Kähler differentials of the cuspidal cubic I want to compute $\Omega^1_{A,\mathbb{C}}$ for $A = \mathbb{C}[X,Y]/(Y^2 - X^3)$, or more precisely, I want to show that the module of Kähler differentials is free of rank 2 at the origin, and free of rank 1 everywhere else.
Following Vakil's "Foundations of Algebraic Geometry" p. 550, I see that $\Omega^1_{A,\mathbb{C}} = \operatorname{d}X \oplus \operatorname{d}Y / (2Y\operatorname{d}Y - 3X^2\operatorname{d}X)$. It is then claimed that the fiber at the origin is computed by setting $X = 0$ and $Y = 0$, but I cannot make sense of this.
Just by the very definitions, I would have to show that $A_{(X,Y)}\operatorname{d}X \oplus A_{(X,Y)}\operatorname{d}Y / A_{(X,Y)}(2Y\operatorname{d}Y - 3X^2\operatorname{d}X)$ is a free $A_{(X,Y)}$-module of rank 2, right? If so, how do I do this?
 A: Vakil's question is about the fibres at various points, not the stalk.  The fibre at points away from the cusp is one-dimensional, while at the cusp, the fibre is two-dimensional.  This shows that the module is not free (it has torsion supported at the cusp, as Martin Brandenburg remarks).  
You are suggesting that you should localize at $(X,Y)$, but to compute the fibre,
you should quotient out by $(X,Y)$, just as Vakil suggests.
A: $\Omega_{A}$ is the free $A$-module generated by the differentials $dX, dY$ subject to the relation $2Y dY = 3X^2 dX$. Since $\Omega$ commutes with localization, it follows that $\Omega_{A_{(X,Y)}}$ is the free $A_{(X,Y)}$-module generated by $dX, dY$ subject to the relation $2Y dY = 3X^2 dX$. You seem to claim that this is free of rank $2$. This is wrong. In fact, after modding out $(Y)$, we get the free $\mathbb{C}[X]_{(X)}$-module generated by $dX,dY$ subject to the relation $X^2 dX = 0$. This has torsion.
However, it is correct that for every other maximal ideal $\mathfrak{m} \subseteq A$ than $(X,Y)$ the $A_\mathfrak{m}$-module $\Omega_{\large A_\mathfrak{m}}$ is free of rank $1$. The reason is that the normalization map $\mathrm{Spec}(\mathbb{C}[t]) \to \mathrm{Spec}(A)$ given by $x \mapsto t^2$, $y \mapsto t^3$, is an isomorphism outside of the origin. Explicitly, if $\mathfrak{m}=(x-a,y-b)$ with $a \neq 0$, then $dY$ is a free generator, and if $b \neq 0$, then $dX$ is a free generator.
