# When is the module of Kahler differentials free?

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 $$l) has the property that the determinant of the matrix $$(\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?

• Freeness of $\Omega_{S/k}$ is very closely related to smoothness of $S$ as a $k$-algebra (one has to assume also that it has the correct rank, to avoid issues with characteristic $p$); this should be in most commutative algebra resources. If $S$ is not smooth, then it seems a bit subtle exactly when $\Omega_{S/k}$ is reflexive, but in general one expects to have both torsion and cotorsion (see, e.g., arxiv.org/pdf/1012.5940.pdf for the failure of reflexivity for some fairly mild singularities). – Devlin Mallory May 23 '20 at 1:15
• For $S$ as above, $\Omega^1_X$ is free of rank $=\dim X$ (X=\operatorname{Spec} S$implies$S=k[x_0,\ldots,x_n, x_{n+1},\ldots x_m]/(g_1,\ldots, g_{l+m})$where the$g_i$form a regular sequence, that is$S$is a complete intersection possibly after adding more variables. – Mohan May 23 '20 at 1:58 • @Mohan Could you explain this a bit further please? Perhaps add some details? This would answer a huge problem I have been facing with these modules! – Plank May 23 '20 at 2:11 • @DevlinMallory Ah yes, I see what you mean that these modules are closely related to smoothness. I mean this is the essence of smooth varities and when the sheaf of differentials is finite locally free, there is a lot of interaction. Thanks for the reference, that covers part of the question nicely! – Plank May 23 '20 at 2:13 • To give one more precise reference (for$\Omega ^1_{X/k}$free of rank$\dim X\Longleftrightarrow \ X$smooth over$k\$): Kunz Kähler differentials, Theorem 8.1. – abx May 23 '20 at 7:59

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.