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Let $X$ be a variety over $\mathbb{C}$, and $X^{an}$ be the analytic space associated to $X$. $X$ has Gorenstein singularity at $x \in X$ iff the local ring $\mathcal{O}_{X,x}$ is a Gorenstein ring. Is this result true:

$\mathcal{O}_{X,x}$ is a Gorenstein $\iff$ $\mathcal{O}_{X^{an},x}$ is a Gorenstein.

I want to show the following thing: Suppose $X,Y$ are varieties over $\mathbb{C}$, and $x \in X^{an}, y\in Y^{an}$. Suppose $U \subset X^{an}, V\subset Y^{an}$ are open sets in the Eucliden topology, and there exists an analytic isomorphism $f : U \to V$ sending $x$ to $y$. Then, if $X$ has Gorenstein singularity at $x$, it also has Gorenstein singularity at $y$. Certainly, if the aforementioned result holds, this result follows from it.

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1 Answer 1

up vote 9 down vote accepted

The key point is that if $X$ is a locally finite type $\mathbf{C}$-scheme and $x \in X(\mathbf{C})$ then $O_{X,x}$ and $O_{X^{\rm{an}},x}$ are both local noetherian rings with the same completion (induced by the evident canonical map from the algebraic local ring to the analytic one). So for any property of local noetherian rings which is equivalent to check on the completion (such as the Gorenstein or CM properties), one gets immediately the equivalence on algebraic and analytic sides by comparison through common complete local rings.

Local rings on complex-analytic spaces and algebraic schemes over a field are excellent, so for more delicate properties which cannot always be checked on completions for general local noetherian rings (such as normality and reducedness) the equivalence still goes through; see Expose XII in SGA1 for a thorough discussion of these matters.

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Thank you so much for your answer!! –  Li Yutong Jul 13 '13 at 3:26

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