I am reading Section 1.1C of Lazarsfeld "Positivity in Algebraic Geometry I" and I need help understanding one line.
On Page 17, Remark 1.1.13(iii), he says the following:
If $D_1,..., D_n$ are effective (Cartier) divisors that meet transversly at smooth points of $X$ (where $X$ is some irreducible complete variety), then $$(D_1 \cdot ... \cdot D_n) = \#\{D_1 \cap \cdots \cap D_n\}$$
I am trying to understand this statement. I guess first I need to understand the intersection of let's say, $2$ Cartier divisors, $D_1$ and $D_2$...and then go from there to $n$ divisors. Can someone please explain to me how the statement bolded above applies to two cartier divisors $D_1=\{(U_i,f_i)\}$ and $D_2 = \{(V_j, g_j)\}$?
First of all think about $n$ general hyperplanes $H_1,\ldots,H_n$ on a vector space of dimension $n$. Being general, the dimension of their intersection is zero, and so the intersection consists of finite points (and obviously only one point). Thus we would say that $(H_1\cdots H_n)=1$.
This intuition basically works in general: the intersection of $n$ general subvarieties of codimension 1 on a variety $X$ is finite. Now, we would not only like to count the amount of points in the intersection, but we want to keep track of multiplicity as well. For example, on $\mathbb{A}^2$, the line $y=0$ intersects the parabola $y=x^2$ at one point, but since it is tangent to the parabola at $0$, we would like to think of this intersection as an intersection with multiplicity 2, and so we would like to have $(\{y=0\}\cdot\{y-x^2=0\})=2$. This is an example in dimension 2.
In general for dimension 2, you can think of two effective divisors as being the formal sum of curves on your surface, and since the intersection product is going to be multilinear, we only have to see what the intersection of two irreducible curves would be. A transversal intersection between two curves is essentially a point where the tangent lines of the curve are different (assuming they meet at a smooth point). If every point of intersection between the two curves satisfies this, then we don't have to worry about multiplicity and we only have to count points in the intersection to define the intersection number of the two curves.
Sometimes $n$ divisors do not intersect in a finite number of points, and a Moving Lemma assures us that we can move the divisors in their linear equivalence classes to divisors that do intersect transversely (that is, at each point of intersection between any two divisors, the sum of the tangent planes of each divisor at that point is the tangent plane of the variety at that point), and so we can count points (be careful because when moving the divisors, effectiveness may be lost).