Question

Let $k$ be an algebraic closure of a finite field of characteristic $p$. Fix an integer $l\neq p$. For a separated $k$-scheme $X$ of finite type, we define the (compactly supported) Euler characteristic of $X$ to be $$e(X) =\sum_i (-1)^i \dim_\mathbf{Q_l} H^i_c(X,\mathbf{Q}_l).$$ Here $H^i_c(-,\mathbf{Q}_l)$ denotes the $l$-adic cohomology with compact support.

For example, if $X$ is smooth and projective over $k$, we have that $e(X)$ equals the degree of the top Chern class of $X$.

Let $X$ and $Y$ be separated $k$-schemes of finite type. Let $\pi:X\longrightarrow Y$ be a finite etale morphism of degree $d$.

Question. Is it true that $e(X) = d \cdot e(Y)$?

Here is what I "know":

For any separated $\mathbf{C}$-scheme $M$ of finite type, we define the (compactly supported) Euler characteristic of $M$ to be $$e(M) =\sum_i (-1)^i \dim_\mathbf{Q} H^i_c(M,\mathbf{Q}).$$ Here $H^i_c(-,\mathbf{Q})$ denotes the cohomology with compact support and coefficients in $\mathbf{Q}$ on the category of para-compact Hausdorff spaces. (Just to be clear, we use the analytification of $M$ here.)

Let $M$ and $N$ be separated $\mathbf{C}$-schemes of finite type. Let $\pi:M\longrightarrow N$ be a finite etale morphism of degree $d$. One can show that $e_c(M) = d \cdot e_c(N)$ quite easily as follows:

We may assume that $M$ and $N$ are connected and we may assume that $\pi$ is Galois. Let $G$ be the Galois group. Let $K_0(\mathbf{Q}[G])$ be the Grothendieck group of finitely generated $\mathbf{Q}[G]$-modules. Since the action of $G$ is free, a nontrivial element $g\in G$ has no fixed points. By the Lefschetz trace formula (see the paper by Deligne-Lusztig), we have that $$\sum (-1)^i \textrm{Tr}(H^i_c(g)) = 0.$$ Therefore, by character theory or some result in loc. cit, we have that the class of $H^\cdot_c(M,\mathbf{Q})$ in $K_0(\mathbf{Q}[G])$, defined to be the alternating sum of the classes of $H^i_c(M,\mathbf{Q})$, is an integer multiple of the regular representation. The result then follows from an easy computation.

Question. The same proof works to answer my above question positively when the cover $\pi:X\longrightarrow Y$ above is tame. In particular, if $p$ does not divide $d$. But what about the wild case? Are there some comparison theorems which allow us to simply reduce to the complex case?

Answer 1

It's not true in general. Over a field of characteristic $p>0$, the map $f:\mathbf{A}^1\to\mathbf{A}^1$ defined by $f(z)=z^p+z$ is etale because its derivative is $1$. The degree of $f$ is $p$, and the Euler characteristic of $\mathbf{A}^1$ is $1$, but $1\neq 1\times p$.