# Local complete intersections which are not complete intersections

The following definitions are standard:

An affine variety $V$ in $A^n$ is a complete intersection (c.i.) if its vanishing ideal can be generated by ($n - \dim V$) polynomials in $k[X_1,\ldots, X_n]$. The definition can also be made for projective varieties.

$V$ is locally a complete intersection (l.c.i.) if the local ring of each point on $V$ is a c.i. (that is, quotient of a regular local ring by an ideal generated by a regular sequence).

What are examples (preferably affine) of l.c.i. which are not c.i. ? I have never seen such one.

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(To supplement Alberto's example)

If $V$ is projective, then the gap between being locally c.i and c.i is quite big. In particular, any smooth $V$ would be locally c.i., but they are not c.i. typically. For instance, take $V$ to be a few points in $\mathbb P^2$ would give simple examples. In higher dimensions, by Grothendick-Lefschetz, if $V$ is smooth, $\dim V\geq 3$, and $V$ is c.i. then $\text{Pic}(V)=\mathbb Z$, so it is a serious restriction.

The affine case is more subtle. Again one can look at smooth varieties. If $V$ is a smooth affine curve and c.i., then the canonical bundle of $V$ is trivial. So it gives the following strategy: start with a projective curve $X$ of genus at least $2$, removing some general points to obtain an (still smooth) affine curve with non-trivial canonical bundle.

For more details on the second paragraph, see this question, especially Bjorn Poonen's comments. This paper contains relevant references, and also an example with trivial canonical bundle.

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So the condition l.c.i. is made to be fulfilled naturally by taking smooth varieties. It would be fine if we have also singular examples at hands... –  Adam K Jun 7 '10 at 14:24

The first example is the twisted cubic in $\mathbb{P}^3$.

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This is not, however, an affine example, since in $\mathbb{A}^3$, the twisted cubic $\{(t, t^2, t^3) : t \in k\}$ is the vanishing set of the two polynomials $y-x^2$, $z - x^3$. –  Charles Staats Jun 5 '10 at 22:06
I was worrying about a very similar problem concerning complete intersections recently, and the twisted cubic provided a counterexample to what I hoped was true. Very depressing when a counterexample is contained in the second exercise in Hartshorne's Alg. Geom... –  Matthew Morrow Jun 5 '10 at 22:49
Matthew, it's not that depressing, Hartshorne's exercises include many good theorems (-: –  Hailong Dao Jun 5 '10 at 22:56
But the cone on a twisted cubic should furnish an affine example... –  Michael Thaddeus Jun 6 '10 at 9:31
Michael: the local ring at the origin is not c.i. –  Hailong Dao Jun 6 '10 at 9:58

EDIT: This is wrong. I haven't deleted it in order that the subsequent comments make sense.

You will never see an example, for the following reason: given a local complete intersection $V$ inside $\mathbb{A}_k^n$, you can always find a global complete intersection $W$ inside $\mathbb{A}_k^n$ such that the reduced varieties associated to $V$ and $W$ are the same.

Proof:

Suppose $I$ is an ideal of $k[X_1,\dots,X_n]$ such that the variety $V(I)$ is a local complete intersection. This forces all the local rings of $V(I)$ to be Cohen-Macaulay, hence equidimensional. So the irreducible components of $V(I)$ all have the same codimension in $\mathbb{A}_k^n$; lets call this codimension $r$.

Since $k[X_1,\dots,X_n]$ is Cohen-Macaulay, the height of $I$ (which is $r$) is the same as its depth, meaning that $I$ contains a regular sequence $f_1,\dots,f_r$ of length $r$. By considering heights we see that the minimal primes over the ideal $J=\langle f_1,\dots,f_r\rangle$ are the same as the minimal primes over $I$. Therefore $J$ and $I$ have the same radical, which implies the claim (with $W=V(J)$). QED

So if you are trying to draw counterexamples, you have to worry about whether that line on the paper has nilpotent elements in the structure sheaf...

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Matthew: the minimal primes over $(f_1,\cdots, f_r)$ are not necessarily those over $I$. Indeed, deciding whether a variety is set-theoretic complete intersection is a subtle problem. –  Hailong Dao Jun 5 '10 at 23:32
Yes, 25 minutes after posting I realised my error and have returned to edit my 'answer'. But I thought that the notion of analytic spread and set-theoretic complete intersection were understood well in the affine case, though not the projective. Are they not? –  Matthew Morrow Jun 5 '10 at 23:56
@Matthew: Is not known whether any irreducible affine curve in $\mathbb C^3$ is a set-theoretic c.i. See problem 5 [here](www.math.lsa.umich.edu/~hochster/Lip.text.pdf ) –  Hailong Dao Jun 6 '10 at 0:06
Hailong, thanks very much for the reference! I've actually been reading an old paper by Hochster on Cohen-Macaulay rings (and I've just noticed from your webpage that he was your supervisor!). Thanks again. –  Matthew Morrow Jun 6 '10 at 8:58
From Hailong's answer, I suppose it is possible to make simpler examples as follows: take $V$ a smooth affine variety which is not equidimensional (so clearly it is l.c.i but not c.i). For instance, $V$ is the union of the plane $z = 0$ and the line $z=1, x=y$ in $\mathbb A^3$. $V$ is smooth (it can be proven that $I(V) = (zx-zy, z^2-z)$).
The disadvantage of this construction is $V$ must be reducible.