10
$\begingroup$

Let $X$ be an algebraic variety over an algebraically closed field. Consider the two subsets $X_0\subseteq X_1 \subseteq X$: $$X_0 = \{a\in X| a \mbox{ is a scheme-theoretic complete intersection in }X\},$$ $$X_1 = \{a\in X| a \mbox{ is a set-theoretic complete intersection in }X\}.$$

Question: what do we know about these sets?

I am looking for any positive information, possibly for some restricted classes of varieties. Here are some precise questions:

(1) Is $X_k$ open, closed, locally closed, constructible, etc.?

(2) What are the varieties with $X_k=\emptyset$?

(3) What are the varieties with $X_k=X$?

Improve this question
$\endgroup$
5
  • $\begingroup$ Let $n=dim(X)$, $I_x$ the sheaf of ideals determining the point $x$. We are looking for $n$ codimension 1 subvarieties with sheaves of ideals $I_1,\ldots I_n$ such that $I_x=I_1+\ldots +I_n$ for the scheme-theoretic c.i. and $I_x=\sqrt{I_1+\ldots +I_n}$ for the set-theoretic c.i. $\endgroup$
    – Bugs Bunny
    Nov 16 '18 at 14:50
  • $\begingroup$ If $X$ is affine, we ask for points defined by exactly $n$ equations. $\endgroup$
    – Bugs Bunny
    Nov 16 '18 at 14:53
  • $\begingroup$ In the affine case, asking for complete intersections of divisors does not guarantee that the ideal is cut out by exactly $n$ equations. Even for $n = 1$ this is false; for example if $E$ is an elliptic curve and $X = E \setminus O$, then no point $x \in X$ is cut out by one equation (since $\operatorname{Pic}(X) \stackrel\sim\to E(k)$ by $x \mapsto x$ extended linearly using the addition on $E$). $\endgroup$
    – R. van Dobben de Bruyn
    Nov 16 '18 at 18:35
  • $\begingroup$ (I thought at first that this would be fixed by looking at very ample divisors only, but any divisor on the same $X = E \setminus O$ is very ample since $x + nO$ is very ample on $E$ for $n \gg 0$, so the situation is no better.) $\endgroup$
    – R. van Dobben de Bruyn
    Nov 16 '18 at 18:37
  • $\begingroup$ I agree, cutting out by $n$ equations makes no sense. If $dim(X)=1$, every point is a complete intersection. $\endgroup$
    – Bugs Bunny
    Nov 17 '18 at 9:53
5
$\begingroup$

This should be difficult in general. However there are some easy remarks to get going:

First, if $p$ is a CI point, then $X_p$ is regular. That is because when you localize, the number of generators can only drops, and it is still have to be at least $n=\dim X$. So they are equal.

Now let's try to answer number 3), when $X_0=X$? The above remark says that $X$ is non-singular. But it is more, it says that the Chow group of points is trivial.

Consider $X$ projective. Clearly the property $X_0=X$ depends on the embedding. For example, with $P^1= \text{Proj} \ k[x,y,z]/(xy-z^2)$, the point $(x,z)$ is not CI. So we just consider $X= \text{Proj}\ S/I$ with $S=k[x_0,...,x_d]$. Then a point $p$ in $\text{Proj} S$ is defined by $d$ linear forms. Modulo $I$, the number of generators drops to $n$, so $I$ must contains $d-n$ forms, and by dimension reasons, $I$ is generated by those. So $X$ must be $P^n$. (indeed, here we only needs to assume that $X_0$ is non-empty, so this also answers Question (2)).

It is harder when $X$ is affine. Except in dimension one, then we are looking for a smooth affine curve with trivial Picard group, so it must be rational curve.

I don't know much about the sCI case, or when $X_1=X$. If $X$ is smooth, we are forcing the Chow group of points to have rank one, and this should be restrictive.

As for number (1), there is a paper by Weibel, where he conjectured that for affine $X$, the set $X_0$ is a countable union of closed subsets, and solved it for dimension at most $3$.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.