# Flatness of the integral closure

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.

• 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 $S$. – Jason Starr Feb 17 at 23:09
• @Jason Starr Sorry, I meant $S$ is a finite etale extension of $R[1/p]$. – 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 $$\dim_{\mathbb{F}_p}\left(S^+/(p,x,y)\right)=3$$).
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 $$t_1\vert{t_2\vert......\vert{t_k}}$$ !