# how does one intersect the diagonal with a graph on the surface $X\times X$

I want to do a concrete example of an intersection product for myself.

Consider the endomorphism $f:\mathbf{P}^1_k\to \mathbf{P}^1_k$ given by $(x:y)\to (y:x)$. It has precisely two fixed points: $(1:1)$ and $(1:-1)$. I want to compute that the intersection product on $\mathbf{P}^1\times \mathbf{P}^1$ is two.

I think I can somehow see that this is the length of the module $k[x,y]/(x-y,x+y)$...Can somebody explain or correct me?

I also wish I was capable of doing something similar with the morphism $(x:y)\mapsto (x^n:y^n)$. Can somebody explain how this works?

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To be perfectly clear, I'll choose a different copy of $\mathbb P^1_k$ for the target, with coordinates $(u:v)$.
The morphism $f$ is then described by $u=y, v=x$ and its graph $\Gamma$ has equation $ux-vy=0$ in $\mathbb P^1_k \times \mathbb P^1_k$.
The fixed points are given by the intersection with the diagonal $\Delta$ of equation $uy-vx=0$ in $\mathbb P^1_k \times \mathbb P^1_k$.

An intersection point clearly satisfies $x\neq 0$ and $v\neq 0$, so that the intersection takes place in $\mathbb A^1_k \times \mathbb A^1_k$, the product of the affine lines with coordinates $y=(1:y)$ and $u=(u:1)$.
There $\Gamma$ has equation $u-y=0$ and $\Delta$ has equation $uy-1=0$.

The intersection $\Gamma\cap \Delta$ is the affine scheme $X=Spec(A)$ of the $k$-algebra $A=k[u,y]/(u-y,uy-1)\simeq k[u]/(u^2-1)$ (of dimension 2 as you correctly said)
You get (if $char.k\neq 2$) two reduced points for the intersection $\Gamma\cap \Delta=\lbrace (1,1), (-1,1)\rbrace$

A family of morphisms
If you want to study the morphism $(x:y)\mapsto (x^n:y^n)$, you will similarly get a curve $\Gamma$ with equation $uy^n-vx^n=0$ to be intersected with the diagonal $\Delta$ with equation $uy-vx=0$.
The intersection $\Gamma\cap \Delta$ can now be decomposed in two parts:

I) In the affine plane $\mathbb A^1_k \times \mathbb A^1_k$ with coordinates $(x,u)$ obtained by setting $y=v=1$ there are $n$ reduced points.
They are given by the equations $u=x$ and $u=x^n$ and consist
of the point $(0,0)$ and of the $n-1$ pointe $(\zeta,\zeta)$ with $\zeta^{n-1}=1$.

II) The reduced point $((1:0),(1:0))\in \mathbb P^1_k$

[I have assumed $n-1$ prime to $char.k$]

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Sorry. I think in your generalization your mixing up $x$ and $y$. The equation should be $uy^n-vx^n=0$ for $\Gamma$. (For $n=1$ it should just be the diagonal.) –  Ali Feb 2 '12 at 22:28
I have corrected my previous calculation in the "family of morphisms". I had not noticed that Ail's second question was actually not a generalization of the first, because he had switched $x$ and $y$. Thanks to Ali for alerting me of my confusion. –  Georges Elencwajg Feb 2 '12 at 23:21
Thanks, Ali, you are absolutely right. Corrected –  Georges Elencwajg Feb 2 '12 at 23:23
No problem Georges. Here is another remark which might be useful. I don't think you need the assumptions on the characteristic if you are interested in the total intersection number. For example, in the first example your ring becomes $k[u]/(u-1)^2$ in characteristic two. This has length two. Thus, you still have "two" intersection points (counted with multiplicity). Similarly, in the second example, if $n-1$ is not coprime with $p$, you get $m$ intersection points of weight $p^a$, where we wrote $n-1 = mp^a$. –  Ali Feb 3 '12 at 4:42
Dear Ali, what you write is correct but, since I described the intersection scheme as a disjoint union of reduced points, the field characteristic condition was indispensable because otherwise the intersection would be non-reduced, as shown by your own calculations in characteristic $2$ and $p$ –  Georges Elencwajg Feb 3 '12 at 7:49