Is the integral closure of a Gorenstein domain still Gorenstein?

Question

Let $(R,\mathfrak{m})$ be an $n$-dimensional local Gorenstein domain. Is it true that the integral closure of $R$ in its fraction field is still an $n$-dimensional local Gorenstein domain?

Answer 1

Let $R=k[[x^3,x^2y,y^3]]$ which is Gorenstein and its integral closure is $R[xy^2]$ which is not Gorenstein.

I add a few more details as a commenter requested.

Let $S=R[xy^2]=k[[x^3,x^2y,xy^2,y^3]]$. Then, $R\subset S$ and $S$ is integral over $R$ is clear, since $(xy^2)^3=x^3y^6\in R$. Also, $xy^2=(x^2y)^2/x^3$ shows that $S$ is birational to $R$. Since $R$ has dimension two and embedding dimension three, it is a hypersurface. Complete intersections are Gorenstein.

So, it remains to show that $S$ is integrally closed and it is not Gorenstein.

I quote a standard theorem, even though verifying that $S$ is normal can also be done by hand.

Theorem: Let $A$ be a normal domain and let a finite group $G$ act on $A$. Then $A^G$, the subring of $G$-invariants is normal.

One easily checks that $S\subset k[[x,y]]=P$ and $G=\mathbb{Z}/3\mathbb{Z}$ acts on $P$ by multiplication on $x,y$ by third root of unity and $S=P^G$.

Lastly, to check $S$ is not Gorenstein, again there are several ways, let me describe one. One checks that $S/(x^3,y^3)$ is isomorphic to $k[u,v]/(u^2,uv,v^2)$ and since this ring has socle dimension two, it is not Gorenstein and thus nor is $S$.