundle is trivial, $\Omega_A^1  = A\,dx\oplus A\,dy /(2y\,dy - (3x^2-1)\,dx)$ is a free $A$-module. But how does one compute a generator?
I can find generators locally which glue, such as $dx/y = 2\,dy/(3x^2-1)$, but I'm interested to know a presentation of the global solution in the form $p\,dx + q\, dy$ for $p,q \in A$.
This is equivalent (take $p = 2P, q = Q$) to:

Write $$1 = P(3x^2-1) + Qy$$ for some $P,Q\in A$.

 A: The problem is reduced to the Euclidean algorithm in $\mathbb C[x]$. This technique works in the case of a nonsingular plane curve given by an equation of the form $y^n = f(x)$, and in particular for an elliptic curve in Weierstrass form.
We wish to write $$1 = P\cdot f' + Q\cdot ny^{n-1}$$ for some $P,Q \in \mathbb C[x,y]/(y^n-f)$.
As in msteve's answer, we make the simplifying assumption that $Q = \frac yn\cdot R$, and that $R, P$ are represented as functions of only $x$. Using the relation $y^n = f$ to rewrite the equation, we wish to solve
$$1 = P\cdot f' + R\cdot f$$
But nonsingularity implies that $f, f'$ are coprime so that this is accomplished even in $\mathbb C[x]$; it is just the problem of finding Bezout coefficients.
UPDATE:
After a little fiddling in Macaulay2, one finds that for a general elliptic curve in Weierstrass form having $f = x^3 + ax + b$, the Bezout coefficients may always be chosen to have degree $\leq 2$.
Writing $P = Ax+B$ and $R = Cx^2 + Dx + E$ and collecting terms one gets
$$x^4(A+3C) + x^3(B+3D) + x^2(Aa + Ca + 3E) + x(Ba + Ab + Da) + (Bb + Ea) = 1.$$
After processing this information one obtains a general solution:
$$R = \frac{-18a\,x + 27b}{27b^2+4a^3}$$
$$P = \frac{6a\,x^2 -9b\,x  +4a^2}{27b^2+4a^3}$$
(setting $a=-1, b=0$ obtains $R = -\frac92x$ and $P = \frac32x^2-1$, like we had before.)

For $A = \mathbb C[x,y]/(x^3+ax+b-y^2)$, a generator for $\Omega_A^1$ as an $A$-module is:
  $$(-18a\, xy + 27b\, y)\,dx + (12a\, x^2 - 18b\, x + 8a^2)\,dy$$

UPDATE:
A similar question was brought to my attention. It is a nice check that setting $a=0,b=-1$ in the above formula obtains $-27y\,dx + 18\,dy$ which agrees with their answer.
A: As is suggested in the question, in order to find a free generator of $\Omega_A^1$, it suffices to find $P,Q \in A$ satisfying $P(3x^2-1) + Qy = 1$. Indeed, a generator is then given by $2P dx + Qdy$.
The solution below finds such $P$ and $Q$, but it is rather ad hoc and I would be very interested in seeing a systematic method, if one exists.
Let's first assume $P$ is of the form $P = g(x,y) - 1$ so that the equation we need to solve is 
$$
1 = (g(x,y) - 1)(3x^2 - 1) + Qy \iff 3x^2 g(x,y) - 3x^2 - g(x,y) + Qy = 0.
$$
If $Q$ contains one factor of $y$ and $g$ is only a function of $x$, then we can turn this into an equation only involving $x$, so assume $Q = yh(x)$. Our equation then becomes
$$
3x^2 g(x) - 3x^2 -g(x) + (x^3 - x)h(x) = 0.
$$
Further, guess that $g(x)$ has degree-2 and $h(x)$ has degree-1 (so that we have terms of the highest degree that can cancel); so, write $g(x) = ax^2 + bx + c$ and $h(x) = \alpha x + \beta$, then the equation becomes
$$
(3a + \alpha)x^4 + (3b+\beta)x^3 + (3c-3-a-\alpha) x^2 + (-b-\beta)x + (-c) = 0.
$$
Equating all of the coefficients to zero, we find that $c=0$, which gives the 2 equations for $a$ and $\alpha$ (from the degree-4 and degree-2 pieces)
$$
\begin{cases}
3a + \alpha = 0\\
-3 - a - \alpha = 0.
\end{cases}
$$
The solution of this system is $a = \frac{3}{2}$ and $\alpha = -\frac{9}{2}$. Similarly, the degree-3 and degree-1 coefficients give that $b = \beta = 0$. Putting all of this together, we have found that $P = \frac{3}{2}x^2 - 1$ and $Q = -\frac{9}{2} xy$.
