# What does it mean geometrically for a variety to be locally a complete intersection?

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!

• 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. Mar 26 '14 at 0:21
• This might also be useful: mathoverflow.net/questions/27197/… Mar 26 '14 at 0:37
• 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. Mar 26 '14 at 0:55
• @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 1:14
• @AlexYoucis so you believe the statement is true? By open neighborhood I meant Zariski open.
– Seth
Mar 26 '14 at 1:14

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.

• @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, Apr 1 '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, Apr 1 '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.