# Can the Kahler differentials of a "good" local ring R be free of rank not equal to dim(R)?

Let $R$ be a local ring containing a field isomorphic to its residue field $k$. Assume $R$ is a localization of a finitely-generated $k$-algebra.

1. Can $\Omega_{R/k}$ be free of rank $r\neq\dim{R}$? If it can, are there bounds on $r$?
2. Does requiring $k$ to be perfect change anything?
• Look at Lemma 17.14.4 here: stacks.math.columbia.edu/tag/01C5. A locally free sheaf has constant rank if the stalks of $\mathcal{O}_X$ are non-zero. So if $R$ is noetherian, the answer is no. Nov 9 '14 at 21:45
• @adrido, I don't see how that Lemma gives a restriction on the rank. Nov 10 '14 at 2:56
• If the rank is constant, it has to be dim R, since $\Omega_{R/k}$ is always locally free of rank dim R at the generic point, no? so is the rank goes up, it can't be locally free. Nov 10 '14 at 6:30
• sorry, what I said is only true in characteristic 0. Nov 10 '14 at 6:48

Yes, this may happen. For example, take $k = {\mathbb F}_p$ and $R := {\mathbb F}_p[X]/(X^p)$. Then $\Omega^1_{R/k}\cong R\cdot\text{d}x$ is free of rank $1$ over $R$. Considering the tensor powers $R\otimes_k ...\otimes_k R$ also shows that there is not even a bound on $r$.

Edit Deleted rubbish. Another try: If $R$ is reduced and $k$ is perfect, then we may realize $R$ as the local ring at a $k$-rational point of a reduced $k$-variety $X$ such that $\Omega_{X/k}$ is free of rank $r$, say. Then $X$ is geometrically reduced, hence generically smooth. Now, for smooth $k$-varieties $X$, the local rank of $\Omega^1_{X/k}$ is the dimension, so we conclude that $X$ is equidimensional of dimension $r$, and in particular $r = \text{dim}(R)$.

• But I think $\mathbb F_p$ is perfect? Also, the bound is supposed to be in terms of $\operatorname{dim}(R),$ right? Nov 9 '14 at 7:28
• @awllower: omg... what a rubbish. Thank you. Nov 9 '14 at 7:30
• No problem; also, I am thinking whether or not the condition every finitely generated extension of $k$ is separably generated suffices? Nov 9 '14 at 7:32
• @awllower: Hm, finitely generated extensions of perfect fields are always separably generated, aren't they? I just gave the answer another try... Nov 9 '14 at 8:00
• @AviSteiner: No, the dimension is $0$ since $(X)$ is the unique prime ideal (it is maximal, and as $X$ is nilpotent, it must be contained in any other prime). Nov 9 '14 at 22:30

I know very little about characteristic $p$ geometry, so I'm assuming $k$ is characteristic zero. Also, I'm taking $R$ to be a domain, so this is more a brainstorm about singularities than about reducible varieties.

Let $K$ be the field of fractions of $R$. Then $\Omega_{R/k} \otimes K$ is a $K$-vector space of dimension $r = \dim(R)$. (Indeed, $\{df_i\}$ form a $K$-basis for this vector space if and only if the $\{f_i\}$ form a transcendence basis for $K/k$.) So right away we see that the only possible rank that $\Omega_{R/k}$ could have, if it were in fact a free $R$-module, is $r$. If it isn't free, it's because $\dim_k (\Omega_{R/k} \otimes k) > r$ as a $k$-vector space.

(Recall that, for finitely-generated modules $M$ over local (noetherian) domains, $M$ is free if and only if $\dim_k (M \otimes k) = \dim_K (M \otimes K)$.)