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We say that an affine variety $X \subset \mathbb{A}^n$ of dimension $n-k$ is a complete intersection if the ideal of $X$, call it $I(X)$ is generated by $k$ polynomials, $f_1,\dots,f_k$.

We say that an affine variety is locally a complete intersection if at every point the local ring is a complete intersection ring.

I find this last definition rather opaque.

If a variety is locally a complete intersection does it mean that every point has an affine open neighborhood which is isomorphic to a variety $Y \subset \mathbb{A}^m$ with $Y$ a complete intersection?

A reference to a reliable source with this statement would suffice as an answer. I'm not sure I know enough to understand a proof of this result yet. If you can give me some intuition that would be great too!

Seth
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  • So the local complete intersection ring is a Noetherian local ring whose completion is the quotient of a regular local ring by an ideal generated by a regular sequence. I think the ideas is similar to being a complete intersection. Here you are working in the completion where you find a similar $I(X)$ generated by $k$ elements which are a regular sequence. This should mean that locally we can write the variety as a complete intersection. By locally however, I don't mean in a Zariski open set, I mean in a formal neighborhood, so you can think of it as taking arbitrarily small neighborhoods. – Sergio Da Silva Mar 26 '14 at 00:21
  • This might also be useful: http://mathoverflow.net/questions/27197/local-complete-intersections-which-are-not-complete-intersections?rq=1 – Sergio Da Silva Mar 26 '14 at 00:37
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    This is similar to what Sergio has told you. A complete intersection is one where it takes precisely the amount of hypersurfaces that you'd guess (from the dimension) to cut it out of affine space. The same intuition applies for local complete intersections, but this is only true "Zariski close". So, if you zoom in on each point (in the sense of the Zariski topology!) your variety does, indeed, look like the intersection of the right number of hypersurfaces--but this is only (Zariski!) locally true. – Alex Youcis Mar 26 '14 at 00:55
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    @SergioDaSilva thanks for the comment. What do you mean by "formal neighborhood"? I guess I do see an analogy between the notion of the local ring being a complete intersection and the variety being a complete intersection, although it's still not completely clear. – Seth Mar 26 '14 at 01:14
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    @AlexYoucis so you believe the statement is true? By open neighborhood I meant Zariski open. – Seth Mar 26 '14 at 01:14
  • Seth: When you look at the completion, it is like looking at formal power series. When doing operations in the completion of the local ring, we want to know what is really happening down below in the local ring. For example, the Whitney Umbrella $x^2+yz^2=0$ is not reducible, but along the y-axis away from the origin, we can split it as $(x+\sqrt{y}z)(x-\sqrt{y}z)$ in the completion. This is supposed to translate into your variety being reducible locally in the Euclidean sense (think of normal crossings for example). I believe the lci is in this sense (stronger than the Zariski close sense). – Sergio Da Silva Mar 26 '14 at 01:33
  • I could be wrong in my understanding though, so perhaps more expert members can comment. – Sergio Da Silva Mar 26 '14 at 01:35
  • Locally a complete intersection means that any point has an open neighborhood which is a complete intersection in some open subvariety $U$ of some $\mathbb A^m$. The definition of complete intersection in $U$ is similar to that of complete intersection in an affine space. – Cantlog Mar 26 '14 at 17:26
  • @Cantlog can you state or give a reference for the definition? – Seth Mar 27 '14 at 19:41
  • @Seth: to state which definition ? – Cantlog Mar 27 '14 at 21:16
  • @Cantlog I mean the definition of being a complete intersection in $U$. – Seth Mar 27 '14 at 21:53
  • Suppose $U$ is affine. Then it means the ideal definining the subvarietiy you are concerned with is generated by the right number of regular functions. – Cantlog Mar 28 '14 at 21:37

1 Answers1

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The answer to your question "If a variety is locally a complete intersection does it mean that every point has an affine open neighborhood which is isomorphic to a variety $Y \subset \mathbb A^m$ with $Y$ a complete intersection?" is yes.

See e.g. this page of the stacks project, especially Lemma 10.128.9.

Matt E
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  • Dear Matt E, you probably meant Lemma 10.128.4: http://stacks.math.columbia.edu/tag/00SC – Cantlog Apr 01 '14 at 07:27
  • Thanks for the answer but I don't understand how anything on that page gives us the result I am looking for. Maybe I just don't have the background to decipher how what is being said there applies to varieties. – Seth Apr 01 '14 at 11:48
  • @Cantlog: Dear Cantlog, No, I meant Lemma 10.128.9, which says that a f.t. $k$-alg. $S$ is a l.c.i. (i.e., is Zariski locally a global c.i., by definition) iff all its local rings are complete intersection rings. Regards, – Matt E Apr 01 '14 at 14:22
  • @Seth: Dear Seth, Did you look at Lemma 10.128.9, and at Definition 10.128.1? Taken together, they exactly say that a f.t. $k$-alg. is, Zariski locally, a global complete intersection iff its local rings are complete intersection rings. This precisely answers your question. Regards, – Matt E Apr 01 '14 at 14:24
  • @MattE Thank you, I'm starting to understand. I just need to figure out what $D(g_i)$ and $S_{g_i}$ mean. Is $D(g_i)$ supposed to be a basic open neighborhood corresponding to a regular function $g_i$ and $S_{g_i}$ the localization at the multiplicative subset of powers of $g_i$? Then I think its true that the ring of regular functions on $D(g_i)$ is $S_{g_i}$. Is this correct? I am new to this subject and I haven't learned things from this perspective yet. If what I said is correct then I do believe this answers my question. – Seth Apr 03 '14 at 00:02
  • Ok, I thought about it some more and I'm 99% sure that what I said before is right so I accepted your answer. – Seth Apr 03 '14 at 00:11
  • @Seth: Dear Seth, Your interpretation of $D(g_i)$ and $S_{g_i}$ are correct, and yes, the functions on $D(g_i)$ are precisely $S_{g_i}$. Cheers, – Matt E Apr 03 '14 at 01:58