Calculating intersection with diagonal in $\mathbb{P}^2 \times \mathbb{P}^2$ The following is example 6.1.2 from Fulton, Intersection Theory.
Denote projective coordinates on $\mathbb{P}^2$ by $[x,y,z]$, and on $Y=\mathbb{P}^2 \times \mathbb{P}^2$ by $([x,y,z],[u,v,w])$. Consider the lines $A=V(x) \subset \mathbb{P}^2$ and $B=V(z) \subset \mathbb{P}^2$. They meet in a point $P$. Define divisors $D_1=2A+B=V(x^2 z)$ and $D_2=A+2B=V(x z^2)$ in $\mathbb{P}^2$. Let $X$ be the closed subscheme $D_1 \times D_2 \subset \mathbb{P}^2 \times \mathbb{P}^2$, so $X=V(x^2 z,uw^2)$. Furthermore, let $f:V=\mathbb{P}^2 \to \mathbb{P}^2 \times \mathbb{P}^2$ be the diagonal.
Fulton claims that the intersection product $X \cdot V= 3\alpha + 3\beta + 3[P]$, where $\alpha,\beta$ are zero cycles of degree $1$ on $A$ resp. $B$.
I am trying to understand how this is calculated. I am aware that one could probably reverse the roles of $V$ and $X$ here to simplify things; I do not want to do that at the moment.
First, one would calculate the intersection $W=V\times_Y X$. One sees that $W=V(x^2 z, xz^2) \subset \mathbb{P}^2$. If we denote $g:W \to X$, we need to calculate the cycle of $C=C_W V \subset g^* (N_X Y)$, and transfer it to $W$ via Gysin.
Because nothing interesting happens at infinity (or does it?), we can only look at the affine picture on $\mathbb{A}^2 =\{y\neq 0\}\subset \mathbb{P}^2$. The normal cone would then be 
$$\mathrm{Spec} \bigoplus_n (x^2z,xz^2)^n/(x^2z,xz^2)^{n+1}\simeq \mathrm{Spec}\ k[x,z,U,T]/(x^2z,xz^2,zT-xU),$$
whereas the pullback of the normal bundle would be $\mathrm{Spec} \bigoplus_n k[x,z]/(xz^2,x^2z)\otimes ((x^2z,uw^2)^n/(x^2z,uw^2)^{n+1})$.
How would one now calculate $[C]$, and in particular, the intersection class $s^*[C]$, where $s:W\to g^* N_X Y$ is the zero section?
 A: The pullback $N = g^*N_XY$ of the normal bundle is given (on the affine chart that you have chosen) by 
$$ \mathrm {Spec} \ k[x,z,U,T]/(x^2z,xz^2).$$
The cone $C = C_WV$ is indeed what you computed. You can see that it has three components:


*

*$C_0: x = z = 0$, which maps to $P$,

*$C_1: x = T = 0$, which maps to $A$,

*$C_2: z = U = 0$, which maps to $B$.


The component $C_0$ has multiplicity $3$. Indeed, if you localize at the ideal $(x,z)$ then you obtain a ring of length $3$. This explains the term $3[P]$ in $X\cdot V$.
The components $C_1$ and $C_2$ are of multiplicity $1$. Here you have to compute the Gysin homomorphism: something interesting does indeed happen at infinity! 
Let us compute the contribution of $C_1$. In the first $\mathbb P^2$ you chose the coordinates $x,z$ near the point $P$. Let $\bar x, \bar y$ be the coordinates near the point $Q$ at infinity on $A$. They are related by $[x:1:z] = [\bar x:\bar y:1]$, so that $x = \bar x/\bar y$ and $z = 1/\bar y$. The ideal of $D_1$ is generated there by $\bar T := \bar x^2$. Changing coordinates in the cone you can see that
$$ T = x^2z = \frac {\bar x^2} {\bar y^2} \cdot\frac 1{\bar y} = \frac {\bar T}{\bar y^3}$$ 
Hence, you see that $[C_1]$ is linearly equivalent to the pullback of the divisor given by $\bar y^3$, which is just $3\alpha$, where $\alpha$ is the class of $Q$ (which is rationally equivalent to any other point on the line $A$). Therefore, by definition of Gysin homomorphism, $s^*[C_1] = 3\alpha$.
Similarly the contribution of $C_2$ is $3\beta$, so $X \cdot V = s^*[C] = 3\alpha + 3\beta + 3[P]$.
