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Could you give me an example (with proof) of a Gorenstein domain that is not a complete intersection?

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I think maybe this paper answers your question (though he does not construct an explicit example, only proves existence):… – Fredrik Meyer Nov 15 '12 at 19:54
Explicit examples can be extracted from his construction. – user18119 Nov 16 '12 at 0:31
up vote 2 down vote accepted

A classical example of Gorenstein domain which is not a complete intersection comes form the Buchsbaum-Eisenbud structure theorem of Gorenstein ideals of grade $3$ (see Bruns and Herzog, Theorem 3.4.1).

Let's take a skew-symmetric matrix $X$ of size $2r+1$ whose entries are indeterminates over a field $K$. The ideal $I_{2r}(X)$ generated by all $2r$-pfaffians of $X$ has height $3$ and is minimally generated by the set of $2r$-pfaffians, so it has $2r+1$ generators and no less. Moreover, the quotient ring $R=K[X]/I_{2r}(X)$ is a Gorenstein domain. (For more details one can look here.) Now, if $R$ would be a complete intersection ring, then the ideal $I_{2r}(X)$ would be generated by a regular sequence (of length at most $3$). If we take $r>1$, this is not possible.

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there should be a simpler example, something that uses only the theorems from matsumura – Chris Nov 15 '12 at 22:17
I invite you to find one. – user26857 Nov 15 '12 at 22:37

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