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 $R[1/p]$. 



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



Flatness of the integral closure
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 $\dim\_{\mathbb{F}\_p}\left(S^+/(p,x,y)\right)=3$).


