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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 $X_k=X$?

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  • $\begingroup$ 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, 2018 at 14:50
  • $\begingroup$ If $X$ is affine, we ask for points defined by exactly $n$ equations. $\endgroup$
    – Bugs Bunny
    Nov 16, 2018 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, 2018 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, 2018 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, 2018 at 9:53

1 Answer 1

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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 $3$.

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