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The title says it all - Suppose you are given a noetherian Gorenstein local ring $(A,m,k)$ of finite Krull dimension.

Does there exist a local complete intersection ring $B$ such that $A$ is a homomorphic image of $B$?

In the answer to this question

A local ring not a quotient of a regular local ring

This paper was given

Link

where there is an example of a Gorenstein ring which is not a quotient of a regular ring, but it is not clear to me if it cannot be expressed as a quotient of a local complete intersection.

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    $\begingroup$ What is your definition of a local complete intersection ring? For me, every such ring is a quotient of a regular local ring. Thus the negative answer to the original question implies a negative answer to this question. $\endgroup$ May 30, 2014 at 17:16
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    $\begingroup$ I thought that c.i means that its completion is a quotient of a regular local ring... $\endgroup$
    – user51400
    May 30, 2014 at 17:18
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    $\begingroup$ "I thought that c.i. means that its completion is a quotient of a regular local ring." I guess that you are correct, at least according to Matsumura. In algebraic geometry, there is a relative notion of local complete intersection, e.g., "LCI morphisms" or an "LCI $R$-algebra" $A$. That definition definitely does require that $A$ is Zariski locally a quotient of a smooth $R$-algebra. So the different usages are somewhat inconsistent. $\endgroup$ May 30, 2014 at 17:30
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    $\begingroup$ And they are not equivalent: arxiv.org/abs/1109.4921 $\endgroup$ May 30, 2014 at 22:33

1 Answer 1

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No. A quotient of a local complete intersection ring has complete intersection formal fibers and so its complete intersection locus is open. In "Greco-Marinari, Nagata's Criterion and Openness of Loci for Gorenstein and Complete Intersection, Math. Z 160 (1978), 207-216", Example 4.3, there is an example of a local Gorenstein ring that do not satisfy this property.

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