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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.

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  • $\begingroup$ 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
    Commented Feb 17, 2019 at 23:09
  • $\begingroup$ @Jason Starr Sorry, I meant $S$ is a finite etale extension of $R[1/p]$. $\endgroup$
    – gdb
    Commented Feb 17, 2019 at 23:10

1 Answer 1

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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 $\dim_{\mathbb{F}_p}\left(S^+/(p,x,y)\right)=3$).

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  • $\begingroup$ Thanks! Nice example. $\endgroup$
    – gdb
    Commented Feb 18, 2019 at 21:24

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