Absolute integral closure of local UFD

Question

Let $R$ be a Nagata Noetherian local UFD, and $K$ be its fraction field. I wonder if its absolute integral closure $R^+$, which is the integral closure of $R$ in $K^\text{sep}$, is flat over $R$. Let $L$ be a finite separable field extension of $K$, and $S$ be the integral closure of $R$ in $L$. We can also ask if $S$ is flat over $R$. It seems unduly good for a UFD, but I can't get any counterexample by myself.

Answer 1

For the second question, let $k$ be a field of characteristic zero and $S$ a quotient of $k[[x_1,\dots,x_m]]$ which is a normal domain but not Cohen-Macaulay (in particular, $\dim(S)=d\geq3$). If $(t_1,\dots,t_d)$ is a system of parameters for $S$, then $S$ is finite over $R:=k[[t_1,\dots,t_d]]$ which is regular, hence a UFD. Clearly, $S$ is the normalization of $R$ in $\mathrm{Frac}(S)$ which is finite separable over $\mathrm{Frac}(R)$. But $S$ is not flat over $R$, otherwise it would be Cohen-Macaulay.

Answer 2

For the finite extensions as you describe. There is no chance they will always be flat even when $R$ is a regular ring of characteristic $p > 0$. Indeed, there are plenty of finite extensions $R \subseteq S$ such that $S$ is not Cohen–Macaulay. The interesting thing is that in the colimit they become Cohen–Macaulay, by work of Sannai–Singh "Galois extensions, plus closure, and maps on local cohomology" in char $p > 0$, see Corollary 3.3. (in mixed characteristic, the $p$-adic or $m$-adic completion of $R^+ = R^{+,sep}$ is Cohen–Macaulay by a recent breakthrough of Bhatt).

For the first question in characteristic $p > 0$, as I already mentioned, ${R^{+,\text{sep}}}$ is Cohen–Macaulay by Sannai–Singh. That will force $R$ to be Cohen–Macaulay.

I think also by mimicking work of Linquan Ma and myself in mixed characteristic (based on earlier work of Ma), it should also force $R$ to have pseudo-rational = klt singularities (I can give details if desired), and probably some variant of $F$-regular. My guess is it should also force $R$ to be regular, but I don't see that right now.

Answer 3

Paul Roberts remarks before Example 1 in https://www.ams.org/journals/proc/1980-078-03/S0002-9939-1980-0553363-8/S0002-9939-1980-0553363-8.pdf that if $R$ is a UFD and if $S$ is a module finite normal extension of $R$ such that the corresponding extension of fraction fields is Galois, has an Abelian Galois group, and is of degree coprime to the characteristic then $S$ is free (and in particular flat) as an $R$ module.

As Laurent and pop1 point out when $R$ is of dimension $3$ or more and has equicharacteristic zero, $R^{+}$ is not flat over $R$ even when $R$ is regular as it is not Cohen-Macaulay. Here are some more details and different perspectives regarding your question about $R^{+}$ when $R$ has equicharacteristic zero.

In fact it is easy to see that there are no excellent rings $R$ such that $R^{+}$ is flat over $R$ when $R$ has dimension $4$ or more - one can simply localize to a prime of height $3$ in the regular locus. This is Lemma 3.3 of https://arxiv.org/pdf/2212.09025.pdf (there is an excellence hypothesis missing in the statement of the lemma). When $R$ is of equicharacteristic zero and has dimension $0$ or $1$ it follows from standard commutative algebra that $R^{+}$ being flat over $R$ is equivalent to $R$ being regular. When $R$ is regular and has dimension $2$ then $R^{+}$ is flat over $R$. Conversely, if $R^{+}$ is flat over $R$ then $R$ is regular assuming further that $R$ is a $\mathbb{N}$ graded domain finitely generated over an equicharactersitic zero field. This is the main result of the above arxiv preprint.

To conclude and to incorporate what Karl is saying, perhaps a better question to ask would be, ``for what $R$ does $R^{+, sep}$ or the finite extension $S$ have finite flat dimension over $R$ ? " This takes care of the issues Laurent and Karl mention. If $R^{+, sep}$ is replaced by $R^{+}$ then this follows from `Kunz's theorem' due to Bhatt, Iyengar and Ma https://arxiv.org/pdf/1803.03229.pdf (Theorem 4.7, 4.13) since $R^{+}$ is a perfect(oid) ring. In equicharacteristic zero (where $R^{+, sep} = R^{+}$) it is an open problem.

Here is a simpler way to see why $R^{+, sep}$ being flat over $R$ implies $R$ has rational singularities. Flat morphisms are pure and see Corollary 3.4 of this older paper of Singh (https://link.springer.com/article/10.1007/s002290050156). Hence it implies $R$ is in fact a splinter, and it is well known by work of Karen Smith (`F-rational rings have rational singularities') that splinters have rational singularities. As Karl says it should probably follow with some more work that $R$ is regular, and one can also ask if vanishing of Tors of $R^{+, sep}$ characterizes regularity.