Fulton's Book on intersection theory (Pg.223, theorem 12.3) asserts the following result:
For r pure dimensional schemes in P^n, whose co-dimensions add to at most n, the product of their degrees is at least as great as the sum of the degrees of the irreducible components of their intersection.
Under what conditions can we say that the product of the degrees of schemes is equal to the degrees of the irreducible components of the scheme?
Thanks.
Call the schemes to be intersected $(X_i)$, where $X_i$ has pure codimension $r_i$ in ${\mathbb P}^n$. Let $R = \sum r_i$. (Edited so as not to restrict to $R = n$ unnecessarily.)
Definitely, every component of the intersection has codimension at most $R$. If the codimensions are all exactly $R$, and the schemes being intersected are Cohen-Macaulay, then the product of the degrees = the degree of the intersection ( = the sum of the degrees of its primary components).
Non-example: let $X$ be the projective completion of a random plane through the origin in $A^4$, and $Y$ the projective completion of the union of two other random planes through the origin (so, not Cohen-Macaulay). Then $X \cap Y$ is a triple point, not a double point as one might hope $(deg\ X = 1,deg\ Y = 2)$. The basic issue is that if we think about intersecting $Y$ first with a $3$-plane $X' \supset X$, we get a union of two lines plus an embedded point we should throw away before we go all the way down to $X$. Then the intersection of $X$ picks up a point for each line in $Y \cap X'$, which is good, but also the embedded point, which is a failure of codimensions adding up.
In this non-example $r_1 = r_2 = 2$, $n = R = 4$.
