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An algebraic variety is called a complete intersection if its defining ideal is generated by codimension many polynomials.

A Noetherian local ring $R$ is called a complete intersection if its completion is the factor ring of a regular local ring by a regular sequence.

What is the connection between these two definitions?

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They are compatible when they both apply. For instance, if $P$ is a point on a variety $V$, then $V$ is a local complete intersection at $P$ -- i.e. there is some affine open neighborhood of $P$ which is isomorphic to a complete intersection in affine space -- iff the local ring at $P$ is a complete intersection ring. Or so I would think: I am leaving this as a comment rather than an answer to give someone with more expertise an opportunity to correct me. –  Pete L. Clark Jul 9 '11 at 16:06
    
@Pete: Could you give me a proof of that, please? –  Jacob Fox Jul 9 '11 at 16:13
    
Not off the top of my head, unfortunately. I would have to pull some books off my shelf (and, you know, it's Saturday). More seriously, there are other regulars on this site who would do a better job. I'll bet they'll show up soon enough: let's see. –  Pete L. Clark Jul 9 '11 at 16:41
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I did a quick search through EGA, and IV-19.3.2 the following is proved: A quotient of a regular local noetherian ring is a complete intersection if and only if the ideal is generated by a regular sequence. So this means that the question reduces to whether a variety is locally a complete intersection (in the classical sense) if and only if it is cut out by a regular sequence. One direction holds (if it's cut out by a regular sequence, then the dimension is what you want), but the other seems much less obvious to me. –  Akhil Mathew Jul 9 '11 at 17:03

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I heard it is open whether a local complete intersection (in your second sense), is actually a complete intersection. That is, if a local ring $R$ is such that $\hat R\cong T/(\underline x)$, $T$ regular, $\underline x$ regular on $T$, then must $R$ be isomorphic to $Q/(\underline y)$, $Q$ regular, $\underline y$ regular on $Q$?

But I found this on MO which is probably more relevant to your actual question.

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Well, not open anymore: arxiv.org/abs/1109.4921 –  user26857 Nov 25 at 12:33

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