When is the module of Kahler differentials free?

Question

As the title says, when is the module of Kahler differentials a free module? In particular, are there known conditions or criterions that could be met that ensures that it will be free?

For example, if one has a finitely generated algebra $S=k[x_0,\cdots,x_n]/(f_1,\cdots, f_l)$ over a field $k$, then one could require that the generators induced from the $f_i$ for $\Omega_{S/k}$ be linearly independent. However, this is a very naive approach. I was curious if there something more interesting. For example, if the ring $S=k[x_0,\cdots,x_n]/(f_1,\cdots, f_l)$ (where $l<n$) has the property that the determinant of the matrix $(\frac{\partial f_i}{\partial x_j})_{i,j=1}^l$ is a unit of $S$. I am not entirely sure if that is accurate on the top of my head, but something along those lines.

Another question is, when is the module of differentials reflexive?

Answer 1

This is just expanding my comment above (which I messed up forgetting dollar signs).

For simplicity, let me assume that $X\subset\mathbb{A}^n$ be a $d$ dimensional smooth variety with $\Omega^1_X$ free of rank $d$ (in characteristic zero, like your situation, it always is smooth, but in positive characteristic, you need to assume smoothness). Then, for a sufficiently large $m$, embed $\mathbb{A}^n\subset \mathbb{A}^{n+m}$ as a linear subspace and then $X\subset\mathbb{A}^{n+m}$ is a complete intersection. Here is a sketch of the proof.

Let $I$ define $X\subset\mathbb{A}^n$. Then one has the Euler sequence, $$0\to I/I^2\to \Omega^1_{\mathbb{A}^n|X}\to\Omega^1_X\to 0.$$

Thus $I/I^2$ is stably free. So, if we emebed $X\subset\mathbb{A}^{n+r}$, for large $r$, and call $I$ as the defining ideal of $X$ in this larger space, one gets $I/I^2$ to be stably free and large rank. A stably free module of sufficiently large rank is free by Bass's theorem. So, we may assume that $I/I^2$ is free (of rank, the codimension of $X$).

Now adding one more variable, say $y$, one can check that $I+(y)$ is in fact generated by the correct number of elements. For this, first pick a set of elements $f_1,\ldots, f_s\in I$ which generate $I/I^2$. Then by Nakayama, it is easy to see that there exists an element $h\in I$ such that $h(1-h)\in (f_1,\ldots, f_s)$ and $I=(f_1,\ldots, f_s,h)$. Then $I+(y)=(f_1,\ldots, f_s,h+y(1-h))$, proving what you want.