Kähler differentials not the same as regular differentials on a singular curve Let $X$ be the affine cubic curve $y^2=x^3$, over a field of characteristic not equal to 2 or 3. Let $A=k[X]$, the ring of regular functions on $X$. Let $\Omega_A$ be the $A$-module of Kähler differentials on $X$, i.e. the $A$-module generated by symbols $\mathrm{d}f$, where $f\in A$, satisfying the properties
$\mathrm{d}\lambda=0$, $\mathrm{d}(f+g)=\mathrm{d}f+\mathrm{d}g$ and $\mathrm{d}(fg)=f\mathrm{d}g+g\mathrm{d}f$, for all $f,g\in A$ and $\lambda\in k$. Let $\Omega[X]$ denote the set of all regular differential forms on $X$. Why is the differential form $3y\mathrm{d}x-2x\mathrm{d}y$ non-zero in $\Omega_A$, but zero in $\Omega[X]$?
 A: Clearly, $\Omega_{k[X]}$ is generated (as an $k[X]$-module) by $dx$ and $dy$. 
Since $k[X]=k[x]+k[x]y$, any Kähler differential $\omega$ may be written as 
$$ (A+By)dx + (C+Dy) dy,$$
with $A,B, C, D\in k[x]$. 
Note that the relations are generated by $2ydy-3x^2dx$ over $k[X]$, so when regarded as a module over $k[x]$, it is generated by 
$$1\cdot (2ydy-3x^2dx)= 2ydy-3x^2dx $$
 and $$y\cdot (2ydy-3x^2dx)=2y^2dy-3x^2ydx= 2x^3dy-3x^2ydx.$$ 
Replacing $ydy$ by $3x^2dx/2$, and $x^2ydx$ by $2x^3dy/3$ respectively, we see that any $\omega$ can be written uniquely as 
$$ Adx +Bydx+Cdy$$
with $\deg B<2$. This shows that $3ydx-2xdy$ is nonzero in $\Omega_{k[X]}$. 
Next we show that it is zero in $\Omega[X]$. Recall that each $\omega'\in \Omega[X]$ is  an assignment of a cotangent vector at each point of $X$ that is "nice" locally.  A differential form vanishes iff it vanishes at every point. However, it is clear in our case we only need to check this for the origin.  Let $m=(x,y)$ be the maximal ideal of $k[X]$ corresponding to the origin. We need to show that the image of $3ydx-2xdy$ under the map $\Omega_A\to m/m^2$ is zero.   But this map is nothing but $dx\mapsto x$ and $dy\mapsto y$. 
