Let $R$ be a $p$-torsion free ring which is integrally closed in $R[1/p]$ and let $S$ be a finite etale extension of $R[1/p]$.

Is it true that an integral closure $S^+$of $R$ in $S$ is flat over $R$?

Remark 1: It suffices to show that $S^+/p$ is flat over $R/p$, but I don't know how to see this.

Remark 2: I am mostly interested in the situation when $R$ is non-noetherian by itself, but $R[1/p]$ is. But I don't know whether this result holds even in the noetherian case.

If it helps, feel free to add extra conditions on a pair $(R, R[1/p])$. For example, in my main case of interest $R$ is $p$-adically complete and $R[1/p]$ is regular. Though I am not sure how useful it is.

  • $\begingroup$ I do not understand the statement. Did you intend to write that $S$ is etale and quasi-finite over $R$? If $S$ is finite over $R$, then the integral closure of $R$ in $S$ equals $S$. $\endgroup$ – Jason Starr Feb 17 at 23:09
  • $\begingroup$ @Jason Starr Sorry, I meant $S$ is a finite etale extension of $R[1/p]$. $\endgroup$ – gdb Feb 17 at 23:10

No. Take $R=\mathbb{Z}_p[x,y]/(xy-p^2)$, $S^+=\mathbb{Z}_p[u,v]/(uv-p)$, and map $R$ into $S^+$ by $x\mapsto u^2$, $y\mapsto v^2$. Note that $R$ is normal, $S^+$ is a finite extension of $R$, étale of degree 2 over $\mathbb{Q}_p$, but not flat, e.g. because $S^+$ is regular and $R$ isn't (or because $\dim_{\mathbb{F}_p}\left(S^+/(p,x,y)\right)=3$).

  • $\begingroup$ Thanks! Nice example. $\endgroup$ – gdb Feb 18 at 21:24

since the etale extension use a concept of open immersion, only closed immersion can preserve surjective. so your algebraic object seems to be Cartesian only in local, in general it still has torsion (for example, use the category of fibre product)! that is what you are puzzling.

the regular property can ensure a direct sum of free bases, which means that we can ignore its intersectional mettings. the completeness of p-adic number allow us to do a lifting of power, which may eliminate its torsion group. since a torsion group is just the standard form of power $t_1\vert{t_2\vert......\vert{t_k}}$ !

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