# When is the module of Kahler volume forms torsion-free?

Let $R$ be a commutative algebra over a field $k$. Denote the $R$-module of Kahler differentials by $\Omega^1_kR$; this is the $R$-module generated by symbols of the form $da$, $a\in R$, and relations $$d(\lambda a+b)=\lambda da+db,\;\;\; dab =adb+bda,\;\;\; d1=0, \;\;\; \forall a,b\in A, \lambda\in k$$ The $R$-module of $i$-forms is the $i$th exterior power $\Omega^i_kR:= \Lambda^i_k\Omega^1_kR$.

When $R$ has Krull dimension $n$, call $\Omega^n_kR$ the module of volume forms. When $R$ is the coordinate ring of a smooth affine variety $X$ of dimension $n$, then elements of $\Omega^nR$ determine volume forms on $X$.

For singular varieties, the behavior of $\Omega^nR$ can be weirder. For example, the simple cusp in the plane $$R=\mathbb{C}[x,y]/x^3-y^2$$ has a volume form $3ydx-2xdy$ which is non-zero, but $$y(3ydx-2xdy) = 3x^3dx-2xydy = xd(x^3)-xd(y^2)=0$$ So, the module $\Omega^1_kR$ of volume forms here has a torsion element.

My question is: Under what conditions on $R$ is $\Omega^n_kR$ torsion-free?

• Do you mind if I ask if there is any intuition behind this particular example of a torsion element? I bumped on your question by trying to find such a torsion element as it is an exercise in Matsumura's book on Commutative Algebra. – Patrick Da Silva May 24 '15 at 4:10

This is an interesting question! When $n=1$ there is a conjecture by Berger that the module of differential is torsion-free iff the curve $X$ is non-singular. Googling Berger conjecture module of differentials" should give you some references. Bernd Ulrich's first paper (by Mathcsinet) was on this topic, so you can look there also. I do not know anything about higher dimensions but it is plausible that torsion-freeness of $\Omega^n_k$ imposes strong conditions on $R$, as tensor products of non-projective modules tend to have torsions.
• Long... thanks for the references! Your intuition in the last sentence is the opposite of what I expected, though. The one positive example I did completely was the cone $xz=y^2$, in which $\Omega^1$ is non-projective, but $\Omega^2$ is torsion-free. Hence, I was hoping for something like 'integrally-closed' implies 'torsion-free volume forms'. – Greg Muller Apr 16 '13 at 23:13