Let $A$ be an integral complete local ring over a field which is complete intersection.
Let $B$ be a normalization of $A$.
Q. Is $B$ Gorenstein?
I guess that even the normalization of Gorenstein local ring should be Gorenstein.
$\begingroup$
$\endgroup$
4
-
8$\begingroup$ No. Every variety is the normalization of a hypersurface. mathoverflow.net/questions/68246/… $\endgroup$– Karl SchwedeCommented May 26, 2016 at 2:15
-
5$\begingroup$ @KarlSchwede: Well, every normal (projective) variety! $\endgroup$– Laurent Moret-BaillyCommented May 26, 2016 at 6:02
-
$\begingroup$ :-) that is true $\endgroup$– Karl SchwedeCommented May 26, 2016 at 6:39
-
3$\begingroup$ Cross-link to meta discussion about why this question was closed. $\endgroup$– Gro-TsenCommented May 26, 2016 at 17:06
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Consider $A=k[[x^3,x^2y,y^3]]\subset k[[x^3, x^2y, xy^2, y^3]]=B$. $B$ is the integral closure of $A$, $A$ is a hypersurface, but $B$ is not Gorenstein.
