Continuity of Intersection Multiplicities

Question

I’m looking for a correct technical version (and in the best case a reference) for a statement of the following type:

Consider a complex algebraic variety $X\subset\mathbb{P}^n$ and a sequence of linear spaces $(L_n)_{n\in\mathbb{N}}$ that is convergent in the Grassmannian $\mathbb{G}(k+1,n+1)$ of k-dimensional linear subspaces of $\mathbb{P}^n$ (in the euclidean topology). Let $L\in\mathbb{G}(k+1,n+1)$ be the limit of this sequence. Now suppose $L_j\cap X$ is 0-dimensional for every $j\in\mathbb{N}$ and the length of the scheme $L_j\cap X$ is equal to m for every $j\in\mathbb{N}$. Then I would like to conclude that the length of $L\cap X$ is also m, given that $L\cap X$ is also 0-dimensional.

Answer 1

Nope; you need $X$ to be Cohen-Macaulay. Consider $X$ a union of two planes in $\mathbb P^4$, meeting at a point $p$. If you intersect $X$ with a general hyperplane, you get two disjoint lines; in the limit that the hyperplane goes through $p$, the lines collide and give you an embedded point at $p$. So far so good.

But now slice everyone with another hyperplane through $p$, i.e. let $(L_n)$ be family of planes, so $L_n \cap X$ is a pair of points until you get to $L \cap X$ which is a fat point of length 3.

The basic problem is that the second hyperplane section doesn't drop the dimension of every component; it shrinks each of the two lines to points, but allows the embedded point in without shrinking it (to be (-1)-dimensional, i.e. the empty set).

(I haven't thought through exactly why $X$ Cohen-Macaulay should be sufficient to cure such problems, but am sure that that's the right condition.)