Intersection degree of noncomplete intersection

Question

Via the theory of hilbert polynomials, I think the following is true:

Suppose $X$ is a closed subscheme of $P^n$ of degree $d$, and $Y$ is also such one but is also a complete intersection and also doesn't share any component with $X$, then (assuming the the sum of dimensions is at least $n$)

$$deg(X \cap Y) = deg(X) \cdot deg(Y)$$

To prove this, it suffices to show this for $Y$ a hypersurface $f$, which is proved via the exact sequence

$$ S_*(X) \to ^f S_* (X) \to S_*(X)/(f)$$

Is this true if $X,Y$ are not complete intersection?

Answer 1

Jake brought an example that illustrates the problems that occur when $Y$ is a complete intersection in a bad way. Let me show that everything goes wrong when both are not complete intersections.

In $P^4$, a union of two generic planes is not a complete intersection (note they meet at a point). Take two pairs (of union of two planes) that all meet the same point. I claim they do the job. Each is certainly of degree $2$, so you'd expect the intersection to be of degree $4$. But, computing the Hilbert polynomial; one easily finds it's $5$ instead!

Answer 2

It’s false.

In $\mathbb{P}^4$ with coordinates $x,y,z,w,t$, let $X$ be the union of the planes $x=y=0$ and $z=w=0$ and let $Y$ be the plane $x-z=y-w=0$, a complete intersection.

Then you can check that $X \cap Y$ is a point of multiplicity $3$, but $\deg(X)\deg(Y)=2 \cdot 1 = 2$.