# Is the intersection of the diagonal with a graph always transverse in characteristic zero

Let X be a projective smooth connected curve over $\mathbf{C}$. Let $f:X\to X$ be a non-constant morphism.

Is the intersection of the diagonal $\Delta_X$ and the graph $\Gamma_f$ on $X\times X$ transversal? That is, do we have that $\Delta_X \cdot \Gamma_f = \# \mathrm{Fix}(f)$, where $\mathrm{Fix}(f)$ is the set of fixed points of $f$?

This is not true in positive characteristic. Consider the morphism $(x:y)\mapsto (y:x)$ from $\mathbf{P}^1$ to itself. In characteristic two, there is precisely one fixed point $(1:1)$ with multiplicity $2$, i.e., $\Gamma_f\cdot \Delta_X = 2$. (In characteristic $\neq 2$, there are precisely two fixed points with multiplicity 1: $(1:1)$ and $(1:-1)$.)

I also tried to do it with an elliptic curve, but the example I wrote down also gave a transversal intersection.

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Your example is not quite right. In characteristic two, your map is the identity! In odd characteristic, the two points you listed are not fixed! – Mariano Suárez-Alvarez Feb 3 '12 at 8:02
Whooooops! I didn't mean to write that down. I meant to write down $(x:y) \mapsto (y:x)$. – Ali Feb 3 '12 at 8:08

The map $(x:y)\mapsto(xy+x^2:y^2)$ on $P^1$ works, no?
More generally, pick any polynomial whose graph is not transverse to the diagonal, and homogenise—I started with $x+x^2$. – Mariano Suárez-Alvarez Feb 3 '12 at 7:53
Thnx for this. The funny thing is I was struggling with $x^2$, because I thought it wasn't transverse to the diagonal. But actually, it is. :D – Ali Feb 3 '12 at 8:18