Local complete intersections which are not complete intersections - MathOverflow

Local complete intersections which are not complete intersections

2010-06-05T21:36:06Z

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.

Answer by Alberto García-Raboso for Local complete intersections which are not complete intersections

2010-06-05T21:38:18Z

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

Answer by Hailong Dao for Local complete intersections which are not complete intersections

2010-06-05T22:26:10Z

(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.

Answer by Matthew Morrow for Local complete intersections which are not complete intersections

2010-06-05T23:28:14Z

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...

Answer by Adam K for Local complete intersections which are not complete intersections

2010-06-06T09:15:07Z

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.

Please correct me if I'm wrong.