# Is the integral closure of a Gorenstein domain still Gorenstein?

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?

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 $$S$$.

• Would you mind adding some citation/justification/explanation on why these rings are/aren't Gorenstein? Perhaps it is obvious to you, but I have yet to develop the skill of looking at a ring and immediately knowing if it's Gorenstein or not. Nov 30, 2022 at 5:29
• And what if we replace the Gorenstein property with the Cohen-Macaulay property?
– Fraz
Nov 30, 2022 at 9:31
• @Fraz There exists Gorenstein domains (of dimension at least three) such that its integral closure is not Cohen-Macaulay. Explicitly writing one down may be messy, but existence can be proven using some algebraic geometry. Nov 30, 2022 at 22:25
• Thankyou! So you say that the integral closure of a $2$-dimensional Gorestein ring is always Cohen-Macaulay? Sorry for the specific question, but in my work I have to handle $2$-dimensional rings
– Fraz
Dec 1, 2022 at 13:49
• @Fraz Look up Serre criterion for normality, which will imply that for dimension two, normality implies Cohen-Macaulay. Dec 1, 2022 at 15:02