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 $\Omega[X]$?
Kähler differentials not the same as regular differentials on a singular curve
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0@Alan I'm guessing it has something to do with $y^2=x^3$ implying $2ydy=3x^2 dx$. Multiplying by $x$ gives $2xydy=3x^3 dx=3y^2 dy$ and dividing by $y$ gives $2xdy=3ydx$. – 2012-05-05
1 Answers
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 $dy\mapsto y$.