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?


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    $\begingroup$ Should probably add the [Fulton] page number. $\endgroup$ Jan 20, 2011 at 18:49

1 Answer 1


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 $r_1 = r_2 = 2$, $n = R = 4$.

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    $\begingroup$ Allen, I don't understand your non-example. Are you saying that $Y$ as a cycle is not the sum of its components? Or are you saying that the intersection of two random planes through a point is $1.5$ points? Finally, are you saying that the degree of the triple point is strictly less than the product of the degrees (=$2$)? I am totally confused. $\endgroup$ Jan 20, 2011 at 8:44
  • $\begingroup$ Thanks Allen, but I am a little confused, you say: "If the co-dimensions are all exactly n..." by this do you mean that the dimension of the intersection is zero? I am sorry if I have not given this enough thought. $\endgroup$ Jan 20, 2011 at 17:38
  • $\begingroup$ Sorry, Sagar: fixed. @Sándor: the intersection of $X$ and $Y$ should be $1*2$ points, not $3$. Note that $3 \geq 1*2$ is in agreement with the statement from [Fulton]. $\endgroup$ Jan 20, 2011 at 18:48
  • $\begingroup$ This is indeed a little confusing, I think because of language. The example Allen gives appears to be a counterexample to the statement from Fulton, since the degree (=multiplicity) of the (irreducible) scheme Z is greater than the product of the degrees of the intersecting schemes. However, I believe the statement the OP refers to is the more elementary fact that the underlying variety Z has degree at most equal to the product of the degrees of the intersecting varieties X and Y, a simple application of the generalized Bezout theorem. $\endgroup$ Jan 21, 2011 at 1:04
  • $\begingroup$ ...In this case, a closely related question is: When is the cycle representing $X\cdot Y$ equivalent to the scheme-theoretic intersection $X\cap Y$? (And here's where the Cohen-Macaulay hypothesis does its work.) $\endgroup$ Jan 21, 2011 at 1:09

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