# The Locus of Complete Intersection Points

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 $$X_k=X$$?

• 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. Commented Nov 16, 2018 at 14:50
• If $X$ is affine, we ask for points defined by exactly $n$ equations. Commented Nov 16, 2018 at 14:53
• 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 $E$). Commented Nov 16, 2018 at 18:35
• (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.) Commented Nov 16, 2018 at 18:37
• I agree, cutting out by $n$ equations makes no sense. If $dim(X)=1$, every point is a complete intersection. Commented Nov 17, 2018 at 9:53

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 $$3$$.