Chow ring of projetive space

Question

The Chow ring of $\mathbb{P}^n$ is $\mathbb{Z}[H]/H^{n+1}$, where $H$ is the linear equivalence lass determined by the zero-set of a linear functional. I have a few (very basic) points of confusion:

1. Firstly, intersection of curves of various degrees in $\mathbb{Q}[H]/H^{n+1}$ should correspond to tensor product of line bundles in the Grothendieck ring. $aH$ is nilpotent in this ring, and therefore not invertible. Yet it corresponds to a line bundle in the grothendieck ring, which is invertible. Where do I go wrong here?

2. It seems like the intersection of a point and a curve could be either a point or $\emptyset$. Why is the intersection in $\mathbb{Z}[H]/H^{n+1}$ equal to $0$?

Answer 1

The Chow ring tries to capture what happens in the generic case. Generically, a point and a curve do not intersect. Neither does a pair of curves if $n > 2$.

To your first point, line bundles correspond to divisors (codimension 1) not to curves. You're referring to the identity that looks something like

$$ D_1 \cdot D_2 \cdots D_n = \deg(\mathcal{O}(D_1) + \dots + \mathcal{O}(D_n)). $$

But as far as I know, the tensor product $\mathcal{O}(D_1) + \dots + \mathcal{O}(D_n)$ does not correspond to the intersection $D_1 \cap \cdots \cap D_n$ which is a finite set of points, it corresponds to the data $(D_1, \dots, D_n)$.

For example, in $\mathbb{P}^2$, given say $6$ points, we can factor $\mathcal{O}(6) = \mathcal{O}(1) + \mathcal{O}(6) = \mathcal{O}(2) + \mathcal{O}(3)$. Given two curves of degrees $(1,6)$ or $(2,3)$, their intersection is (generically) 6 points, but given 6 points, you can't just construct two curves whose intersection is those 6 points. If the 6 points are generic there are simply no lines or conics that pass through them in the first place.

So the symbol "$\mathcal{O}(2) + \mathcal{O}(3)$" contains more information than just the number "6". In particular, it contains more information than the intersection product $C_1 \cdot C_2$ of two curves of degrees 2 and 3. That's why we need to lose information by taking the degree for these to be equal.

Answer 2

To your specific questions,

1. It's not true that there's a correspondence of curves to line bundles taking intersection to tensoring. There's a correspondence of divisors (co-dimension one subvarieties) to line bundles, but line bundle tensoring corresponds to taking a union of divisors.

2. A point intersecting a curve in a big-dimensional space is not a transverse intersection. For a heuristic picture of what intersection theory in algebraic geometry is trying to capture that explains why examples like this ought to be zero, you might consider reading Guillemin and Pollack's book on Differential Topology. Or, an algebro-geometric way to say it is that if the curve is passing thru a specified point, the curve is not "generic" in its equivalence class w.r.t. the point.